Estimates of operator moduli of continuity
aa r X i v : . [ m a t h . F A ] A p r ESTIMATES OF OPERATOR MODULI OF CONTINUITY
A.B. ALEKSANDROV AND V.V. PELLER
Abstract.
In [AP2] we obtained general estimates of the operator moduli of conti-nuity of functions on the real line. In this paper we improve the estimates obtained in[AP2] for certain special classes of functions.In particular, we improve estimates of Kato [Ka] and show that (cid:13)(cid:13) | S | − | T | (cid:13)(cid:13) ≤ C k S − T k log (cid:18) k S k + k T kk S − T k (cid:19) for every bounded operators S and T on Hilbert space. Here | S | def = ( S ∗ S ) / . Moreover,we show that this inequality is sharp.We prove in this paper that if f is a nondecreasing continuous function on R thatvanishes on ( −∞ ,
0] and is concave on [0 , ∞ ), then its operator modulus of continuityΩ f admits the estimate Ω f ( δ ) ≤ const Z ∞ e f ( δt ) dtt log t , δ > . We also study the problem of sharpness of estimates obtained in [AP2] and [AP3].We construct a C ∞ function f on R such that k f k L ∞ ≤ k f k Lip ≤
1, andΩ f ( δ ) ≥ const δ r log 2 δ , δ ∈ (0 , . In the last section of the paper we obtain sharp estimates of k f ( A ) − f ( B ) k in thecase when the spectrum of A has n points. Moreover, we obtain a more general result interms of the ε -entropy of the spectrum that also improves the estimate of the operatormoduli of continuity of Lipschitz functions on finite intervals, which was obtained in[AP2]. Contents
1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22. Schur multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43. Remarks on absolutely convergent Fourier integrals . . . . . . . . . . . . . . . . . . . . . . . . 84. Estimates of certain multiplier norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125. Operator Lipschitz functions and operator modulus of continuity . . . . . . . . . 156. The operator Lipschitz norm of the function | x | on subsets of R . . . . . . . . . . 237. The operator modulus of continuity of a certain piecewise linear function . 258. Operator moduli of continuity of concave functions on R + . . . . . . . . . . . . . . . . 269. Lower estimates for operator moduli of continuity . . . . . . . . . . . . . . . . . . . . . . . . 34 The first author is partially supported by RFBR grant 11-01-00526-a ; the second author is partiallysupported by NSF grant DMS 1001844.
0. Lower estimates in the case of unitary operators . . . . . . . . . . . . . . . . . . . . . . . . . . 4011. Self-adjoint operators with finite spectrum. Estimates in terms of the ε -entropyof the spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481. Introduction
In this paper we study operator moduli of continuity of functions on subsets of thereal line. For a closed subset F of the real line R and for a continuous function f on F ,the operator modulus of continuity Ω f, F is defined byΩ f, F ( δ ) def = sup k f ( A ) − f ( B ) k , δ > , where the supremum is taken over all self-adjoint operators A and B such that σ ( A ) ⊂ F , σ ( B ) ⊂ F , and k A − B k ≤ δ. If F = R , we use the notation Ω f def = Ω f, R . Recall that a continuous function f on F iscalled operator Lipschitz if Ω f, F ( δ ) ≤ const δ , δ > Lipschitz function f on R , i.e., a function f satisfying | f ( x ) − f ( y ) | ≤ const | x − y | , x, y ∈ R , does not have to be operator Lipschitz. This was established for the first time by Far-forovskaya [Fa1]. It was shown later in [Ka] that the function x
7→ | x | on R is notoperator Lipschitz. The paper [Ka] followed the paper [Mc], in which it was shown thatthe function x
7→ | x | is not commutator Lipschitz. We refer the reader to § f belongs tothe Besov class B ∞ ( R ), then f is operator Lipschitz (we refer the reader to [Pee] and[Pe5] for the definition of Besov classes).In our joint papers [AP1] and [AP2] we obtain the following upper estimate for con-tinuous functions f on R :Ω f ( δ ) ≤ const δ Z ∞ δ ω f ( t ) t dt = const Z ∞ ω f ( tδ ) t ds, δ > , (1.1)where ω f is the modulus of continuity of f , i.e., ω f ( δ ) def = sup (cid:8) | f ( x ) − f ( y ) | : x, y ∈ R , | x − y | ≤ δ (cid:9) , δ > . We deduced from (1.1) in [AP2] that for a Lipschitz function f on [ a, b ], the followingestimate for the operator modulus of continuity Ω f holds:Ω f, [ a,b ] ( δ ) ≤ const δ (cid:18) (cid:18) b − aδ (cid:19)(cid:19) k f k Lip , here k f k Lip def = sup x = y | f ( x ) − f ( y ) || x − y | . A similar estimate was obtained earlier in [Ka] in the very special case f ( x ) = | x | .Namely, it was shown in [Ka] that for bounded self-adjoint operators A and B on Hilbertspace, the following inequality holds: (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≤ π k A − B k (cid:18) k A k + k B kk A − B k (cid:19) . It turns out, however, that for the function x
7→ | x | the operator modulus of continuityadmits a much better estimate. Namely, we show in § (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≤ const k A − B k log (cid:18) k A k + k B kk A − B k (cid:19) . We also prove in this paper that this estimate is sharp.Note that in [NiF] an estimate slightly weaker than (1.1) was obtained by a differentmethod.In § f is a continuous nondecreasing function on R such that f ( x ) = 0for x ≤ f to [0 , ∞ ) is a concave function, then estimate (1.1) canalso be improved considerably:Ω f ( δ ) ≤ const Z ∞ e f ( δt ) dtt log t , δ > . We also obtain other estimates of operator moduli of continuity in § f is a function on R such that k f k L ∞ ≤ k f k Lip ≤
1, thenΩ f ( δ ) ≤ const δ (cid:18) δ (cid:19) , δ ∈ (0 , . We construct in § C ∞ function f on R such that k f k L ∞ ≤ k f k Lip ≤
1, andΩ f ( δ ) ≥ const δ r log 2 δ , δ ∈ (0 , . To construct such a function f , we use necessary conditions for operator Lipschitznessfound in [Pe2]. We do not know whether the results of § §
10 we obtain lower estimates in the case of functions on the unit circle and unitaryoperators.Finally, we obtain in §
11 the following sharp estimate for the norms k f ( A ) − f ( B ) k for Lipschitz functions f and self-adjoint operators A and B on Hilbert space such thatthe spectrum σ ( A ) of A has n points: k f ( A ) − f ( B ) k ≤ C (1 + log n ) k f k Lip k A − B k . (1.2)Moreover, we obtain in §
11 an upper estimate in the general case (see Theorem 11.5) interms of the ε -entropy of the spectrum of A , where ε = k A − B k . It includes inequalities(1.1) and (1.2) as special cases. Note that (1.2) improves earlier estimates in [Fa1] and[Fa2]. n § § § § § R .2. Schur multipliers
In this section we define Schur multipliers and discuss their properties. Note that thenotion of a Schur multiplier can be defined in the case of two spectral measures (see e.g.,[Pe2]). In this section we give define Schur multipliers in the case of two scalar measures.This corresponds to the case of spectral measures of multiplicity 1.Let ( X , µ ) and ( Y , ν ) be σ -finite measure spaces. Let k ∈ L ( X × Y , µ ⊗ ν ). Then k induces the integral operator I k = I µ,νk from L ( Y , ν ) to L ( X , µ ) defined by( I k f )( x ) = Z Y k ( x, y ) f ( y ) dν ( y ) , f ∈ L ( Y , ν ) . Denote by k k k B = k k k B µ,ν X , Y the operator norm of I k . Let Φ be a µ ⊗ ν -measurablefunction defined almost everywhere on X × Y . We say that Φ is a
Schur multiplier withrespect to µ and ν if k Φ k M µ,ν X , Y def = sup (cid:8) k Φ k k B : k , Φ k ∈ L ( X × Y , µ ⊗ ν ) , k k k B ≤ (cid:9) < ∞ . We denote by M µ,ν X , Y the space of Schur multipliers with respect to µ and ν . It can beshown easily that M µ,ν X , Y ⊂ L ∞ ( X × Y , µ ⊗ ν ) and k Φ k L ∞ ( X ×Y ,µ ⊗ ν ) ≤ k Φ k M µ,ν X , Y . Thus ifΦ ∈ M µ,ν X , Y , then k Φ k M µ,ν X , Y = sup (cid:8) k Φ k k B : k ∈ L ( X × Y , µ ⊗ ν ) , k k k B ≤ (cid:9) . Note that M µ,ν X , Y is a Banach algebra: k Φ Φ k M µ,ν X , Y ≤ k Φ k M µ,ν X , Y k Φ k M µ,ν X , Y . It is easy to see that k Φ k M µ,ν X , Y = k Ψ k M ν,µ Y , X for Ψ( y, x ) = Φ( x, y ).If X is a µ -measurable subset of X , then we denote by ( X , µ ) the correspondingmeasure space on the σ -algebra of µ -measurable subsets of X .Let X = ∞ S n =1 X n and Y = ∞ S n =1 Y n , where the X n are µ -measurable subsets of X , andthe Y n are ν -measurable subsets of Y . It is easy to see thatsup m,n ≥ k k k B µ,ν X n, Y n ≤ k k k B µ,ν X , Y ≤ ∞ X m =1 ∞ X n =1 k k k B µ,ν X n, Y n or every k ∈ L ( X × Y , µ ⊗ ν ), andsup m,n ≥ k Φ k M µ,ν X n, Y n ≤ k Φ k M µ,ν X , Y ≤ ∞ X m =1 ∞ X n =1 k Φ k M µ,ν X n, Y n (2.1)for every Φ ∈ L ∞ ( X × Y , µ ⊗ ν ).We state the following elementary theorem: Theorem 2.1.
Let ( X , µ ) , ( X , µ ) , ( Y , ν ) and ( Y , ν ) be σ -finite measure spaces.Suppose that µ is absolutely continuous with respect to µ and ν is absolutely continuouswith respect to ν . Let Φ ∈ M µ,ν X , Y . Then Φ ∈ M µ ,ν X , Y and k Φ k M µ ,ν X , Y ≤ k Φ k M µ,ν X , Y . Proof.
By the Radon–Nikodym theorem, dµ = ϕdµ and dν = ψdν for nonnegativemeasurable functions ϕ and ψ on X and Y . Let k ∈ L ( X × Y , µ ⊗ ν ). Put( T k )( x, y ) def = k ( x, y ) p ϕ ( x ) ψ ( y ) . Clearly, T is an isometric embedding from L ( X × Y , µ ⊗ ν ) in L ( X × Y , µ ⊗ ν ).Moreover, k T k k B µ,ν X , Y = k k k B µ ,ν X , Y . We have k Φ k k B µ ,ν X , Y = k T (Φ k ) k B µ,ν X , Y = k Φ T k k B µ,ν X , Y ≤k Φ k M µ,ν X , Y k T k k B µ,ν X , Y = k Φ k M µ,ν X , Y k k k B µ ,ν X , Y for every k ∈ L ( X × Y , µ ⊗ ν ). Hence, Φ ∈ M µ ,ν X , Y and k Φ k M µ ,ν X , Y ≤ k Φ k M µ,ν X , Y . (cid:4) Note that if X and Y coincide with the set Z + of nonnegative integers and µ and ν are the counting measure, the above definition coincides with the definition of Schurmultipliers on the space of matrices: a matrix A = { a jk } j,k ≥ is called a Schur multiplieron the space of bounded matrices if A ⋆ B is a matrix of a bounded operator, whenever B is . Here we use the notation
A ⋆ B = { a jk b jk } j,k ≥ (2.2)for the Schur–Hadamard product of the matrices A = { a jk } j,k ≥ and B = { b jk } j,k ≥ .Let X and Y be closed subsets of R . We denote by M X , Y the space of Borel Schurmultipliers on X × Y , i.e., the space of Borel functions Φ defined everywhere on
X × Y such that k Φ k M X , Y def = sup k Φ k M µ,ν X , Y < ∞ , where the supremum is taken over all regular positive Borel measures µ and ν on X and Y . It can be shown easily that sup ( x,y ) ∈X ×Y | Φ( x, y ) | ≤ k Φ k M X , Y . It is also easy to verify that if Φ n ∈ M X , Y , Φ is a bounded Borel function on X × Y , andΦ n ( x, y ) → Φ( x, y ) for all ( x, y ) ∈ X × Y , then k Φ k M X , Y ≤ lim inf n →∞ k Φ n k M X , Y . n particular, Φ ∈ M X , Y if lim inf n →∞ k Φ n k M X , Y < ∞ .We are going to deal with functions f on X × Y that are continuous in each variable.It must be a well-known fact that such a function f has to be a Borel function. Indeed,one can construct an increasing sequence {Y n } ∞ n =1 of discrete closed subsets of Y suchthat ∞ S n =1 Y n is dense in Y . Let us consider the function f n : X × R → C such that f (cid:12)(cid:12) ( X × Y n ) = f n (cid:12)(cid:12) ( X × Y n ) and f n ( x, · ) is a piecewise linear function with nodes in Y n for all x ∈ X . Clearly, the function f n is defined uniquely if we require that f n ( x, · ) isconstant on each unbounded complimentary interval of Y n . It is easy to see that f n iscontinuous on X × R and lim n →∞ f n ( x, y ) = f ( x, y ) for all ( x, y ) ∈ X × Y . Thus, f belongsto the first Baire class, and so it is Borel. Theorem 2.2.
Let X and Y be closed subsets of R and let Φ be a function on X × Y that is continuous in each variables. Suppose that µ and µ are positive regular Borelmeasures on X , and ν and ν are positive regular Borel measures on Y . If supp µ ⊂ supp µ and supp ν ⊂ supp ν , then k Φ k M µ ,ν X , Y ≤ k Φ k M µ,ν X , Y . We need two lemmata.
Lemma 2.3.
Let X and Y be compact subsets of R and let µ and ν be finite positiveBorel measures on X and Y . Suppose that { ν j } ∞ j =1 is a sequence of finite positive Borelmeasures on Y that converges to ν in the weak- ∗ topology σ (cid:0) ( C ( Y )) ∗ , C ( Y ) (cid:1) . If k is abounded Borel function on X × Y such that k ( x, · ) ∈ C ( Y ) for every x ∈ X , then lim j →∞ (cid:13)(cid:13) I µ,ν j k (cid:13)(cid:13) B µ,νj X , Y = (cid:13)(cid:13) I µ,νk (cid:13)(cid:13) B µ,ν X , Y . Proof.
Clearly, I µ,ν j k (cid:0) I µ,ν j k (cid:1) ∗ is an integral operator on L ( X , µ ) with kernel l j ( x, y ) = R Y k ( x, t ) k ( y, t ) dν j ( t ). Besides, the sequence { l j } converges in L ( X × X , µ ⊗ µ ) to thefunction l defined by l ( x, y ) = R Y k ( x, t ) k ( y, t ) dν ( t ), which is the kernel of the integraloperator I µ,νk (cid:0) I µ,νk (cid:1) ∗ . Hence,lim j →∞ (cid:13)(cid:13) I µ,ν j k (cid:13)(cid:13) B µ,νj X , Y = lim j →∞ (cid:13)(cid:13) I µ,ν j k (cid:0) I µ,ν j k (cid:1) ∗ (cid:13)(cid:13) B µ,νj X , Y = (cid:13)(cid:13) I µ,νk (cid:0) I µ,νk (cid:1) ∗ (cid:13)(cid:13) B µ,ν X , Y = (cid:13)(cid:13) I µ,νk (cid:13)(cid:13) B µ,ν X , Y . (cid:4) Corollary 2.4.
Let X and Y be compact subsets of R , and let µ and ν be finite positiveBorel measures on X and Y . Suppose that { ν j } ∞ j =1 is a sequence of finite positive Borelmeasures on Y that converges to ν in σ (cid:0) ( C ( Y )) ∗ , C ( Y ) (cid:1) . If Φ is a Borel function on X × Y such that Φ( x, · ) ∈ C ( Y ) for all x ∈ X , then k Φ k M µ,ν X , Y ≤ lim inf j →∞ k Φ k M µ,νj X , Y . Proof.
It is easy to see that k Φ k M µ,ν X , Y = sup (cid:8) k Φ k k B µ,ν X , Y : k ∈ C ( X × Y ) , k k k B µ,ν X , Y ≤ (cid:9) . et k ∈ C ( X × Y ) with k k k L ( µ ⊗ ν ) >
0. Then k Φ k k B µ,ν X , Y = lim j →∞ k Φ k k B µ,νj X , Y ≤ lim inf j →∞ (cid:16) k Φ k M µ,νj X , Y k k k B µ,νj X , Y (cid:17) = lim inf j →∞ k Φ k M µ,νj X , Y lim j →∞ k k k B µ,νj X , Y = k k k B µ,ν X , Y lim inf j →∞ k Φ k M µ,νj X , Y which implies the result. (cid:4) We are going to use the following notation: for a measure µ and an integrable function ϕ , we write ν = ϕµ if ν is the (complex) measure defined by dν = ϕ dµ .The following fact can be proved very easily. Lemma 2.5.
Let ν and ν be finite Borel measures on R with compact supports.Suppose that supp ν ⊂ supp ν . Then there exists a sequence { ϕ j } ∞ j =1 in C ( R ) such that ϕ j ≥ everywhere on R for all j and ν = lim j →∞ ϕ j ν in σ (cid:0)(cid:0) C (supp ν ) (cid:1) ∗ , C (supp ν ) (cid:1) . Proof of Theorem 2.2.
Put X n def = [ − n, n ] ∩ X and Y n def = [ − n, n ] ∩ Y . Clearly, n k Φ k M µ,ν X n, Y n o is a nondecreasing sequence andlim n →∞ k Φ k M µ,ν X n, Y n = k Φ k M µ,ν X , Y . This allows us to reduce the general case to the case when X and Y are compact. Besides,it suffices to consider the case where µ = µ . Indeed, the case ν = ν can be reduced tothe case µ = µ , and we have k Φ k M µ ,ν X , Y ≤ k Φ k M µ,ν X , Y ≤ k Φ k M µ,ν X , Y . Let X and Y be compact, and µ = µ . Applying Lemma 2.5, we can take a sequence { ϕ j } ∞ j =1 of nonnegative functions in C ( R ) such that ν = lim j →∞ ϕ j ν in the weak topology σ (( C ( Y ) ∗ , C ( Y )). Put ν j def = ϕ j ν . By Theorem 2.1, k Φ k M µ,νj X , Y ≤ k Φ k M µ,ν X , Y for every j ≥ (cid:4) Theorem 2.2 implies the following fact:
Theorem 2.6.
Let X and Y be closed subsets of R and let Φ be a function on X × Y that is continuous in each variables. Suppose that µ and ν are positive regular Borelmeasures on X and Y such that supp µ = X and supp ν = Y . Then k Φ k M X , Y = k Φ k M µ,ν X , Y . The following result is well known.
Let f ∈ C ( R ) . Put Φ( x, y ) def = f ( x − y ) . Then Φ ∈ M R , R if and only if f is the Fouriertransform of a complex measure on R . Moreover, k Φ k M R , R = | µ | ( R ) . A similar statement holds for any locally compact abelian group. In particular, it istrue for the group Z : Let f be a function defined on Z . Put Φ( m, n ) def = f ( m − n ) . Then Φ ∈ M Z , Z if andonly if { f ( n ) } n ∈ Z are the Fourier coefficients of a complex Borel measure µ on the unitcircle T . Moreover, k Φ k M Z , Z = | µ | ( T ) . We need the following well-known fact. emma 2.7. Let H ( m, n ) def = ( m − n , if m, n ∈ Z , m = n, , if m = n ∈ Z . Then k H k M Z , Z = π . Proof.
It suffices to observe that H ( n,
0) = 12 π Z π i( π − t ) e − i nt dt and 12 π Z π | π − t | dt = π . (cid:4) Remarks on absolutely convergent Fourier integrals
In this section we collect elementary estimates of certain functions in the space ofabsolutely convergent Fourier integrals. Such estimates will be used in the next sectionfor estimates of certain functions in the space of Schur multipliers.We are going to deal with the space b L = b L ( R ) def = F ( L ( R )) , k f k b L = k f k b L ( R ) def = (cid:13)(cid:13) F − f (cid:13)(cid:13) L . Here we use the notation F for Fourier transform :( F f )( x ) def = Z R f ( t ) e − i xt dt, f ∈ L ( R ) . Unless otherwise stated, an interval throughout the paper means a closed nondegen-erate (not necessarily finite) interval. For such an interval J , we consider the class b L ( J )defined by b L ( J ) def = (cid:8) f (cid:12)(cid:12) J : f ∈ b L (cid:9) . If f ∈ C ( J ), we put k ϕ k b L ( J ) def = inf (cid:8) k f k b L : f (cid:12)(cid:12) J = ϕ (cid:9) . For ϕ ∈ C ( R ), we put k ϕ k b L ( J ) def = (cid:13)(cid:13) ϕ (cid:12)(cid:12) J (cid:13)(cid:13) b L ( J ) . Clearly, k ϕ k L ∞ ( J ) ≤ k ϕ k b L ( J ) .For an interval J , we use the notation | J | for its length .It is easy to see that the nonzero constants belong to the space b L ( J ) for boundedintervals J and k k b L ( J ) = 1. Moreover, b L ( J ) = (cid:8) ( F µ ) | J : µ ∈ M ( R ) (cid:9) and k f k b L ( J ) = inf (cid:8) k µ k M : ( F µ ) (cid:12)(cid:12) J = f (cid:9) for every bounded interval J , where M ( R ) denotes the space of (complex) Borel measureson R .In this section we are going to discuss (mostly known) estimates for k · k b L ( J ) .First, we recall the P´olya theorem, see [Po]. Let f be an even continuous functions such that f (cid:12)(cid:12) [0 , ∞ ) is decreasing convex functionvanishing at the infinity. Then f ∈ b L and k f k b L = f (0) . This theorem readily implies the following fact. emma 3.1. Let f be a continuous function on a closed ray J that vanishes at infinity.Suppose that f is monotone and convex (or concave). Then f ∈ b L ( J ) and k f k b L ( J ) =max J | f | . In what follows by a locally absolutely continuous function on R we mean a functionwhose restriction to any compact interval is absolutely continuous. Lemma 3.2.
Let f be a locally absolutely continuous function in L ( R ) such that f ′ ∈ L ( R ) . Then f ∈ b L ( R ) and k f k b L ≤ k f k L k f ′ k L . Proof.
Put a = k f k L , b = k f ′ k L . By Plancherel’s theorem, (cid:13)(cid:13) F − f (cid:13)(cid:13) L = a π and (cid:13)(cid:13) x F − f (cid:13)(cid:13) L = b π . Hence, (cid:13)(cid:13)(cid:13)p b + a x F − f (cid:13)(cid:13)(cid:13) L = a b π . and by the Cauchy–Bunyakovsky inequality, (cid:13)(cid:13) F − f (cid:13)(cid:13) L ≤ ab √ π (cid:13)(cid:13)(cid:13)(cid:13) √ a x + b (cid:13)(cid:13)(cid:13)(cid:13) L = √ ab. (cid:4) Corollary 3.3.
Let a > . Put f a ( x ) def = ( a − x, if | x | ≤ a,x − , if | x | ≥ a. Then f a ∈ b L ( R ) and k f a k b L ≤ a . Proof.
It suffices to observe that k f a k L = a , k f ′ a k L = a , and q ≤ (cid:4) Lemma 3.4.
Let J be a bounded interval and let f be a Lipschitz function on R suchthat supp f ⊂ J . Then f ∈ b L and k f k b L ≤ √ | J | · k f ′ k L ∞ . Proof.
Let J = [ − a, a ]. Clearly, | f ( x ) | ≤ ( a − | x | ) k f ′ k L ∞ for all x ∈ J . Hence, k f k L ≤ k f ′ k L ∞ Z a ( a − t ) dt = 112 k f ′ k L ∞ | J | . Using the obvious inequality k f ′ k L ≤ k f ′ k L ∞ | J | , we get the desired estimate. (cid:4) Corollary 3.5.
Let f be a Lipschitz function on R such that f (0) = 0 . Then k f k b L ( J ) ≤ √ | J | · k f ′ k L ∞ for every bounded interval J that contains . Proof.
Put 2 J def = { x : x ∈ J } . Clearly, there exists a function f J in C ( R ) suchthat f J = f on J , supp f J ⊂ J , and k f ′ J k L ∞ ≤ k f ′ k L ∞ . (cid:4) emma 3.6. Let f be a locally absolutely continuous function on R such that (1 + | x | ) f ′ ( x ) ∈ L ( R ) . Suppose that lim x →−∞ f ( x ) = 0 and lim x →∞ f ( x ) = 1 . Then k f k b L ( −∞ ,a ] ≤ √ π k f ′ k L + r π k xf ′ k L + 72 π + 2 π log a. for every a ≥ . Proof.
Put f a ( x ) def = f ( x ) − a − Z x −∞ χ [ a, a ] ( t ) dt. Clearly, k f k b L ( −∞ ,a ] ≤ k f a k b L .We have − i x F − f a = F − ( f ′ a ) = F − ( f ′ ) − e a i x − e a i x πa i x Put h def = F − ( f ′ ). Then k f a k b L = Z R (cid:12)(cid:12)(cid:12)(cid:12) h ( x ) − e a i x − e a i x πa i x (cid:12)(cid:12)(cid:12)(cid:12) · dx | x |≤ Z − | h ( x ) − h (0) || x | dx + 12 π Z − (cid:12)(cid:12)(cid:12)(cid:12) e a i x − e a i x a i x − (cid:12)(cid:12)(cid:12)(cid:12) · dx | x | + Z {| x |≥ } | h ( x ) || x | dx + 12 πa Z {| x |≥ } | e a i x − | x dx. We have Z | h ( x ) − h (0) | x dx ≤ Z x (cid:18)Z x | h ′ ( t ) | dt (cid:19) dx = Z | h ′ ( t ) | · | log t | dt. Hence, Z − | h ( x ) − h (0) || x | dx ≤ Z − | h ′ ( t ) | · (cid:12)(cid:12) log | t | (cid:12)(cid:12) dt ≤ k h ′ k L (cid:18)Z − log | t | dt (cid:19) / = r π k xf ′ ( x ) k L because h ′ = F − (i xf ′ ).By Taylor’s formula for the function e x − e i x , we have (cid:12)(cid:12) e x − e i x − i x (cid:12)(cid:12) ≤ x . Thus 12 π Z − (cid:12)(cid:12)(cid:12)(cid:12) e a i x − e a i x a i x − (cid:12)(cid:12)(cid:12)(cid:12) · dx | x | = 12 π Z a − a (cid:12)(cid:12)(cid:12)(cid:12) e x − e i x i x − (cid:12)(cid:12)(cid:12)(cid:12) · dx | x |≤ π Z a − a min (cid:26) , | x | (cid:27) dx ≤ π (5 + 4 log a ) . inally, Z | x |≥ | h ( x ) || x | dx ≤ √ k h k L = 1 √ π k f ′ k L by the Cauchy–Bunyakovsky inequality and12 πa Z {| x |≥ } | e a i x − | x dx = 12 π Z {| x |≥ a } | e i x − | x dx ≤ πa ≤ π for a ≥
2. This implies the desired inequality. (cid:4)
Theorem 3.7.
Let J be a bounded interval containing . Then (cid:13)(cid:13)(cid:13)(cid:13) e x − e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( J ) ≤ √ | J | ≤ | J | . (3.1) Proof.
Ii suffices to observe that (cid:13)(cid:13) ( e x − e x +1 ) ′ (cid:13)(cid:13) L ∞ = and apply Corollary 3.5. (cid:4) Theorem 3.7 gives a sufficiently sharp estimate of b L -norm for little intervals J . Forbig intervals J , this estimate will be improved in Corollary 3.9. Theorem 3.8.
Let a ≥ . Then (cid:13)(cid:13)(cid:13)(cid:13) e x e x (cid:13)(cid:13)(cid:13)(cid:13) b L ( −∞ ,a ] ≤ π log a. Proof.
We have (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) e x e x + 1 (cid:19) ′ (cid:13)(cid:13)(cid:13)(cid:13) L = Z R e x dx ( e x + 1) = Z ∞ t dt ( t + 1) = 16 , and (cid:13)(cid:13)(cid:13)(cid:13) x (cid:18) e x e x + 1 (cid:19) ′ (cid:13)(cid:13)(cid:13)(cid:13) L = 2 Z ∞ x e x dx ( e x + 1) ≤ Z ∞ x e − x dx = 12 , whence for a ≥ (cid:13)(cid:13)(cid:13)(cid:13) e x e x (cid:13)(cid:13)(cid:13)(cid:13) b L ( −∞ ,a ] ≤ √ π + 1 √ π + 72 π + 2 π log a ≤ π log a by Lemma 3.6. (cid:4) Remark.
Lemma 3.1 implies that (cid:13)(cid:13)(cid:13)(cid:13) e x e x (cid:13)(cid:13)(cid:13)(cid:13) b L ( −∞ ,a ] ≤ e a e a ≤ e a for a ≤ Corollary 3.9.
Let J be a bounded interval containing . Then (cid:13)(cid:13)(cid:13)(cid:13) e x − e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( J ) ≤ π log (cid:18) | J | (cid:19) if | J | ≥ . In fact, k x ( e x e x +1 ) ′ k L = π − . roof. We may assume that the center of J is nonpositive. Then J ⊂ (cid:0) − ∞ , | J | (cid:3) .We have (cid:13)(cid:13)(cid:13)(cid:13) e x − e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( J ) ≤ (cid:13)(cid:13)(cid:13)(cid:13) e x e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( J ) ≤ π log a = 5 + 4 π log (cid:18) | J | (cid:19) . (cid:4) Estimates of certain multiplier norms
In this section we are going to obtain sharp estimates for the Schur multiplier norms (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J = (cid:13)(cid:13)(cid:13)(cid:13) e x − y − e x − y + 1 (cid:13)(cid:13)(cid:13)(cid:13) M J ,J (4.1)for every intervals J and J . First, we consider two special cases. In the first case J = J while in the second case J and J do not overlap, i.e., their intersection has atmost one point. Theorem 4.1.
Let J and J be nonoverlapping intervals. Then (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ . Proof.
Clearly, either J − J ⊂ ( −∞ ,
0] or J − J ⊂ [0 , ∞ ). It suffices to considerthe case when J − J ⊂ ( −∞ , (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ (cid:13)(cid:13)(cid:13)(cid:13) e x − y e x − y + 1 (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ (cid:13)(cid:13)(cid:13)(cid:13) e x e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( −∞ , = 2by the P´olya theorem [Po], see also Lemma 3.1. (cid:4) Theorem 4.2.
Let J be a bounded interval. Then (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≤ min (cid:26) | J | , π log + | J | (cid:27) and so (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≤ | J | ) . Proof.
We have (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≤ (cid:13)(cid:13)(cid:13)(cid:13) e x − e x + 1 (cid:13)(cid:13)(cid:13)(cid:13) b L ( J − J ) . Note that | J − J | = 2 | J | and 0 ∈ J − J . The result follows now from Theorem 3.7 andCorollary 3.9. (cid:4) heorem 4.3. Let J and J be nonoverlapping intervals and let J be the convex hullof J ∪ J . Then e − e + 1 min (cid:8) , | J | (cid:9) ≤ (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ min (cid:26) , | J | (cid:27) . Proof.
The upper estimate follows readily from Theorems 4.1 and 4.2. Let us provethe lower estimate. We have (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≥ sup x ∈ J , y ∈ J (cid:12)(cid:12)(cid:12)(cid:12) e x − e y e x + e y (cid:12)(cid:12)(cid:12)(cid:12) ≥ e | J | − e | J | + 1 ≥ e − e + 1 min { , | J |} because the function t e t − t ( e t +1) decreases on [0 , ∞ ), while the function t e t − e t +1 in-creases. (cid:4) Theorem 4.4.
Let J be a bounded interval. Then (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≥
19 min n | J | , + | J | o . Proof.
Put Q ε ( t ) def = π tt + ε , where ε >
0. Let us consider the convolution operator C Q ε on L ( R ), C Q ε f def = f ∗ Q ε . Clearly, kC Q ε k = k F Q ε k L ∞ = 1, see, for example, [Ga],Ch. III, §
1. Note that C Q ε is an integral operator with kernel Q ε ( x − y ). We can definethe integral operator X J,ε on L ( J ) with kernel1 π x − y ( x − y ) + ε e x − e y e x + e y . We have | J | · k X J,ε k ≥ ( X J,ε χ J , χ J ) = 1 π Z Z J × J x − y ( x − y ) + ε e x − e y e x + e y dxdy = 2 √ π Z | J | tt + ε e t − e t + 1 ( | J | − t ) dt and k X J,ε k ≤ kC Q ε k · (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J = (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J . Hence, (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≥ √ π · | J | Z | J | tt + ε e t − e t + 1 ( | J | − t ) dt for every ε >
0, whence (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≥ √ π Z | J | e t − t ( e t + 1) (cid:18) − t | J | (cid:19) dt ≥ √ π Z | J | e t − t ( e t + 1) dt ecause the function t e t − t ( e t +1) decreases on (0 , ∞ ). It follows that (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≥ √ π · e − e + 1 Z | J | min { , t − } dt. This implies the desired estimate. (cid:4)
Remark 1.
Every rectangle J × J is the union at most of three rectangles, each ofwhich satisfies the hypotheses of either Theorem 4.2 or Theorem 4.3. This allows us toobtain a sharp estimate for the norms in (4.1) for every rectangle J × J . Remark 2.
Remark 1 and the change of variables x log x , y log y allow us toobtain a sharp estimate for (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J , where J and J are intervals containing in(0 , ∞ ).We proceed now to estimates of multiplier norms that will be used in this paper. Theorem 4.5.
There exists a positive number C such that (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M [ a, ∞ ) , ( −∞ ,b ] ≤ C log(2 + ( b − a ) + ) for all a, b ∈ R . Proof.
The result follows from Theorems 4.1 if a ≥ b . If a < b , then[ a, ∞ ) × ( −∞ , b ] = ([ a, b ] × [ a, b ]) ∪ ([ a, b ] × ( −∞ , a ]) ∪ ([ b, ∞ ) × ( −∞ , b ]) , and we can apply Theorem 4.2 to the first rectangle and Theorem 4.1 to the remainingrectangles. (cid:4) Theorem 4.6.
There exists a positive number C such that (cid:13)(cid:13)(cid:13)(cid:13) e x − e y e x + e y (cid:13)(cid:13)(cid:13)(cid:13) M R , [ a,b ] ≤ C log(2 + b − a ) for every a, b ∈ R satisfying a < b . Proof.
We have R × [ a, b ] = ([ a, b ] × [ a, b ]) ∪ (( −∞ , a ] × [ a, b ]) ∪ ([ b, ∞ ) × [ a, b ]) . It remains to apply Theorem 4.2 to the first rectangle and Theorem 4.1 to the remainingrectangles. (cid:4)
Theorem 4.7.
There exists a positive number c such that (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [ a, ∞ ) , [0 ,b ] ≤ c log (cid:18) + ba (cid:19) for all a, b ∈ (0 , ∞ ) . roof. Theorem 4.5 with the help of the change of variables x log x and y log y yields (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [ a, ∞ ) , [ ε,b + ε ] ≤ c log (cid:18) + b + εa (cid:19) for every ε >
0, whence (cid:13)(cid:13)(cid:13)(cid:13) x − y − εx + y + ε (cid:13)(cid:13)(cid:13)(cid:13) M [ a, ∞ ) , [0 ,b ] ≤ c log (cid:18) + b + εa (cid:19) for every ε >
0. It remains to pass to the limit as ε → (cid:4) Theorem 4.8.
There exists a positive number c such that (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [ a,b ] , [0 , ∞ ) ≤ c log (cid:18) ba (cid:19) whenever a, b ∈ (0 , ∞ ) and a < b . Proof.
The result follows from Theorem 4.6 in the same way as Theorem 4.7 followsfrom Theorem 4.5. (cid:4)
Theorem 4.9.
There exists a positive number c such that (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [ a,b ] , [ a,b ] ≥ c log (cid:18) ba (cid:19) whenever a, b ∈ (0 , ∞ ) and a < b . Proof.
The result follows from Theorem 4.4 with the help of the change of variables x log x and y log y . (cid:4) Operator Lipschitz functions and operator modulus of continuity
In this section we study operator Lipschitz functions on closed subsets of the realline. It is well known that a function f on R is operator Lipschitz if and only if it is commutator Lipschitz , i.e., k f ( A ) R − Rf ( A ) k ≤ const k AR − RA k for an arbitrary bounded operator R and an arbitrary self-adjoint operator A .It turns out that the same is true for functions on closed subsets of R ; moreoverthe operator Lipschitz constant coincides with the commutator Lipschitz constant. Thefollowing theorem was proved in [AP2] (Th. 10.1) in the case F = R . The general caseis analogous to the case F = R . See also [KS] where similar results for symmetric idealnorms are considered. Theorem 5.1.
Let f be a continuous function defined on a closed subset F of R andlet C ≥ . The following are equivalent: (i) k f ( A ) − f ( B ) k ≤ C k A − B k for arbitrary self-adjoint operators A and B withspectra in F ; ii) k f ( A ) R − Rf ( A ) k ≤ C k AR − RA k for all self-adjoint operators A with σ ( A ) ⊂ F and all bounded operators R ; (iii) k f ( A ) R − Rf ( B ) k ≤ C k AR − RB k for arbitrary self-adjoint operators A and B with spectra in F and for an arbitrary bounded operator R . A function f ∈ C ( F ) is said to be operator Lipschitz if it satisfies the equivalentstatements of Theorem 5.1. We denote the set of operator Lipschitz functions on F byOL( F ). For f ∈ OL( F ), we define k f k OL( F ) to be the smallest constant satisfying theequivalent statements of Theorem 5.1. Put k f k OL( F ) = ∞ if f OL( F ).It is well known that every f in OL( F ) is differentiable at every non-isolated point of F , see [JW]. Moreover, the same argument gives differentiability at ∞ in the followingsense: there exists a finite limit lim | x |→ + ∞ x − f ( x ) provided F is unbounded.Let f ∈ OL( F ). Suppose that F has no isolated points. Put (cid:0) D f (cid:1) ( x, y ) def = ( f ( x ) − f ( y ) x − y , if x, y ∈ F , x = y,f ′ ( x ) , if x ∈ F , x = y. The following equality holds: k f k OL( F ) = k D f k M F , F . (5.1)The inequality k f k OL( F ) ≤ k D f k M F , F is an immediate consequence of the formula f ( A ) − f ( B ) = Z Z ( D f (cid:1) ( x, y ) dE A ( x )( A − B ) dE B ( y ) , (5.2)where A and B are self-adjoint operators with bounded A − B whose spectra are in F ,and E A and E B are the spectral measures of A and B . The expression on the right iscalled a double operator integral. We refer the reader to [BS1], [BS2], and [BS3] for thetheory of double operator integrals elaborated by Birman and Solomyak. The validityof formula (5.2) under the assumption D f ∈ M F , F and the inequality (cid:13)(cid:13)(cid:13)(cid:13)Z Z ( D f (cid:1) ( x, y ) dE A ( x )( A − B ) dE B ( y ) (cid:13)(cid:13)(cid:13)(cid:13) ≤ k D k M F , F k A − B k was proved in [BS3]. The opposite inequality in (5.1) is going to be proved in Corollary5.4.In the general case for f ∈ OL( F ) we can define the function (cid:0) D f (cid:1) ( x, y ) def = ( f ( x ) − f ( y ) x − y , if x, y ∈ F , x = y, , if x ∈ F , x = y. The following inequalities hold: k f k OL( F ) ≤ k D f k M F , F ≤ k f k OL( F ) . (5.3)The first inequality in (5.3) follows from the formula f ( A ) − f ( B ) = Z Z ( D f (cid:1) ( x, y ) dE A ( x )( A − B ) dE B ( y ) , (5.4) hose validity can be verified in the same way as the validity of (5.2). The secondinequality in (5.3) is going to be verified in Corollary 5.5.Let f be a continuous function on a closed set F , F ⊂ R . We define the operatormodulus of continuity Ω f, F as followsΩ f, F ( δ ) def = sup (cid:8) k f ( A ) − f ( B ) k : A = A ∗ , B = B ∗ , σ ( A ) , σ ( B ) ⊂ F , k A − B k ≤ δ (cid:9) , and the commutator modulus of continuity as followsΩ ♭f, F ( δ ) def = sup (cid:8) k f ( A ) R − Rf ( A ) k : A = A ∗ , σ ( A ) ⊂ F , k R k ≤ , k AR − RA k ≤ δ (cid:9) . One can prove that we get the same right-hand side if we require in addition that R isself-adjoint. On the other hand, k f ( A ) R − Rf ( B ) k ≤ Ω ♭f, F (cid:0) k AR − RB k (cid:1) for every self-adjoint operators with σ ( A ) , σ ( B ) ⊂ F and for every bounded operator R with k R k ≤ f, F ≤ Ω ♭f, F ≤ f, F .These results were obtained in [AP2] in the case F = R . The same reasoning works inthe general case. Lemma 5.2.
Let F be a closed subset of R and let µ and ν be regular positive Borelmeasures on F . Suppose that k is a function in L ( F × F , µ ⊗ ν ) such that k = 0 on thediagonal ∆ F def = { ( x, x ) : x ∈ F } almost everywhere with respect to µ ⊗ ν . Then k k D f k B µ,ν F , F ≤ k f k OL( F ) k k k B µ,ν F , F for every continuous function f on F . Proof.
Let F n def = F ∩ [ − n, n ], and let µ n and ν n be the restrictions of µ and ν to F n .Clearly, lim n →∞ k k k B µn,νn F n, F n = k k k B µ,ν F , F for every k ∈ L ( F × F , µ ⊗ ν )and lim n →∞ k f k OL( F n ) = k f k OL( F ) for every f ∈ C ( F ) . Thus we may assume that F is compact. It suffices to consider the case when k vanishesin a neighborhood of the diagonal ∆ F . Put l ( x, y ) def = ( x − y ) − k ( x, y ). Denote by A and B multiplications by the independent variable on L ( F , µ ) and L ( F , ν ). Then I µ,νk = A I µ,νl − I µ,νl B and I µ,νk D f = f ( A ) I µ,νl − I µ,νl f ( B ). It remains to observe that (cid:13)(cid:13) f ( A ) I µ,νl − I µ,νl f ( B ) (cid:13)(cid:13) ≤ k f k OL( F ) (cid:13)(cid:13) A I µ,νl − I µ,νl B (cid:13)(cid:13) , (cid:13)(cid:13) A I µ,νl − I µ,νl B (cid:13)(cid:13) = k k k B µ,ν F , F , and (cid:13)(cid:13) f ( A ) I µ,νl − I µ,νl f ( B ) (cid:13)(cid:13) = k k D f k B µ,ν F , F . (cid:4) Corollary 5.3.
Let F be a closed subset of R with no isolated points, and let µ and ν be finite positive Borel measures on F . Suppose that f is a differentiable function on F and k ∈ L ( F × F , µ ⊗ ν ) . If k vanishes µ ⊗ ν -almost everywhere on the diagonal ∆ F def = { ( x, x ) : x ∈ F } , then k k D f k B µ,ν F , F ≤ k f k OL( F ) k k k B µ,ν F , F . roof. It suffices to observe that k D f = k D f almost everywhere with respect to µ ⊗ ν . (cid:4) Corollary 5.4.
Let F be a closed subset of R with no isolated points, and let µ and ν be finite positive Borel measures on F . If f is a differentiable function on F , then k D f k M F , F ≤ k f k OL( F ) . Proof.
Let µ be a regular Borel measure on F with no atoms and such that supp µ = F .Then ( µ ⊗ µ )(∆ F ) = 0 and Corollary 5.3 implies that k k D f k B µ,µ F , F ≤ k f k OL( F ) k k k B µ,µ F , F for all k ∈ L ( F × F , µ ⊗ µ ). Hence, k D f k M µ,µ F , F ≤ k f k OL( F ) . It remains to apply Theorem2.6. (cid:4) Corollary 5.5.
Let F be a closed subset of R . Then k D f k M F , F ≤ k f k OL( F ) for every f ∈ C ( F ) . Proof.
Let µ and ν be regular Borel measures on F . We have to verify that k k D f k B µ,ν F , F ≤ k f k OL( F ) k k k B µ,ν F , F for every k ∈ L ( F × F , µ ⊗ ν ). Put k = χ ∆ F k and k = k − k . We have k k k B µ,ν F , F ≤ k k k B µ,ν F , F . This inequality can be verified easily. We leave the verification to the reader.It follows that k k k B µ,ν F , F ≤ k k k B µ,ν F , F + k k k B µ,ν F , F ≤ k k k B µ,ν F , F . It remains to observe that k k D f k B µ,ν F , F = k k D f k B µ,ν F , F ≤ k f k OL( F ) k k k B µ,ν F , F ≤ k f k OL( F ) k k k B µ,ν F , F . (cid:4) Let F and F be closed subsets of R . We define the space OL( F , F ) as the space offunctions f in C ( F ∪ F ) such that k f ( A ) R − Rf ( B ) k ≤ C k AR − RB k (5.5)for all bounded operator R and all self-adjoint operators A and B such that σ ( A ) ⊂ F and σ ( B ) ⊂ F with some positive number C . Denote by k f k OL( F , F ) the minimalconstant satisfying (5.5). Clearly, k f k OL( F , F ) = k f k OL( F , F ) and k f k OL( F , F ) = k f k OL( F ) .As in the case F = F , we can prove that k f k OL( F , F ) ≤ k D f k M F , F ≤ k f k OL( F , F ) . (5.6)(cf. (5.3)). Remark.
In the case where F = F we cannot claim that the inequality k f ( A ) − f ( B ) k ≤ C k A − B k (5.7)for all self-adjoint A and B such that σ ( A ) ⊂ F and σ ( B ) ⊂ F implies (5.5). ndeed, in the case f ( t ) = | t | , F = ( −∞ , F = [0 , ∞ ), inequality (5.7) holdswith C = 1 because k A − B k ≤ k A + B k for positive self-adjoin operators A and B . However, inequality (5.5) does not hold withany positive C . Indeed, (cid:13)(cid:13)(cid:13)(cid:13) | x | − | y | x − y (cid:13)(cid:13)(cid:13)(cid:13) M ( −∞ , , [1 , ∞ ) = (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [1 . ∞ ) , [1 , ∞ ) = ∞ by Theorem 4.9. Theorem 5.6.
Suppose that inequality (5.5) holds for every bounded operator R andarbitrary self-adjoint operators A and B with simple spectra such that σ ( A ) ⊂ F and σ ( B ) ⊂ F . Then f ∈ OL( F , F ) and k f k OL( F , F ) ≤ C . Proof.
We have to prove inequality (5.5) for arbitrary self-adjoint operators A and B with σ ( A ) ⊂ F and σ ( B ) ⊂ F . It is convenient to think that the operators A and B act in different Hilbert spaces. Let A act in H and B in H . Then R acts from H into H . We are going to verify that (cid:12)(cid:12) ( f ( A ) Ru, v ) − ( Rf ( B ) u, v ) (cid:12)(cid:12) = (cid:12)(cid:12) ( Ru, f ( A ) v ) − ( f ( B ) u, R ∗ v ) (cid:12)(cid:12) ≤ C k AR − RB k for every unit vectors u ∈ H and v ∈ H . Denote by H and H the invariant subspacesof A and B generated by v and u . Clearly, A = A |H and B = B |H are self-adjointoperators with simple spectra. Consider the operator R : H → H , R h def = P Rh for h ∈ H , where P is the orthogonal projection from H onto H . Note that for h ∈ H ,we have A R h = AP Rh = P ARh and R B h = P RBh . Clearly, k A R − R B k ≤k AR − RB k . Applying (5.5) to the operators A , B , and R , we obtain | ( f ( A ) Ru, v ) − ( Rf ( B ) u, v ) | = | ( Ru, f ( A ) v ) − ( Rf ( B ) u, v ) | = | ( R u, f ( A ) v ) − ( R f ( B ) u, v ) | = | ( f ( A ) R u, v ) − ( R f ( B ) u, v ) |≤ C k A R − R B k ≤ C k AR − RB k . (cid:4) Remark.
Theorem 5.6 allows us to give alternative the proofs of (5.1), (5.3) and (5.6)that do not use double operator integrals.
Theorem 5.7.
Let f be a function defined on Z . Then Ω ♭f, Z ( δ ) = δ k f k OL( Z ) for δ ∈ (cid:0) , π (cid:3) . Proof.
The inequality Ω ♭f, Z ( δ ) ≤ δ k f k OL( Z ) , δ > , s a consequence of Theorem 5.1. Let us prove the opposite inequality for δ ∈ (cid:0) , π (cid:3) .Fix ε >
0. There exists a self-adjoint operator A and a bounded operator R such that k AR − RA k = 1, σ ( A ) ⊂ Z , and k f ( A ) R − Rf ( A ) k ≥ k f k OL( Z ) − ε . Put R A def = X j = k E A ( { j } ) RE A ( { k } ) = R − X j ∈ Z E A ( { j } ) RE A ( { j } ) . Clearly, AR − RA = AR A − R A A and f ( A ) R − Rf ( A ) = f ( A ) R A − R A f ( A ). Thus wemay assume that R = R A . Note that AR − RA = X j = k ( j − k ) E A ( { j } ) RE A ( { k } ) . Since R = R A = X j = k j − k ( j − k ) E A ( { j } ) RE A ( { k } ) , we have R = H ⋆ ( AR − RA ), where H ( j, k ) def = ( j − k , if j = k, , if j = k, where ⋆ denotes Schur–Hadamard multiplication, see (2.2). It follows that k R k ≤ k H k M Z , Z k AR − RA k = k H k M Z , Z = π δ ∈ (cid:0) , π (cid:3) . Then k A ( δR ) − ( δR ) A k = δ and k δR k ≤
1. Hence,Ω ♭f, Z ( δ ) ≥ δ k f ( A ) R − Rf ( A ) k ≥ δ (cid:0) k f k OL( Z ) − ε (cid:1) . Passing to the limit as ε →
0, we obtain the desired result. (cid:4)
Let ω f, F denote the usual scalar modulus of continuity of a continuous function f defined on F . Clearly, ω f, F ≤ Ω f, F . Put ω f def = ω f, R and Ω f def = Ω f, R . We are going to getsome estimates for the commutator modulus of continuity Ω ♭f, F . We consider first thecase when F = R . The following theorem is contained implicitly in [NiF]. Theorem 5.8.
Let f be a continuous function on R . Then Ω ♭f ( δ ) ≤ ω f ( δ/
2) + 2 k f ( δx ) k OL( Z ) . Proof.
Let k AR − RA k ≤ δ with k R k = 1. We can take a self-adjoint operator A δ such that A δ A = AA δ , k A − A δ k ≤ δ/ σ ( A δ ) ⊂ δ Z . Then k f ( A ) − f ( A δ ) k ≤ ω f ( δ/ k A δ R − RA δ k ≤ k A δ R − AR k + k AR − RA k + k RA − RA δ k ≤ δ Hence, k f ( A ) R − Rf ( A ) k ≤ k f ( A ) R − f ( A δ ) R k + k f ( A δ ) R − Rf ( A δ ) k + k Rf ( A δ ) − Rf ( A ) k≤ ω f ( δ/
2) + k A δ R − RA δ k · k f k OL( δ Z ) ≤ ω f ( δ/
2) + 2 δ k f k OL( δ Z ) = 2 ω f ( δ/
2) + 2 k f ( δx ) k OL( Z ) . (cid:4) heorem 5.9. Let f be a continuous function on R . Then Ω ♭f ( δ ) ≥ max (cid:26) ω f ( δ ) , π k f ( δx ) k OL( Z ) (cid:27) for all δ > . Proof.
Clearly, ω f ≤ Ω f ≤ Ω ♭f . It remains to prove that k f ( δx ) k OL( Z ) ≤ π Ω ♭f ( δ ). Wehave Ω ♭f ( δ ) ≥ Ω ♭f,δ Z ( δ ) = Ω ♭f ( δx ) , Z (1) ≥ Ω ♭f ( δx ) , Z (cid:16) π (cid:17) = 2 π (cid:13)(cid:13) f ( δx ) (cid:13)(cid:13) OL( Z ) by Theorem 5.7. (cid:4) We consider now similar estimates of Ω ♭f, F for an arbitrary closed subset F of R . Recallthat a subset Λ of R is called a δ net for F if F ⊂ S t ∈ Λ [ t − δ, t + δ ]. Theorem 5.10.
Let f be a continuous function on a closed subset F of R . Supposethat F δ is a subset of F that forms a δ/ net of F . Then Ω ♭f, F ( δ ) ≤ ω f, F ( δ/
2) + 2 δ k f k OL( F δ ) . Proof.
The proof is similar to the proof of Theorem 5.8. It suffices to replace the δ/ δ Z of R with the δ/ F δ of F . (cid:4) Theorem 5.11.
Let f be a continuous function on a closed subset F of R and let δ > . Suppose that Λ and M are closed subsets of F such that (Λ − M) ∩ ( − δ, δ ) ⊂ { } .Then Ω ♭f, F ( δ ) ≥ max (cid:26) ω f, F ( δ ) , δ k D f k M Λ , M (cid:27) . Proof.
Clearly, ω f, F ≤ Ω f, F ≤ Ω ♭f, F . Note that k D f k M Λ , M = sup a> k D f k M Λ ∩ [ − a,a ]) , M ∩ [ − a,a ] . Thus it suffices to prove that Ω ♭f, F ( δ ) ≥ δ k D f k M Λ , M in the case when Λ and M are bounded.Let ε >
0. There exist positive regular Borel measures λ on Λ, µ on M, and a function k in L (Λ × M , λ ⊗ µ ) such that k k k B λ,µ Λ , M = 1 and k k D f k B λ,µ Λ , M ≥ k D f k M Λ , M − ε . Wedefine the function k in L (Λ × M , λ ⊗ µ ) by k ( x, y ) def = ( k ( x, y ) , if x = y, , if x = y. Then k D f = k D f and k k k B λ,µ Λ , M ≤
2. Put Φ( x, y ) def = f δ ( x − y ) where f δ denotes thesame as in Corollary 3.3. We define the self-adjoint operators A : L (Λ , λ ) → L (Λ , λ )and B : L (M , µ ) → L (M , µ ) by ( Af )( x ) def = xf ( x ) and ( Bg )( y ) def = yg ( y ). Put h ( x, y ) def = Φ( x, y ) k ( x, y ) = Φ( x, y ) k ( x, y ) . learly, k h k B λ,µ Λ , M ≤ k Φ k M λ,µ Λ , M k k k B λ,µ Λ , M ≤ k Φ k M R , R ≤ δ by Corollary 3.3.Clearly, A I h − I h B = I k and f ( A ) I h − I h f ( B ) = I k D f . (Recall that I ϕ is theintegral operator from L (M , µ ) into L (Λ , λ ) with kernel ϕ ∈ L (Λ × M , λ ⊗ ν ).) Then (cid:13)(cid:13)(cid:13)(cid:13) δ I h (cid:13)(cid:13)(cid:13)(cid:13) = δ k h k B λ,µ Λ , M ≤ , (cid:13)(cid:13)(cid:13)(cid:13) A (cid:18) δ I h (cid:19) − (cid:18) δ I h (cid:19) B (cid:13)(cid:13)(cid:13)(cid:13) = δ k k k B λ,µ Λ , M ≤ δ, and (cid:13)(cid:13)(cid:13)(cid:13) f ( A ) (cid:18) δ I h (cid:19) − (cid:18) δ I h (cid:19) f ( B ) (cid:13)(cid:13)(cid:13)(cid:13) = δ k k D f k B λ,µ Λ , M ≥ δ (cid:0) k D f k M Λ , M − ε (cid:1) . Hence, Ω ♭f, F ( δ ) ≥ δ (cid:0) k D f k M Λ , M − ε (cid:1) for every ε > (cid:4) Theorem 5.11 allows us to obtain another proof of Theorem 4.17 in [AP3].
Theorem 5.12.
Let f be a continuous function on an unbounded closed subset F of R . Suppose that Ω f, F ( δ ) < ∞ for δ > . Then the function t t − f ( t ) has a finite limitas | t | → ∞ , t ∈ F . Proof.
Assume the contrary. Then there exists a sequence { λ n } ∞ n =1 in F such that | λ n +1 | − | λ n | > n ≥
1, lim n →∞ | λ n | = ∞ and the sequence { λ − n f ( λ n ) } ∞ n =1 hasno finite limit. Denote by Λ the image of the sequence { λ n } ∞ n =1 . Then k f k OL(Λ) = ∞ .This fact is contained implicitly in [JW]. Indeed, Theorem 4.1 in [JW] implies thatevery operator Lipschitz function f is differentiable at every non-isolated point. It iswell known that the same argument gives us the differentiability at ∞ in the followingsense: the function t t − f ( t ) has a finite limit as | t | → ∞ , provided the domain of f is unbounded. Applying Theorem 5.11 for M = Λ and δ = 1, we find that Ω f, F (1) = ∞ . (cid:4) We need the following known result, see [KST]. We give the proof for the reader’sconvenience.
Theorem 5.13.
Let f be a bounded continuous function on a closed subset F of R .Suppose that f ∈ OL (cid:0) ( −∞ , ∩ F (cid:1) and f ∈ OL (cid:0) [ − , ∞ (cid:1) ∩ F ) . Then f ∈ OL( F ) and k f k OL( F ) ≤ C (cid:0) k f k OL(( −∞ , ∩ F ) + k f k OL([ − , ∞ ) ∩ F ) + sup F | f | (cid:1) , where C is a numerical constant. Proof.
Put F = F ∩ ( −∞ , − F = F ∩ [ − , F = F ∩ [1 , ∞ ). We have k f k OL( F ) ≤ k D f k M F , F ≤ X j =1 3 X k =1 k D f k M F j, F k = X j =1 k D f k M F j, F j + 2 k D f k M F , F + 2 k D f k M F , F + 2 k D f k M F , F . ach term k D f k M F j, F k except k D f k M F , F can be estimated in terms of2 k f k OL( F ∪ F ) or 2 k f k OL( F ∪ F ) .Let us estimate k D f k M F , F . We have k D f k M F , F = (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M F , F ≤ (cid:18) sup F | f | (cid:19) (cid:13)(cid:13)(cid:13)(cid:13) x − y (cid:13)(cid:13)(cid:13)(cid:13) M F , F + (cid:18) sup F | f | (cid:19) (cid:13)(cid:13)(cid:13)(cid:13) x − y (cid:13)(cid:13)(cid:13)(cid:13) M F , F ≤ (cid:18) sup F | f | (cid:19) (cid:13)(cid:13)(cid:13)(cid:13) x − y (cid:13)(cid:13)(cid:13)(cid:13) M F , F ≤ F | f | because by Corollary 3.3, (cid:13)(cid:13)(cid:13)(cid:13) x − y (cid:13)(cid:13)(cid:13)(cid:13) M F , F ≤ k f ( x − y ) k M R , R ≤ , where f means the same as in Corollary 3.3.Thus k f k OL( F ) ≤ k f k OL( F ∪ F ) + 4 k f k OL( F ∪ F ) + 4 sup F | f | . (cid:4) The operator Lipschitz norm of the function x
7→ | x | on subsets of R In this section we obtain sharp estimates of the operator modulus of continuity ofthe function x
7→ | x | on certain subsets of the real line. This allows us to obtain sharpestimates of (cid:13)(cid:13) | S | − | T | (cid:13)(cid:13) for arbitrary bounded linear operators S and T . Note that ourestimates considerably improve earlier results of [Ka].Put Abs( x ) def = | x | . For J ⊂ [0 , ∞ ), we put log( J ) def = { log t : t ∈ J, t > } . Theorem 6.1.
There exist positive numbers C and C such that C log (cid:0) | log( J ∩ J ) | (cid:1) ≤ k Abs k OL(( − J ) ∪ J ) ≤ C log (cid:0) | log( J ∩ J ) | (cid:1) for all intervals J and J in (0 , ∞ ) . Proof.
Put J = J ∩ J . Let us first establish the lower estimate. Note that k Abs k OL(( − J ) ∪ J ) ≥ k Abs k OL( J ) = 1. This proves the lower estimate in the case | log( J ) | ≤
1. In the case | log( J ) | > k Abs k OL(( − J ) ∪ J ) ≥ k Abs k OL(( − J ) ∪ J ) ≥ (cid:13)(cid:13)(cid:13)(cid:13) | x | − | y | x − y (cid:13)(cid:13)(cid:13)(cid:13) M − J,J = (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M J,J ≥ c log(1 + | log( J ) | ) ≥ c log(2 + | log( J ) | )by Theorem 4.9. e proceed now to the upper estimate. We consider first the case when J = J . Then k Abs k OL(( − J ) ∪ J ) ≤ k Abs k OL(( − J ) ∪ [0 , ∞ )) ≤ (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M J, [0 , ∞ ) and we can apply Theorem 4.8. The case J = J is similar. Suppose that J = J and J = J . Then inf J = inf J . Let inf J > inf J . Put a def = inf J and b def = sup J . Then k Abs k OL(( − J ) ∪ J ) ≤ k Abs k OL(( −∞ − a ] ∪ [0 ,b )) ≤ (cid:13)(cid:13)(cid:13)(cid:13) x − yx + y (cid:13)(cid:13)(cid:13)(cid:13) M [ a, ∞ ) , [0 ,b ) and the result follows from Theorem 4.7. (cid:4) Let us state two special cases of Theorem 6.1.
Theorem 6.2.
There exist positive constants C and C such that C log(2 + log( ba − )) ≤ k Abs k OL(( −∞ , ∪ [ a,b ]) ≤ C log(2 + log( ba − )) for every a, b ∈ (0 , ∞ ) with a < b . Theorem 6.3.
There exist positive constants C and C such that C log(2 + log + ( ba − )) ≤ k Abs k OL(( − b, ∪ [ a, ∞ )) ≤ C log(2 + log + ( ba − )) for every a, b ∈ (0 , ∞ ) . Theorem 6.4.
Let ξ a = Abs (cid:12)(cid:12) [ − a, ∞ ) and η a = Abs (cid:12)(cid:12) [ − a, a ] , where a > . Then thereexist positive numbers C and C such that C δ log(2 + log + ( aδ − )) ≤ Ω η a ( δ ) ≤ Ω ξ a ( δ ) ≤ C δ log(2 + log + ( aδ − )) for every δ > . Proof.
Put F δ def = [ − a, ∞ ) \ (0 , δ ). Clearly, F δ is a δ -net of ( −∞ , a ]. Hence, byTheorem 5.10 we have Ω ξ a ( δ ) ≤ Ω ♭ξ a ( δ ) ≤ δ + 2 δ k ξ a k OL( F δ ) . Applying Theorem 6.3, we obtain the desired upper estimate.To obtain the lower estimate, we use Theorem 5.11. Clearly, Ω η a ( δ ) ≥ δ for all δ ∈ (0 , a ]. Thus it suffices to consider the case δ ∈ (0 , a ). Put Λ = [ − a,
0] and M = [ δ, a ].By Theorem 5.11, Ω η a ( δ ) ≥
12 Ω ♭η a ( δ ) ≥ δ k D η a k M Λ , M . It remains to apply Theorem 4.9. (cid:4)
Theorem 6.5.
There exists a positive number C such that (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≤ C k A − B k log (cid:18) k A k + k B kk A − B k (cid:19) for every bounded self-adjoint operators A and B . Proof.
This is a special case of Theorem 6.4 that corresponds to a = k A k + k B k . (cid:4) Theorem also 6.4 allows us to prove that the upper estimate in Theorem 6.5 is sharp. heorem 6.6. Let a > . There is a positive number c such that for every δ ∈ (0 , a ) ,there exist self-adjoint operators A and B such that k A + B k ≤ a , k A − B k ≤ δ , but (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≥ cδ log (cid:16) aδ (cid:17) . We proceed now to the case of arbitrary (not necessarily self-adjoint) operators. Recallthat for a bounded operator S on Hilbert space, its modulus S is defined by | S | def = ( S ∗ S ) / . Theorem 6.7.
There exists a positive number C such that (cid:13)(cid:13) | S | − | T | (cid:13)(cid:13) ≤ C k S − T k log (cid:18) k S k + k T kk S − T k (cid:19) for every bounded operators S and T . Proof.
Put A = S ∗ S ! and B = T ∗ T ! . Clearly, A and B are self-adjoint operators with | A | = | S | | S ∗ | ! and | B | = | T | | T ∗ | ! . Hence, (cid:13)(cid:13) | S | − | T | (cid:13)(cid:13) ≤ (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≤ C k A − B k log (cid:18) k A k + k B kk A − B k (cid:19) = C k S − T k log (cid:18) k S k + k T kk S − T k (cid:19) . (cid:4) Remark.
Theorem 6.7 significantly improves Kato’s inequality obtained in [Ka]: (cid:13)(cid:13) | S | − | T | (cid:13)(cid:13) ≤ π k S − T k (cid:18) k S k + k T kk S − T k (cid:19) . The operator modulus of continuity of a certain piecewise linear function
In this section we obtain a sharp estimate for the operator modulus of continuity ofthe piecewise linear function κ defined by κ ( t ) def = , if t ≥ ,t, if − < t ≤ . − , if t > . The results obtained in this section will be used in the next section to estimate theoperator modulus of continuity of functions concave on R + .It is easy to see that κ ( t ) = (cid:0) | t | − | − t | (cid:1) . heorem 7.1. There exist positive numbers C and C such that C log | log δ | ≤ k κ k OL(( −∞ , − − δ ] ∪ [ − , ∪ [1+ δ, ∞ )) ≤ C log | log δ | for every δ ∈ (0 , ) . Proof.
Put κ = κ (cid:12)(cid:12) (( −∞ , − − δ ] ∪ [ − , κ = κ (cid:12)(cid:12) ([ − , ∪ [1 + δ, ∞ )). Notethat κ ( t ) = 12 (cid:0) | t | − t (cid:1) and κ ( t ) = 12 (cid:0) t − | t − | (cid:1) . It follows from Theorem 6.3 that C log | log δ | ≤ k κ k OL ≤ C log | log δ | and C log | log δ | ≤ k κ k OL ≤ C log | log δ | . Thus the desired lower estimate is evident and the required upper estimate follows fromTheorem 5.13. (cid:4)
Theorem 7.2.
There exist positive numbers c and c such that c δ log (cid:0) δ − ) (cid:1) ≤ Ω κ ( δ ) ≤ c δ log (cid:0) δ − ) (cid:1) for every δ > . Proof.
Note that lim t →∞ t log (cid:0) t − ) (cid:1) = 1. Thus it suffices to consider thecase when 0 < δ ≤ . Put F δ def = ( −∞ , − − δ ] ∪ [ − , ∪ [1 + δ, ∞ ). Clearly, F δ is a δ -net for R . Hence, by Theorem 5.10, we haveΩ κ ( δ ) ≤ Ω ♭ κ ( δ ) ≤ δ + 2 δ k κ k OL( F δ ) . The desired upper estimate follows now from Theorem 7.1.To obtain the lower estimate we can apply Theorem 6.4 because κ ( t ) = ( | t |− t )for t ≤ (cid:4) Operator moduli of continuity of concave functions on R + . Recall that in [AP2] we proved that if f is a continuous function on R , then its operatormodulus of continuity Ω f admits the estimateΩ f ( δ ) ≤ const δ Z ∞ δ ω f ( t ) t dt = const Z ∞ ω f ( tδ ) t ds, δ > . In this section we show that if f vanishes on ( −∞ ,
0] and is a concave nondecreasingfunction on [0 , ∞ ), then the above estimate can be considerably improved.We also obtain several other estimates of operator moduli of continuity. heorem 8.1. Suppose that f ′′ = µ ∈ M ( R ) (in the distributional sense), µ ( R ) = 0 ,and Z R log(log( | t | + 3)) d | µ | ( t ) < ∞ . Then Ω f ( δ ) ≤ c k µ k M ( R ) δ log(log( δ − + 3)) , where c is a numerical constant. Proof.
Put ϕ s ( t ) def = 12 (cid:0) | t | + | s | (cid:1) − | t − s | , s, t ∈ R . (8.1)It is easy to see that ϕ s ( t ) def = | s | κ (cid:18) ts − (cid:19) + | s | s = 0 . Clearly, ϕ ′′ s = δ − δ s and ϕ s (0) = 0 . (8.2)Theorem 7.2 implies thatΩ ϕ s ( t ) ≤ const t log (cid:18) (cid:18) | s | t (cid:19)(cid:19) ≤ const t log (cid:18) (cid:18) | s | t (cid:19)(cid:19) , t > . (8.3)It is easy to see that t log (cid:16) (cid:0) t − | s | (cid:1)(cid:17) ≤ const (cid:0) log(log( | s | + 3)) (cid:1) t log (cid:16) log (cid:0) t − + 3 (cid:1)(cid:17) . To complete the proof, it suffices to observe that f ( t ) = at + b − Z R ϕ s ( t ) dµ ( s ) , for some a, b ∈ C , which follows easily from (8.2). (cid:4) The assumption that µ ( R ) = 0 in the hypotheses of Theorem 8.1 is essential. More-over, the following result holds. Theorem 8.2.
Suppose that f ′′ = µ ∈ M ( R ) and µ ( R ) = 0 . Then Ω f ( t ) = ∞ forevery t > . Proof.
Indeed, it is easy to see that f ′ ( t ) = c + µ (( −∞ , t )) for almost all t ∈ R .Hence, lim t →∞ f ( t ) t = lim t →∞ f ′ ( t ) = c + µ ( R ) and lim t →−∞ f ( t ) t = lim t →−∞ f ′ ( t ) = c. The result follows from Theorem 5.12. (cid:4)
Let G be an open subset of R . Denote by M loc ( G ) the set of all distributions on G that are locally (complex) measures. heorem 8.3. Let f ∈ C ( R ) . Put µ def = f ′′ in the sense of distributions. Suppose that lim | t |→∞ t − f ( t ) = 0 , µ (cid:12)(cid:12) ( R \ { } ) ∈ M loc ( R \ { } ) and Z R \{ } log(1 + log(1 + | s | )) d | µ | ( s ) < ∞ . Then Ω f ( δ ) ≤ const δ Z R \{ } log (cid:16) (cid:0) | s | δ − (cid:1)(cid:17) d | µ | ( s ) . Proof.
Put g ( t ) = − Z R \{ } ϕ s ( t ) dµ ( s ) , where ϕ s is defined by (8.1). Inequality (8.3) implies thatΩ g ( δ ) ≤ const δ Z R \{ } log (cid:16) (cid:0) | s | δ − (cid:1)(cid:17) d | µ | ( s ) . (8.4)In particular, g is continuous on R . Clearly, g ′′ = f ′′ on R \ { } . Hence, f ( x ) − g ( x ) = a | x | + bx + c for some a, b, c ∈ C . It follows from (8.4) thatlim | t |→∞ (cid:12)(cid:12)(cid:12)(cid:12) g ( t ) t (cid:12)(cid:12)(cid:12)(cid:12) ≤ lim t →∞ ω g ( t ) t ≤ lim t →∞ Ω g ( t ) t = 0 = lim | t |→∞ f ( t ) t which implies that f − g = const. (cid:4) Corollary 8.4.
Let a > and let f be a continuous function on R that is constanton R \ ( − a, a ) . Put µ def = f ′′ in the sense of distributions. Suppose that µ (cid:12)(cid:12) ( R \ { } ) ∈ M loc ( R \ { } ) and C def = sup s> | µ | (cid:0) [ s, s ] ∪ [ − s, − s ] (cid:1) < ∞ . (8.5) Then Ω f ( δ ) ≤ C const δ (cid:16) log aδ (cid:17) log (cid:16) log aδ (cid:17) for δ ∈ (cid:16) , a (cid:17) . Proof.
By Theorem 8.3,Ω f ( δ ) ≤ const δ (cid:18)Z a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) d | µ ( s ) | + Z a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) d | µ ( − s ) | (cid:19) = const δ X n ≥ Z − n a − n − a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) d | µ | ( s )+ const δ X n ≥ Z − n a − n − a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) d | µ | ( − s ) . It follows now from (8.5) and the inequalitylog(1 + log(1 + αx )) ≤ x )) , < x < ∞ , < α ≤ , hatΩ f ( δ ) ≤ const δ X n ≥ Z − n a − n − a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) dss = const δ Z a log (cid:0) (cid:0) sδ − (cid:1)(cid:1) dss = const δ Z a/δ log (cid:0) s ) (cid:1) dss ≤ const δ + const δ Z a/δ log (cid:0) s ) (cid:1) dss = const δ (cid:0) log (cid:0) s ) (cid:1) log s (cid:1)(cid:12)(cid:12)(cid:12) a/δ − Z a/δ log s ds (1 + s ) log (cid:0) s ) (cid:1) ! ≤ const δ + const δ (cid:0) log(1 + log(1 + s ) (cid:1) log s (cid:1)(cid:12)(cid:12)(cid:12) a/δ ≤ const δ (cid:16) log aδ (cid:17) log (cid:16) log aδ (cid:17) for sufficiently small δ . (cid:4) Corollary 8.5.
Let f be a continuous function on R that is constant on R \ ( − a, a ) .Suppose that f is twice differentiable on R \ { } and C def = sup s =0 (cid:12)(cid:12) sf ′′ ( s ) (cid:12)(cid:12) < ∞ . Then Ω f ( δ ) ≤ const C δ (cid:16) log aδ (cid:17) log (cid:16) log aδ (cid:17) for δ ∈ (cid:16) , a (cid:17) . The following result shows that in a sense Theorem 8.1 cannot be improved.
Theorem 8.6.
Let h be a positive continuous function on R . Suppose that for every f ∈ C ( R ) such that f ′′ = µ ∈ M ( R ) , µ ( R ) = 0 , and Z R h ( t ) d | µ | ( t ) < ∞ , we have Ω f ( δ ) < ∞ , δ > . Then for some positive number c , h ( t ) ≥ c log(log( | t | + 3)) , t ∈ R . We need the following lemma, in which ϕ s is the function defined by (8.1). Lemma 8.7.
There is a positive number c such that for every s ≥ , there existself-adjoint operators A and B satisfying the conditions: σ ( A ) , σ ( B ) ⊂ (cid:18) s , s (cid:19) , k A − B k ≤ , and k ϕ s ( A ) − ϕ s ( B ) k ≥ c log log s. roof. Clearly, it suffices to prove the lemma for sufficiently large s . By Theorem 6.4,there exist self-adjoint operators A and B such that k A k , k B k < k A − B k ≤ /s ,and (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≥ const s − log (cid:0) s (cid:1) . Put A def = sI + s A and B def = sI + s B .Then σ ( A ) , σ ( B ) ⊂ (cid:0) s , s (cid:1) and k A − B k ≤
1. Let us estimate k ϕ s ( A ) − ϕ s ( B ) k . Clearly, ϕ s ( A ) − ϕ s ( B ) = s A − B ) − s | A | − | B | ) . Hence, k ϕ s ( A ) − ϕ s ( B ) k ≥ s (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) − s k A − B k≥ const log log s − ≥ const log log s for sufficiently large s . (cid:4) Proof of Theorem 8.6.
Assume the contrary. Then there exists a sequence { s n } ofreal numbers such that lim n →∞ | s n | = ∞ and lim n →∞ (log(log( | s n | ))) − h ( s n ) = 0. Passing toa subsequence, we can reduce the situation to the case when s n > n or s n < n . Without loss of generality we may assume that s n > n . Moreover,we may also assume that s ≥ s n +1 ≥ s n and log log s n ≥ n (1 + h ( s n )) for every n ≥
1. Put α n def = n (log log s n ) − for n ≥ f ( t ) def = P n ≥ α n ϕ s n ( t ). Note that theseries converges for every t because σ def = P n ≥ α n < ∞ . Moreover, f ′′ = σδ − X n ≥ α n δ s n and σh (0) + X n ≥ α n h ( s n ) < ∞ . By Lemma 8.7, there exist two sequences { A n } n ≥ and { B n } n ≥ of self-adjoint operatorssuch that σ ( A n ) , σ ( B n ) ⊂ (cid:18) s n , s n (cid:19) , k A n − B n k ≤ , and k ϕ s n ( A n ) − ϕ s n ( B n ) k ≥ c log log s n . Note that ϕ s k ( A n ) = ϕ s k ( B n ) = s k I for k < n . Also, ϕ s k ( A n ) = A n and ϕ s k ( B n ) = B n for k > n . Hence, f ( A n ) − f ( B n ) = α n ( ϕ s n ( A n ) − ϕ s n ( B n )) + X k>n α k ( A n − B n ) , and so k f ( A n ) − f ( B n ) k ≥ α n (cid:13)(cid:13) ϕ s n ( A n ) − ϕ s n ( B n ) (cid:13)(cid:13) − X k>n α k k A n − B n k≥ Cα n log log s n − X k>n α k → ∞ as n → ∞ . Thus Ω f (1) = ∞ and we get a contradiction. (cid:4) n [AP2] it was proved that Ω f ( δ ) ≤ Z ∞ ω f ( δs ) s ds for every f ∈ C ( R ). The following theorem shows that this estimate can be improvedessentially for functions f concave on a ray. Theorem 8.8.
Let f be a continuous nondecreasing function such that f ( t ) = 0 for t ≤ , lim t →∞ t − f ( t ) = 0 , and f is concave on [0 , ∞ ) . Then Ω f ( δ ) ≤ c Z ∞ e f ( δs ) dss log s , where c as a numerical constant. Proof.
Let µ = − f ′′ (in the distributional sense). Clearly, µ = 0 on ( −∞ , µ is a positive regular measure on (0 , ∞ ) because f is concave on (0 , ∞ ). Hence, µ ∈ M loc ( R \ { } ). By Theorem 8.3, we haveΩ f ( δ ) ≤ const δ Z ∞ log(1 + log(1 + sδ − )) dµ ( s ) . To estimate this integral, we use the equality f ′ ( t ) = µ ( t, ∞ ) for almost all t > δ Z ∞ log(1 + log(1+ sδ − )) dµ ( s ) = Z ∞ (cid:18)Z s dt (1 + log(1 + tδ − ))(1 + tδ − ) (cid:19) dµ ( s )= Z ∞ f ′ ( t ) dt (1 + log(1 + tδ − ))(1 + tδ − )= δ − Z ∞ (cid:18)Z ∞ t (2 + log(1 + sδ − )) ds (1 + log(1 + sδ − )) (1 + sδ − ) (cid:19) f ′ ( t ) dt = δ − Z ∞ sδ − )(1 + log(1 + sδ − )) (1 + sδ − ) f ( s ) ds = Z ∞ s )(1 + log(1 + s )) (1 + s ) f ( sδ ) ds ≤ Z ∞ s ))(1 + s ) f ( sδ ) ds. It remains to observe that Z e s ))(1 + s ) f ( sδ ) ds ≤ f ( eδ ) Z e s ))(1 + s ) ds ≤ f ( eδ ) Z ∞ ds (1 + s ) = f ( eδ ) ≤ const Z ∞ e f ( sδ ) dss log s nd Z ∞ e s ))(1 + s ) f ( sδ ) ds ≤ Z ∞ e f ( sδ ) dss log s . (cid:4) Corollary 8.9.
Suppose that under the hypotheses of Theorem , the function f isbounded and has finite right derivative at . Then Ω f ( δ ) ≤ const aδ log (cid:16) log Maδ (cid:17) for δ ∈ (cid:16) , M a (cid:17) , where a = f ′ + (0) and M = sup f . Proof.
Since f ( t ) ≤ min { at, M } , t >
0, the result follows from Theorem 8.8 and thefollowing obvious facts: Z Maδ e aδdss log s = aδ log (cid:18) log Maδ (cid:19) and Z ∞ Maδ
M dss log s ≤ Z ∞ Maδ
M dss = aδ. (cid:4) In [AP2] we proved that if f belongs to the H¨older class Λ α ( R ), 0 ≤ α <
1, thenΩ f ( δ ) ≤ const(1 − α ) − k f k Λ α δ α , δ > , (8.6)where k f k Λ α def = sup x = y | f ( x ) − f ( y ) || x − y | . The next result shows that if in addition to this f satisfies the hypotheses of Theorem 8.8,then the factor (1 − α ) − on the right-hand side of (8.6) can considerably be improved. Corollary 8.10.
Suppose that under the hypotheses of Theorem , the function f belongs to Λ α ( R ) , ≤ α < . Then Ω f ( δ ) ≤ const (cid:16) log 21 − α (cid:17) k f k Λ α δ α for every δ > . Proof.
Indeed, Z ∞ e dss − α log s = Z ∞ e ( α − t dtt = Z ∞ − α e − t dtt ≤ const log 21 − α . (cid:4) Remark.
The function x κ ( x −
1) satisfies the hypotheses of Corollary 8.9with a = 1 and M = 2, and Corollary 8.9 yields a sharp result in this case. That meansthat Theorem 8.8 is also sharp in a sense.The following theorem is a symmetrized version of Theorem 8.8. Theorem 8.11.
Let f be a continuous function on R such that f is convex or con-cave on each of two rays ( −∞ , and [0 , ∞ ) . Suppose that there exists a finite limit lim | t |→∞ t − f ( t ) def = a . Then Ω f ( δ ) ≤ aδ + c Z ∞ e | f ( δs ) − f (0) − δas | + | f ( − δs ) − f (0) + δas | s log s ds, where c as a numerical constant. roof. It suffices to consider the case where f (0) = a = 0. We assume first that f ( t ) = 0 for t ≤
0. To be definite, suppose that f is concave on [0 , ∞ ). Then f is anondecreasing function because lim | t |→∞ t − f ( t ) = 0, and so the result reduces to Theorem8.8. The case f ( t ) = 0 for t ≥ t
7→ − t . It remains to observe that each function f with a = f (0) = 0can be represented in the form f = g + h in such way that g ( t ) = 0 for t ≤ h ( t ) = 0for t ≥
0, and the cases of the function g and h have been treated above. (cid:4) Theorem 8.12.
Let f be a nonnegative continuous function on R such that f ( x ) = 0 for all x ≤ and the function x x − f ( x ) is nonincreasing on (0 , ∞ ) . Suppose that Ω f ( δ ) < ∞ for δ > . Then f ( x ) ≤ const x log log x for every x ≥ . Proof.
By Theorem 5.11, Ω ♭f (1) ≥ k D f k M [1 , ∞ ) , ( −∞ , . Making the change of variables y
7→ − y we get (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) x + y (cid:13)(cid:13)(cid:13)(cid:13) M [1 , ∞ ) , [0 , ∞ ) ≤ ♭f (1) . Thus for every a > (cid:13)(cid:13)(cid:13)(cid:13) xx + y (cid:13)(cid:13)(cid:13)(cid:13) M [1 ,a ] , [1 ,a ] ≤ max [1 ,a ] (cid:12)(cid:12)(cid:12)(cid:12) xf ( x ) (cid:12)(cid:12)(cid:12)(cid:12) · (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) x + y (cid:13)(cid:13)(cid:13)(cid:13) M [1 ,a ] , [1 ,a ] ≤ af ( a ) (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) x + y (cid:13)(cid:13)(cid:13)(cid:13) M [1 , ∞ ) , [0 , ∞ ) ≤ a Ω ♭f (1) f ( a ) . It remains to apply Theorem 4.9. (cid:4)
Remark.
Let x > e and let g α be a continuous function such that g α ( x ) = x log α (log x ) , if x ≥ x > , , if x ≤ . Then Ω g α ( δ ) < ∞ for α >
1. Indeed, in this case g α coincides with a function satisfyingTheorem 8.8 outside a compact subset of R . On the other hand, Ω g α ( δ ) = ∞ for α < R the function g α coincides with a function f , for which the function x x − f ( x ) is nonincreasing on(0 , ∞ ). The case α = 1 is an open problem. . Lower estimates for operator moduli of continuity
Recall that it follows from (1.1) that if f is a function on R such that k f k L ∞ ≤ k f k Lip ≤
1, then Ω f ( δ ) ≤ const δ (cid:18) δ (cid:19) , δ ∈ (0 , . It is still unknown whether this estimate is sharp. In particular, the question whetherone can replace the factor (cid:0) δ (cid:1) on the right-hand side with (cid:0) (1 + log δ (cid:1) s for some s < § x
7→ | x | on finite intervals.The main purpose of this section is to construct a C ∞ function f on R such that k f k L ∞ ≤ k f k Lip ≤
1, andΩ f ( δ ) ≥ const δ r log 2 δ , δ ∈ (0 , . Let σ >
0. Denote by E σ the set of entire functions of exponential type at most σ .Let F ∈ E σ ∩ L ( R ). Then F ( z ) = X n ∈ Z sin( σz − πn ) σz − πn F (cid:16) πnσ (cid:17) , see, e.g., [L], Lect. 20.2, Th. 1. Let f ∈ E σ ∩ L ∞ ( R ). Then f ( z ) sin (cid:0) σ ( z − a ) (cid:1) σ ( z − a ) ∈ E σ ∩ L ( R ) . Hence, f ( z ) sin( σ ( z − a )) σ ( z − a ) = X n ∈ Z sin(2 σz − πn )2 σz − πn · sin (cid:0) σ ( πn σ − a ) (cid:1) σ (cid:0) πn σ − a (cid:1) f (cid:16) πn σ (cid:17) = 2 X n ∈ Z sin(2 σz − πn ) sin (cid:0) σa − πn (cid:1) (2 σz − πn )(2 σa − πn ) f (cid:16) πn σ (cid:17) . Substituting z = a , we obtain f ( z ) = 2 X n ∈ Z sin(2 σz − πn ) sin( σz − πn )(2 σz − πn ) f (cid:16) πn σ (cid:17) = X n ∈ Z sin ( σz − πn ) cos( σz − πn )( σz − πn ) f (cid:16) πn σ (cid:17) (9.1)for f ∈ E σ ∩ L ∞ ( R ).Denote by E σ (cid:0) C (cid:1) the set of all entire functions f on C such that the functions z f ( z, ξ ) and z f ( ξ, z ) belong to E σ for every ξ ∈ R (or, which is the same, for all ξ ∈ C ). Equality (9.1) implies the following identity: ( z, w ) = X ( m,n ) ∈ Z sin ( σz − πm )cos( σz − πm ) sin ( σw − πn ) cos( σw − πn )( σz − πm ) ( σw − πn ) f (cid:16) πm σ , πn σ (cid:17) (9.2)for every f ∈ E σ ( C ) ∩ L ∞ ( R ). Theorem 9.1.
Let σ > and Φ ∈ E σ ( C ) . Suppose that Φ( πm σ + α, πn σ + β ) ∈ M Z , Z for some α, β ∈ R . Then Φ ∈ M R , R and k Φ( x, y ) k M R , R ≤ (cid:13)(cid:13)(cid:13) Φ (cid:16) πm σ + α, πn σ + β (cid:17)(cid:13)(cid:13)(cid:13) M Z , Z . Proof.
Clearly, it suffices to consider the case when α = β = 0, σ = π/ k Φ( m, n ) k M Z , Z = 1. Then (see [Pi], Theorem 5.1) there exist two sequences { ϕ m } m ∈ Z and { ψ n } n ∈ Z of vectors in the closed unit ball of a Hilbert space H such that ( ϕ m , ψ n ) =Φ( m, n ). Put g x def = 4 π X m ∈ Z sin (cid:0) π ( x − m ) (cid:1) cos (cid:0) π ( x − m ) (cid:1) ( x − m ) ϕ m and h y def = 4 π X n ∈ Z sin (cid:0) π ( y − n ) (cid:1) cos (cid:0) π ( y − n ) (cid:1) ( y − n ) ψ n . We have k g x k H ≤ π X m ∈ Z sin (cid:0) π ( x − m ) (cid:1) | cos (cid:0) π ( x − m ) (cid:1) | ( x − m ) = 4 π X n ∈ Z sin πx (cid:12)(cid:12) cos πx (cid:12)(cid:12) ( x − n ) + 4 π X n ∈ Z sin (cid:0) πx − π (cid:1)(cid:12)(cid:12) cos (cid:0) πx − π (cid:1)(cid:12)(cid:12) ( x − n − = (cid:12)(cid:12)(cid:12) cos πx (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) sin πx (cid:12)(cid:12)(cid:12) ≤ √ . In the same way, k h y k H ≤ √ x ∈ R . Clearly | Φ | ≤ Z . The Cartwrighttheorem (see [L], Lecture 21, Theorem 4) implies that Φ is bounded on R × Z . Applyingonce more the Cartwright theorem, we find that Φ ∈ L ∞ ( R ). Hence, we can applyformula (9.2) to the function Φ, whence Φ( x, y ) = ( g x , h y ) for all x, y ∈ R . It remains toobserve that by Theorem 5.1 in [Pi], k Φ( x, y ) k M R , R ≤ sup x ∈ R k g x k H · sup y ∈ R k h y k H ≤ . (cid:4) Theorem 9.2.
Let f ∈ E σ . Then Ω ♭f ( δ ) ≥ δ (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R for every δ ∈ (cid:0) , σ (cid:3) . roof. The general case easily reduces to the case σ = π/
4. By Theorem 9.1, wehave (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R ≤ (cid:13)(cid:13)(cid:13)(cid:13) f (2 m + 1) − f (2 n )2 m − n + 1 (cid:13)(cid:13)(cid:13)(cid:13) M Z , Z ≤ k f k OL( Z ) . Hence, by Theorem 5.7,Ω ♭f ( δ ) ≥ Ω ♭f, Z ( δ ) = δ k f k OL( Z ) ≥ δ (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R ) for δ ∈ (cid:0) , π (cid:3) . (cid:4) Theorem 9.3.
Let f ∈ E σ . Then Ω f ( δ ) ≥ δ (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R for every δ ∈ (cid:0) , σ (cid:3) . Proof.
It suffices to observe that Ω ♭f ( δ ) ≤ f ( δ ) by Theorem 10.2 in [AP2]. (cid:4) Theorem 9.4.
For every δ ∈ (0 , , there exists an entire function f ∈ E /δ such that k f k L ∞ ( R ) ≤ , k f ′ k L ∞ ( R ) ≤ and Ω f ( δ ) ≥ C δ q log δ , where C is a positive numericalconstant. We need some lemmata.
Lemma 9.5.
For every positive integer n , there exists a trigonometric polynomial f of degree n such that k f k L ∞ ≤ , k f ′ k L ∞ ≤ , and (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ c p log n. Proof.
It follows from the results of [Pe2] that for every function h in C ( T ), (cid:13)(cid:13)(cid:13)(cid:13) h ( e i x ) − h ( e i y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ const k h k B , (9.3)where B is a Besov space (see [Pe5] for the definition) of functions on T . Note that thisresult was deduced in [Pe2] from the nuclearity criterion for Hankel operators (see [Pe1]and [Pe5], Ch. 6). It is easy to see from the definition of B ( T ) (see e.g., [Pe5]) that k h k B ≥ const X j ≥ j | ˆ h (2 j ) | . (9.4)It is well known (see, for example, [Fo]) that for every positive integer n , there existsan analytic polynomial h such that h (0) = 0 , deg h = n, k h ′ k L ∞ ( T ) = 1 , and X j ≥ j | ˆ h (2 j ) | ≥ d p log n, here d is a positive numerical constant. Then inequality (9.3) implies that (cid:13)(cid:13)(cid:13)(cid:13) h ( e i x ) − h ( e i y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ const p log n. Put f ( x ) def = h ( e i x ). It remains to observe that k f ′ k L ∞ = k h ′ k L ∞ ( T ) = 1 and k f k L ∞ = k h k L ∞ ( T ) ≤ (cid:4) Lemma 9.6.
Let n ∈ Z . Then (cid:13)(cid:13)(cid:13)(cid:13) x − y − πne i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ √ π for every intervals J and J with J − J ⊂ (cid:2) (2 n − ) π, (2 n + ) π (cid:3) . Proof.
We can restrict ourselves to the case n = 0. We have (cid:13)(cid:13)(cid:13)(cid:13) x − ye i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J = (cid:13)(cid:13)(cid:13)(cid:13) x − ye i( x − y ) − (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ (cid:13)(cid:13)(cid:13)(cid:13) te i t − (cid:13)(cid:13)(cid:13)(cid:13) b L ([ − π , π ]) = (cid:13)(cid:13)(cid:13)(cid:13) t t/ (cid:13)(cid:13)(cid:13)(cid:13) b L ([ − π , π ]) . Consider the 3 π -periodic function ξ such that ξ ( t ) = t t/ for t ∈ [ − π , π ]. We canexpand ξ in Fourier series ξ ( t ) = X n ∈ Z a n e n i t . Note that a n = a − n ∈ R for all n ∈ Z because ξ is even and real. Moreover, ξ is convexon [ − π , π ]. Hence, by Theorem 35 in [HR], ( − n a n ≥ n ∈ Z . It follows that (cid:13)(cid:13)(cid:13)(cid:13) t t/ (cid:13)(cid:13)(cid:13)(cid:13) b L ([ − π , π ]) ≤ X n ∈ Z | a n | = ξ (cid:18) π (cid:19) = 3 √ π . (cid:4) Corollary 9.7.
Let J = [ πj, πj + π )] and J = [ πk − π , πk + π ] , where j, k ∈ Z .Then (cid:13)(cid:13)(cid:13)(cid:13) x − y − πne i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M J ,J ≤ √ π for some n ∈ Z . Proof.
We have J − J = [ π ( j − k ) − π , π ( j − k ) + π ]. If j − k is even, then we canapply Lemma 9.6 with n = ( j − k ). If j − k is odd, then we can apply Lemma 9.6 with n = ( j − k + 1). (cid:4) Lemma 9.8.
Let g be a π -periodic function in C ( R ) . Then (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≤ √ π (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R . roof. Note that (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R = (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) x − y − πn (cid:13)(cid:13)(cid:13)(cid:13) M R , R for all n ∈ Z and (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] = (cid:13)(cid:13)(cid:13)(cid:13) g ( x ) − g ( y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [ − π , π . Now we can represent the square [0 , π ] × [ − π , π ] as the union of four squares with sidesof length π , each of which satisfies the hypotheses of Corollary 9.7. (cid:4) Proof of Theorem 9.4.
It suffices to consider the case when δ ∈ (cid:0) , (cid:3) . Then δ ∈ (cid:2) n +1 , n (cid:3) for an integer n ≥
2. By Lemma 9.5, there exists a trigonometric polynomial f of degree n such that k f k L ∞ ≤ k f ′ k L ∞ ≤ (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ c p log n. Hence, (cid:13)(cid:13)(cid:13)(cid:13) f ( x ) − f ( y ) x − y (cid:13)(cid:13)(cid:13)(cid:13) M R , R ≥ c p log n by Lemma 9.8. Clearly, g ∈ E n ⊂ E /δ . Applying Theorem 9.3, we obtainΩ f ( t ) ≥ const p log n t, < t ≤ n . Hence, Ω f ( δ ) ≥ Ω f (cid:18) n (cid:19) ≥ C √ log nn ≥ Cδ s log (cid:18) δ (cid:19) for some positive numbers C and C . (cid:4) Theorem 9.9.
There exist a positive number c and a function f ∈ C ∞ ( R ) such that k f k L ∞ ≤ , k f ′ k L ∞ ≤ , and Ω f ( δ ) ≥ c δ q log δ for every δ ∈ (0 , . Proof.
Applying Theorem 9.4 for δ = 2 − n , we can construct a sequence functions { f n } n ≥ and two sequences of bounded self-adjoint operators { A n } n ≥ and { B n } n ≥ suchthat k f n k L ∞ ≤ k f ′ n k L ∞ ≤ k A n − B n k ≤ − n and k f n ( A n ) − f n ( B n ) k ≥ C √ n − n for all n ≥
1. Denote by ∆ n the convex hull of σ ( A n ) ∪ σ ( B n ). Using the translations f n f n ( x − a n ), A n A n + a n I , B n B n + a n I and ∆ n a n + ∆ n for a suitablesequence { a n } ∞ n =1 in R , we can achieve the condition that the intervals ∆ n are disjointand dist(∆ m , ∆ n ) > m = n . We can construct a function f ∈ C ∞ ( R ) such that k f k L ∞ ≤ k f ′ k L ∞ ≤ f (cid:12)(cid:12) ∆ n = f n (cid:12)(cid:12) ∆ n for all n ≥
1. Clearly, Ω f (2 − n ) ≥ C √ n − n for all n ≥ C which easily implies the result. (cid:4) To obtain the lower estimate in Theorem 9.9, we used the inequality (cid:13)(cid:13)(cid:13)(cid:13) f ( e i x ) − f ( e i y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ const X j ≥ j | ˆ f (2 j ) | , (9.5) hich in turn implies that there exists a positive number C such that for every positiveinteger n there exists a polynomial f of degree n such that (cid:13)(cid:13)(cid:13)(cid:13) f ( e i x ) − f ( e i y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ C p log n k f k Lip . (9.6)We do not know whether Theorem 9.9 can be improved. It would certainly be naturalto try to improve (9.6). The best known lower estimate for the norm of divided differencesin the space of Schur multipliers was obtained in [Pe2]. To state it, we need somedefinitions.Let f ∈ L ( T ). Denote by P f the Poisson integral of f ,( P ) f ( z ) def = Z T (1 − | z | ) f ( ζ ) | z − ζ | d m ( ζ ) , z ∈ D , where m is normalized Lebesgue measure on T .For t ∈ R and δ ∈ (0 , Q ( t, δ ) by Q ( t, δ ) def = { re i s : 0 < − r < h, | s − t | < δ } . A positive Borel measure on µ on D is said to be a Carleson measure if C ( µ ) def = µ ( D ) + sup (cid:8) δ − µ ( Q ( t, δ )) : t ∈ R , δ ∈ (0 , (cid:9) < ∞ . If ψ is a nonnegative measurable function on D , we put C ( ψ ) def = C ( µ ) , where dµ def = ψ d m . Here m is planar Lebesgue measure.It follows from results of [Pe2] (see also [Pe4]) that (cid:13)(cid:13)(cid:13)(cid:13) f ( e i x ) − f ( e i y ) e i x − e i y (cid:13)(cid:13)(cid:13)(cid:13) M [0 , π ] , [0 , π ] ≥ const k f k L , (9.7)where k f k L def = C (cid:0)(cid:13)(cid:13) Hess( P f ) (cid:13)(cid:13)(cid:1) , where for a function ϕ of class C , its Hessian Hess( ϕ ) is the matrix of its second orderpartial derivatives.It turns out, however, that for a trigonometric polynomial f of degree n , k f k L ≤ const p log(1 + n ) k f k Lip , (9.8)and so even if instead of inequality (9.5) we use inequality (9.7), we cannot improveTheorem 9.9.Inequality (9.8) is an immediate consequence of the following fact: Theorem 9.10.
For a trigonometric polynomial f of degree n , n ≥ , the followinginequality holds: C (cid:0)(cid:12)(cid:12) ∇ ( P f ) (cid:12)(cid:12)(cid:1) ≤ const p log n k f k L ∞ . e are going to use the well-known fact that a function f in L ( T ) belongs to thespace BMO( T ) if and only if the measure µ defined by d µ = |∇ ( P f ) | (1 − | z | ) d m isa Carleson measure. We refer to [Ga] for Carleson measures and the space BMO. Proof of Theorem 9.10.
Suppose that k f k L ∞ = 1. We have to prove that Z Q ( t,δ ) |∇ ( P f ) | dxdy ≤ const δ p log n . (9.9)Note that |∇ ( P f ) | ≤ n by Bernstein’s inequality. Hence, Z { − n − < | ζ | < }∩ Q ( t,δ ) |∇ ( P f ) | d m ≤ n m (cid:0) { ζ : 1 − n − < | ζ | < } ∩ Q ( t, δ ) (cid:1) = 2 nδ (1 − (1 − n − ) ) ≤ δ. This proves (9.9) in the case δ ≥ − n − . In the case δ < − n − it remains to estimatethe integral over the set { ζ : | ζ | < − n − } ∩ Q ( t, δ ). Note that k f k BMO ≤ const k f k L ∞ .Hence, there exists a constant C such that Z Q ( t,δ ) |∇ ( P f ) | (1 − | ζ | ) d m ( ζ ) ≤ Cδ.
Thus Z {| ζ | < − n − }∩ Q ( t,δ ) |∇ ( P f ) | d m ≤ Z Q ( t,δ ) |∇ ( P f ) | (1 − | ζ | ) d m ( ζ ) ! / Z {| ζ | < − n − }∩ Q ( t,δ ) (1 − | ζ | ) − d m ( ζ ) ! / ≤ const δ (log( nδ )) / ≤ const δ (log n ) / . (cid:4) Lower estimates in the case of unitary operators
The purpose of this section is to obtain lower estimates for the operator modulus ofcontinuity for functions on the unit circle.We define an operator modulus of continuity of a continuous function f on T byΩ f ( δ ) def = sup (cid:8) k f ( U ) − f ( V ) k : U and V are unitary , k U − V k ≤ δ (cid:9) . As in the case of self-adjoint operators (see [AP2]), one can prove that k f ( U ) R − Rf ( V ) k ≤ f ( k U R − RV k )for every unitary operators U , V and an operator R of norm 1. We define the spaceOL( T ) as the set of f ∈ C ( T ) such that k f k OL( T ) def = sup δ> δ − Ω f ( δ ) < ∞ . iven a closed subset F of T , we can also introduce the operator modulus of continuityΩ f, F and define the space OL( F ) of operator Lipschitz functions on F .For closed subsets F and F of T , the space M F , F of Schur multipliers can bedefined by analogy with the self-adjoint case. Note that the analogues of (5.1) and (5.3)for functions on closed subsets of T can be proved as in § f ∈ C ( T ). We put f ♠ ( t ) def = f ( e i t ). It is clear that Ω f ♠ ≤ Ω f . Hence, k f ♠ k OL( R ) ≤k f k OL( T ) . Lemma 9.8 implies that k f k OL( T ) ≤ √ π k f ♠ k OL( R ) . One can prove thatΩ f ≤ const Ω f ♠ .Recall that it follows from results of [Pe2] that for f ∈ C ( T ), k f k OL( T ) ≥ const k f k B ;actually we used this estimate in §
9, see inequality (9.3).We would like to remind also that for each positive integer n , there exists an analyticpolynomial f such that deg f = n , k f ′ k L ∞ ( T ) = 1, and k f k OL( T ) ≥ const √ log n ; seeLemma 9.5.Put d n ( z ) def = 1 n z n − z − n n − X k =0 z k . It is easy to see that d n ( ζz − ) = z − n z n − ζ n n ( z − ζ ) = z − n ζ n − d n ( zζ − ) . Denote by T n the set of n th roots of 1, i.e., T n def = { ζ ∈ T : ζ n = 1 } .Let f be an analytic polynomial of degree less that n . Then f ( z ) = X ζ ∈ τ T n f ( ζ ) d n ( zζ − ) for every τ ∈ T . If f is a trigonometric polynomial and deg f ≤ n , then for every ξ ∈ T , the function z n f ( z ) d n ( zξ − ) is an analytic polynomial of degree less than 4 n . Hence, z n f ( z ) d n ( zξ − ) = X ζ ∈ τ T n f ( ζ ) d n ( ζξ − ) d n ( zζ − ) . Substituting ξ = z we get f ( z ) = z − n X ζ ∈ τ T n f ( ζ ) d n ( ζz − ) d n ( zζ − ) = X ζ ∈ τ T n f ( ζ ) F n ( z, ζ ) (10.1)for every τ ∈ T , where F n ( z, ζ ) def = z − n ζ − n ( z n − ζ n )( z n − ζ n )8 n ( z − ζ ) . Denote by P n (cid:0) T (cid:1) the set of all trigonometric polynomial f on T such that thefunctions z f ( z, ξ ) and z f ( ξ, z ) are trigonometric polynomials on T of degree at ost n for every ξ ∈ T . Equality (10.1) implies the following identity: f ( z, w ) = X ζ ∈ τ T n X ξ ∈ τ T n f ( ζ, ξ ) F n ( z, ζ ) F n ( w, ξ ) (10.2)for every f ∈ P n (cid:0) T (cid:1) and for arbitrary τ and τ in T . Theorem 10.1.
Let Φ ∈ P n (cid:0) T (cid:1) . Then k Φ k M T , T ≤ k Φ k M τ T n,τ T n for every τ , τ ∈ T . Proof.
Clearly, it suffices to consider the case when τ = τ = 1. Then (see [Pi],Theorem 5.1) there exist two sequences { ϕ ζ } ζ ∈ T n and { ψ ξ } ξ ∈ T n of vectors in the closedunit ball of a Hilbert space H such that ( ϕ ζ , ψ ξ ) = Φ( ζ, ξ ). Put g z def = X ζ ∈ T n F n ( z, ζ ) ϕ ζ and h w def = X ξ ∈ T n F n ( w, ξ ) ψ ξ . Taking into account that for z ∈ T ,12 n X ζ ∈ T n (cid:12)(cid:12)(cid:12)(cid:12) z n − ζ n z − ζ (cid:12)(cid:12)(cid:12)(cid:12) = 12 n X ζ ∈ T n \ T n (cid:12)(cid:12)(cid:12)(cid:12) z n − ζ n z − ζ (cid:12)(cid:12)(cid:12)(cid:12) = Z T (cid:12)(cid:12)(cid:12)(cid:12) z n − ζ n z − ζ (cid:12)(cid:12)(cid:12)(cid:12) d m ( ζ ) = 2 n, we obtain k g z k H ≤ X ζ ∈ T n | F n ( z, ζ ) |≤ | z n + 1 | n X ζ ∈ T n (cid:12)(cid:12)(cid:12)(cid:12) z n − ζ n z − ζ (cid:12)(cid:12)(cid:12)(cid:12) + | z n − | n X ζ ∈ T n \ T n (cid:12)(cid:12)(cid:12)(cid:12) z n − ζ n z − ζ (cid:12)(cid:12)(cid:12)(cid:12) = | z n + 1 | + | z n − | ≤ √ . In the same way, k h w k H ≤ √ w ∈ T . By (10.2), we have Φ( z, w ) = ( g z , h w )for all z, w ∈ T . It remains to observe that by Theorem 5.1 in [Pi], k Φ( z, w ) k M T , T ≤ sup z ∈ T k g z k H · sup w ∈ T k h w k H ≤ . (cid:4) We need the following version of Theorem 5.7:
Theorem 10.2.
Let f be a function on T n . Then Ω ♭f, T n ( δ ) = δ k f k OL( T n ) for every δ ∈ (0 , n ] . o prove Theorem 10.2, we need a lemma. Put λ ( z ) def = ( z − , if z ∈ C , z = 0 , , if z = 0 . Lemma 10.3.
Let n be a positive integer. Then k λ ( z − w ) k M T n, T n = ( n , if n is even , n − n , if n is odd . Proof.
It is easy to verify that n X k =1 (cid:18) k − n + 12 (cid:19) z k = nz n z − − z n − z − − n + 12 z z n − z − nλ ( z − z ∈ T n . Hence, λ ( z − w ) = w − λ ( zw − −
1) = 1 n n X k =1 (cid:18) k − n + 12 (cid:19) z k w − k − . (10.3)Thus k λ ( z − w ) k M T n, T n ≤ n n X k =1 (cid:12)(cid:12)(cid:12)(cid:12) k − n + 12 (cid:12)(cid:12)(cid:12)(cid:12) = ( n , if n is even , n − n , if n is odd . The opposite inequality is also true. It can be deduced from the observation that equality(10.3) means that that the function λ ( z −
1) on the group T n is the Fourier transformof the n -periodic sequence { a k } k ∈ Z defined by a k = k − n +12 for k = 1 , , . . . , n . Here weidentify the group dual to T n with the group Z /n Z . We omit details because we needonly the upper estimate. (cid:4) Proof of Theorem 10.2.
The inequalityΩ ♭f, T n ( δ ) ≤ δ k f k OL( T n ) , δ > , is a consequence of a unitary version of Theorem 5.1, which can be proved in the sameway as the self-adjoint version, see also Theorem 4.13 in [AP3].Let us prove the opposite inequality for δ ∈ (cid:0) , n (cid:3) . Fix ε >
0. There exists a unitaryoperator U and bounded operator R such that k U R − RU k = 1, σ ( U ) ⊂ T n , and k f ( U ) R − Rf ( U ) k ≥ k f k OL( T n ) − ε . Put R U def = X ζ,ξ ∈ T n , ζ = ξ E U ( { ζ } ) RE U ( { ξ } ) = R − X ζ ∈ T n E U ( { ζ } ) RE U ( { ζ } ) . Clearly,
U R − RU = U R U − R U U and f ( U ) R − Rf ( U ) = f ( U ) R U − R U f ( U ). Thus wemay assume that R = R U . Note that U R − RU = X ζ,ξ ∈ T n , ζ = ξ ( ζ − ξ ) E U ( { ζ } ) RE U ( { ξ } ) . Since R = R U = X ζ,ξ ∈ T n , ζ = ξ ( ζ − ξ ) λ ( ζ − ξ ) E U ( { ζ } ) RE U ( { ξ } ) , e have R = H n ⋆ ( U R − RU ), where H n ( ζ, ξ ) = λ ( ζ − ξ ), where ζ, ξ ∈ T n . Thus byLemma 10.3, k R k ≤ k H n k M T n, T n k U R − RU k = k H n k M T n, T n ≤ n . Let δ ∈ (cid:0) , n (cid:3) . Then k U ( δR ) − ( δR ) U k = δ and k δR k ≤
1. Hence,Ω ♭f, T n ( δ ) ≥ δ k f ( U ) R − Rf ( U ) k ≥ δ (cid:0) k f k OL( T n ) − ε (cid:1) . Passing to the limit as ε →
0, we obtain the desired result. (cid:4)
Theorem 10.4.
Let f be a trigonometric polynomial of degree n ≥ . Then Ω ♭f, T ( δ ) ≥ δ k f k OL( T ) for δ ∈ (0 , n ] . Proof.
Applying Theorems 10.1 and 10.2, we obtain (cid:13)(cid:13)(cid:13)(cid:13) f ( z ) − f ( w ) z − w (cid:13)(cid:13)(cid:13)(cid:13) M T , T ≤ (cid:13)(cid:13)(cid:13)(cid:13) f ( z ) − f ( w ) z − w (cid:13)(cid:13)(cid:13)(cid:13) M T n, T n = 2 δ − Ω ♭f, T n ( δ ) ≤ δ − Ω ♭f, T ( δ )for δ ∈ (0 , n ]. (cid:4) Theorem 10.5.
Let f be a trigonometric polynomial of degree n ≥ . Then Ω f, T ( δ ) ≥ δ k f k OL( T ) for δ ∈ (0 , n ] . Proof.
It suffices to observe that Ω ♭f, T ( δ ) ≤ f, T ( δ ). (cid:4) Theorem 10.6.
Let f ∈ C ( T ) . Then Ω f (2 − n ) ≥ C − n n − X k =0 k (cid:16)(cid:12)(cid:12) b f (2 k ) (cid:12)(cid:12) + (cid:12)(cid:12) b f ( − k ) (cid:12)(cid:12)(cid:17) , where C is a positive constant. Proof.
Applying the convolution with the de la Vall´ee Poussin kernel, we can find ananalytic polynomial f n such that deg f n < n , b f n ( k ) = b f ( k ) for k ≤ n − and Ω f n ≤ f .Applying inequalities (9.3) and (9.4), we obtain k f n k OL( T ) ≥ const n − X k =0 k (cid:0) | b f (2 k ) | + | b f ( − k ) | (cid:1) . It remains to apply Theorem 10.5 for δ = 2 − n . (cid:4) In the following theorem we use the notation C A for the disk-algebra: C A def = (cid:8) f ∈ C ( T ) : ˆ f ( n ) = 0 for n < (cid:9) . heorem 10.7. Let ω : (0 , → R be a positive continuous function. Suppose that ω (2 t ) ≤ const ω ( t ) , the function t t − (log t ) − ω ( t ) is nondecreasing, and Z ω ( t ) dtt log t < ∞ . (10.4) Then there exists a function f ∈ C A such that f ′ ∈ C A and Ω f ( δ ) ≥ ω ( δ ) for all δ ∈ (0 , . Proof.
Note that the inequality Ω f ( δ ) ≥ ω ( δ ) for δ = 2 − n implies that Ω f ( δ ) ≥ const ω ( δ ) for all δ ∈ (0 , δ = 2 − n .Taking Theorem 10.6 into account, we can reduce the result to the problem to constructa function g ∈ C A such that a n def = 2 n ω (2 − n ) n ≤ n n − X k =0 (cid:12)(cid:12)b g (2 k ) (cid:12)(cid:12) for all nonnegative integer n .Indeed, in this case the function f defined by f ( z ) = Z z g ( ζ ) − g (0) ζ dζ satisfies the inequality a n ≤ n n − X k =0 k (cid:12)(cid:12) b f (2 k ) (cid:12)(cid:12) . Condition (10.4) implies that { a n } n ≥ ∈ ℓ . Moreover, { a n } n ≥ is a nonincreasing se-quence because the function t t − (log t ) − ω ( t ) is nondecreasing.We can find a function g ∈ C A such that b g (2 k ) = a k for all k ≥
0, see, for example,[Fo]. Then 1 n n − X k =0 (cid:12)(cid:12)b g (2 k ) (cid:12)(cid:12) = 1 n n − X k =0 a k ≥ a n − ≥ a n . (cid:4) Remark.
Theorem 10.7 remains valid if we replace the assumption that the function t t − (log t ) − ω ( t ) is nondecreasing with the assumption that there exists a positiveconstant C such that ω ( t ) t log t ≤ C ω ( s ) s log s , whenever 0 < t < s ≤ . Self-adjoint operators with finite spectrum.Estimates in terms of the ε -entropy of the spectrum In this section we obtain sharp estimates of the quasicommutator norms k f ( A ) R − Rf ( B ) k in the case when A has finite spectrum. This allows us to obtainsharp estimates of the operator Lipschitz norm in terms of the Lipschitz norm in thecase of operators on finite-dimensional spaces in terms of the dimension. oreover, we obtain a more general result (see Theorem 11.5) in terms of ε -entropy ofthe spectrum of A , where ε = k AR − RA k . This leads to an improvement of inequality(1.1).Note that the results of this section improve some results of [Fa2] and [Fa3].Let F be a closed subset of R . Denote by Lip( F ) the set of Lipschitz functions on F .Put k f k Lip( F ) def = inf (cid:8) C > | f ( x ) − f ( y ) | ≤ C | x − y | ∀ x, y ∈ F (cid:9) . Let { s j ( T ) } ∞ j =0 be the sequence of singular values of a bounded operator T . We usethe notation S ω for the Matsaev ideal, S ω def = (cid:8) T : k T k S ω def = ∞ X j =0 (1 + j ) − s j ( T ) < ∞ (cid:9) . We need the following statement which is contained implicitly in [NaP].
Theorem 11.1.
Let f be a Lipschitz function on a closed subset F of R . Then forevery nonempty finite subset Λ in F , k D f k M Λ , F ≤ C (cid:0) (cid:1) k f k Lip( F ) , where C is a numerical constant. Proof.
Let k ∈ L ( µ ⊗ ν ), where µ and ν are Borel measures on Λ and F . Clearly,rank I µ,νk ≤ card(Λ). Hence, kI µ,νk k S ω ≤ (cid:0) (1 + log(card(Λ)) (cid:1) kI µ,νk k . Now Theorem 2.3in [NaP] implies that (cid:13)(cid:13) I µ,νk D f (cid:13)(cid:13) ≤ const (cid:0) (1 + log(card(Λ)) (cid:1)(cid:13)(cid:13) I µ,νk (cid:13)(cid:13) · k f k Lip( F ) . (cid:4) Theorem 11.2.
Let A and B be self-adjoint operators. Suppose that σ ( A ) is finite.Then k f ( A ) R − Rf ( B ) k ≤ C (cid:0) σ ( A ))) (cid:1) k f k Lip( σ ( A ) ∪ σ ( B )) k AR − RB k for all bounded operators R and f ∈ Lip (cid:0) σ ( A ) ∪ σ ( B ) (cid:1) , where C is a numerical constant. Proof.
The result follows from Theorem 11.1 if we take into account the followinggeneralizations of (5.2) and (5.4) (see [BS4]): f ( A ) R − Rf ( B ) = Z Z σ ( A ) × σ ( B ) ( D f (cid:1) ( x, y ) dE A ( x )( AR − RB ) dE B ( y )and (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Z Z σ ( A ) × σ ( B ) ( D f (cid:1) ( x, y ) dE A ( x )( AR − RB ) dE B ( y ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ k D f k M ( σ ( A ) × σ ( B )) k AR − RB k which proves the result. (cid:4) orollary 11.3. Let A , B be self-adjoint operators and let R be a linear operator on C n . Then k f ( A ) R − Rf ( B ) k ≤ C (1 + log n ) k f k Lip( σ ( A ) ∪ σ ( B )) k AR − RB k (11.1) for every function f on σ ( A ) ∪ σ ( B ) , where C is a numerical constant. Remark 1.
Note that in the special case f ( t ) = | t | inequality (11.1) is well-known,see, e.g., [Da]. This special case also follows from Matsaev’s theorem, see [GK], Ch. III,Th. 4.2 (see also [Go] where a finite dimensional improvement of Matsaev’s theorem wasobtained). Remark 2.
We also would like to note that inequality (11.1) is sharp. Indeed, itfollows immediately from Lemma 15 of [Da] that for each positive integer n there exist n × n self-adjoint matrices A and R such that (cid:13)(cid:13) | A | R − R | A | (cid:13)(cid:13) ≥ const log(1 + n ) k AR − RA k and AR − RA = . (11.2)We also refer the reader to [Mc] where inequality (11.2) is essentially contained. More-over, (11.2) can be deduced from the results of Matsaev and Gohberg mentioned above.The following result is a special case of Corollary 11.3 that corresponds to R = I . Theorem 11.4.
Let A , B be self-adjoint operators on C n . Then k f ( A ) − f ( B ) k ≤ C (1 + log n ) k f k Lip( σ ( A ) ∪ σ ( B )) k A − B k for every function f on σ ( A ) ∪ σ ( B ) , where C is an absolute constant. Remark.
The estimate in Theorem 11.4 is also sharp. Indeed, for each positiveinteger n there exist n × n self-adjoint matrices A and B such that A = B and (cid:13)(cid:13) | A | − | B | (cid:13)(cid:13) ≥ const log(1 + n ) k A − B k , This follows easily from (11.2), see the proof of Theorem 10.1 in [AP2].
Definition.
Let F be a nonempty compact subset of R . Recall that for ε >
0, the ε - entropy K ε ( F ) of F is defined as K ε ( F ) def = inf log (cid:0) card(Λ) (cid:1) , where the infimum is taken over all Λ ⊂ R such that Λ is an ε net of F . The followingresult is an generalization of Theorem 11.2. On the other hand, it improves inequality(1.1) obtained in [AP2]. Theorem 11.5.
Let A and B be self-adjoint operators and let R be bounded operatorwith k R k ≤ . Suppose that σ ( A ) ⊂ F , where F is a closed subset of R . Then for every f ∈ Lip (cid:0) σ ( A ) ∪ σ ( B ) (cid:1) , k f ( A ) R − Rf ( B ) k ≤ const (cid:0) K ε ( F ) (cid:1) k f k Lip( σ ( A ) ∪ σ ( B )) k AR − RB k , where ε def = k AR − RB k . roof. We repeat the argument of the proof of Theorem 5.8. We can find a self-adjointoperator A ε such that A ε A = AA ε , k A − A ε k ≤ ε , σ ( A ε ) ⊂ F , and log (cid:0) card (cid:0) σ ( A ε ) (cid:1)(cid:1) ≤ K ε ( F ). Then k f ( A ε ) R − Rf ( B ) k ≤ const (cid:0) K ε ( F ) (cid:1) k f k Lip( σ ( A ) ∪ σ ( B )) k A ε R − RB k≤ δ (cid:0) K ε ( F ) (cid:1) k f k Lip( σ ( A ) ∪ σ ( B )) by Theorem 11.2. It remains to observe that since A commutes wit A ε , we have k f ( A ) R − Rf ( B ) k ≤ k f ( A ) − f ( A ε ) k + k f ( A ε ) R − Rf ( B ) k≤ ε k f k Lip( σ ( A )) + k f ( A ε ) R − Rf ( B ) k . (cid:4) Corollary 11.6.
Let A and B be self-adjoint operators and let σ ( A ) ⊂ F , where F isa closed subset of R . Then for every f ∈ Lip (cid:0) σ ( A ) ∪ σ ( B ) (cid:1) , k f ( A ) − f ( B ) k ≤ const (cid:0) K ε ( F ) (cid:1) k f k Lip( σ ( A ) ∪ σ ( B )) k A − B k , where ε def = k A − B k . Proof.
It suffices to put R = I . (cid:4) If we apply Theorem 11.5 to the case K = [ a, b ], we obtain the following estimate,which improves inequality (1.1) in the special case R = I . Corollary 11.7.
Let f ∈ Lip( R ) . Let A be a self-adjoint operator with σ ( A ) ⊂ [ a, b ] .and k R k ≤ . Then for every self-adjoint operators B , k f ( A ) R − Rf ( B ) k ≤ const k f k Lip log (cid:18) b − a k AR − RB k (cid:19) k AR − RB k . Corollary 11.8.
Let f ∈ Lip( R ) . Let A be a self-adjoint operator with σ ( A ) ⊂ [ a, b ] .and k R k ≤ . Then k f ( A ) R − Rf ( B ) k ≤ const k f k Lip log (cid:18) b − a k A − B k (cid:19) k A − B k . References [AP1]
A.B. Aleksandrov and
V.V. Peller , Functions of perturbed operators , C.R. Acad. Sci. Paris,S´er I (2009), 483 – 488.[AP2]
A.B. Aleksandrov and
V.V. Peller , Operator H¨older–Zygmund functions , Advances in Math., (2010), no. 3, 910 – 966.[AP3]
A.B. Aleksandrov and
V.V. Peller , Functions of perturbed unbounded self-adjoint operators.Operator Bernstein type inequalities , Indiana Univ. Math. J., (2010), 1451–1490.[BS1] M.S. Birman and
M.Z. Solomyak , Double Stieltjes operator integrals , Problems of Math. Phys.,Leningrad. Univ. (1966), 33 – 67 (Russian).English transl.: Topics Math. Physics (1967), 25 – 54, Consultants Bureau Plenum PublishingCorporation, New York.[BS2] M.S. Birman and
M.Z. Solomyak , Double Stieltjes operator integrals. II , Problems of Math.Phys., Leningrad. Univ. (1967), 26 – 60 (Russian).English transl.: Topics Math. Physics (1968), 19 – 46, Consultants Bureau Plenum PublishingCorporation, New York. BS3]
M.S. Birman and
M.Z. Solomyak , Double Stieltjes operator integrals. III , Problems of Math.Phys., Leningrad. Univ. (1973), 27 – 53 (Russian).[BS4] M.S. Birman and
M.Z. Solomyak , Double operator integrals in Hilbert space , Int. Equat. Oper.Theory (2003), 131–168.[Da] E. B. Davies , Lipschitz continuity of functions of operators in the Schatten classes , J. LondonMath. Soc. (1988) 148 – 157[Fa1] Yu.B. Farforovskaya , The connection of the Kantorovich-Rubinshtein metric for spectral reso-lutions of selfadjoint operators with functions of operators , Vestnik Leningrad. Univ. (1968), 94– 97. (Russian).[Fa2] Yu.B. Farforovskaya , Lipschitz functions of selfadjoint operators in perturbation theory . Zap.Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) (1985), 176 – 182 (Russian).[Fa3]
Yu.B. Farforovskaya , Double operator integrals and their estimates in the uniform norm . Zap.Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (LOMI) (1996), 148 – 173 (Russian).[Fo]
John J. F. Fournier , An interpolation problem for coefficients of H” functions , Proc.Amer. Math. Soc. (1974), no. 2, 402 – 408.[Ga] G.B. Garnett , Bounded analytic functions , Academic Press, 1981.[Go]
I. C. Gohberg , On connections between Hermitian components of nilpotent matrices and on anintegral of triangular truncation , Bul. Akad. Stiince RSS Moldoven. (1963) no. 1 27 - 37 (Russian).[GK]
I.C. Gohberg, M.G. Krein , Theory of Volterra operators in Hilbert space and its applications ,Nauka, Moscow, 1965; English translation: American Mathematical Society, Providence, R.I. 1970.[HR]
G.H. Hardy and
W.W. Rogosinski , Forier series , Cambridge Tracts in Mathematics and Math-ematical Physics, No. 38, Cambridge at the Universitu Press, 1956.[JW]
B.E. Johnson and
J.P. Williams , The range of a normal derivation , Pacific J. Math. (1975),105 – 122.[Ka] T. Kato , Continuity of the map S
7→ | S | for linear operators, Proc. Japan Acad. (1973), 157 –160.[KS] E. Kissin and V.S. Shulman,
Classes of operator-smooth functions. I. Operator-Lipschitz functions ,Proc. Edinb. Math. Soc. (2) (2005), 151–173.[KST] E. Kissin, V.S. Shulman , and L. B. Turowska,
Extension of Operator Lipschitz and CommutatorBounded Functions , Operator Theory: Advances and Applications, (2006), 225 - 244.[L]
B. Ya. Levin
Lectures on entire functions , Translation of Math. Monogr., vol. 150, 1996.[Mc]
A. McIntosh , Counterexample to a question on commutators , Proc. Amer. Math. Soc. , 337 –340 (1971).[NaP] F. Nazarov and
V. Peller , Lipschitz functions of perturbed operators , C.R. Acad. Sci. Paris,S´er I (2009) 857 - 862.[NiF]
L. Nikolskaya and
Yu.B. Farforovskaya , Operator H¨olderness of H¨older functions , Algebra iAnaliz. (2010), 198 – 213 (Russian).[Pee]
J. Peetre , New thoughts on Besov spaces , Duke Univ. Press., Durham, NC, 1976.[Pe1]
V.V.Peller , Hankel operators of class S p and their applications (rational approximation, Gauss-ian processes, the problem of majorizing operators) , Mat. Sbornik, (1980), 538 – 581.English Transl. in Math. USSR Sbornik, (1982), 443-479.[Pe2] V.V. Peller , Hankel operators in the theory of perturbations of unitary and self-adjoint operators ,Funktsional. Anal. i Prilozhen. (1985), 37 – 51 (Russian).English transl.: Funct. Anal. Appl. (1985) , 111 – 123.[Pe3] V.V. Peller
Hankel operators in the perturbation theory of of unbounded self-adjoint operators .Analysis and partial differential equations, 529 – 544, Lecture Notes in Pure and Appl. Math., ,Dekker, New York, 1990.[Pe4]
V.V. Peller , Functional calculus for a pair of almost commuting selfadjoint operators , J. Funct.Anal., (1993), 325-345.[Pe5]
V.V. Peller , Hankel operators and their applications,
Springer-Verlag, New York, 2003.[Pi]
G. Pisier , Similarity problem and completely bounded maps , 2nd, expanded ed., Lecture Notes inMath. , Springer-Verlag, Berlin, 2001. Po]
G. P´olya , Remarks on characteristic functions , Proc. Berkeley Sympos. Math. Statist. and Prob-ability (August, 1945 and January, 1946), (1949), 115 – 123.