Projecting Lipschitz functions onto spaces of polynomials
aa r X i v : . [ m a t h . F A ] F e b PROJECTING LIPSCHITZ FUNCTIONSONTO SPACES OF POLYNOMIALS
PETR HÁJEK AND TOMMASO RUSSO
Abstract.
The Banach space P ( X ) of -homogeneous polynomials on the Banach space X can be naturally embedded in the Banach space Lip ( B X ) of real-valued Lipschitz func-tions on B X that vanish at . We investigate whether P ( X ) is a complemented subspaceof Lip ( B X ) . Especially important for us is when X is the separable Hilbert space, inwhich case the question is related to an open question posed by Gilles Godefroy askingwhether Lip ( ℓ ) has the approximation property. Moreover, our results can be inter-preted as failure of a polynomial counterpart to a classical result by Joram Lindenstrauss,asserting that P ( X ) = X ∗ is complemented in Lip ( B X ) . Introduction
Given a pointed metric space ( M, M ) , Lip ( M ) denotes the Banach space of all scalar-valued Lipschitz functions on M that vanish on M . The study of Banach spaces ofLipschitz functions, and even more of their canonical preduals, the Lipschitz-free spaces,has been one of the most active fields of research within Banach space theory in the lasttwo decades. We refer, e.g. , to [AACD, AlP, GK, Go] and [CCD, CK, We, We1] forrecent results and additional references on Lipschitz-free spaces and spaces of Lipschitzfunctions, respectively. Starting from the result [GK] that a Banach space X has thebounded approximation property if and only if the Lipschitz-free space F ( X ) over X hasit, approximation properties of Lipschitz-free Banach spaces have been widely investigated,[AmP, GO, Go, LP, PS]. For the definitions and basic properties of the various approx-imation properties we refer to [Ca]. We just recall here that the approximation property(AP for short) passes from the dual of a Banach space to the Banach space itself. Thisnaturally suggests to study approximation properties of spaces of Lipschitz functions, sincethe results would strengthen the corresponding results for Lipschitz-free spaces. However,the study of approximation properties of spaces of Lipschitz functions seems to be a largelyunexplored topic, [J-V]. As it was pointed out by Gilles Godefroy, it is not even known Date : February 5, 2021.2020
Mathematics Subject Classification.
Key words and phrases.
Banach spaces of Lipschitz functions, polynomials, complemented subspaces,Euclidean spaces, type and cotype.Research of P. Hájek was supported in part by OPVVV CAAS CZ.02.1.01/0.0/0.0/16_019/0000778.Research of T. Russo was supported by the GAČR project 20-22230L; RVO: 67985840 and by GruppoNazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of Istituto Nazionaledi Alta Matematica (INdAM), Italy. whether
Lip ( ℓ ) has the approximation property. This question was the original motiva-tion for the research presented in our paper.Since disproving the approximation property of a Banach space is a formidable task[En, Sz], a natural approach to show that a certain Banach space X fails the AP is tostand on the shoulders of giants and derive this from known results. In light of the stan-dard fact that the AP passes to complemented subspaces, it is sufficient to find in X acomplemented subspace which is known to fail the AP. Building on [Sz], Floret [Fl, p. 173]proved that the Banach space P ( ℓ ) of -homogeneous polynomials on ℓ fails to havethe approximation property (see also [DM]). This is very relevant to the above problem,because P ( ℓ ) naturally embeds in Lip ( B ℓ ) , by restricting a polynomial to B ℓ (see § 2.2for details). Moreover, Lip ( B ℓ ) is isomorphic to Lip ( ℓ ) , [Ka]. In conclusion, if P ( ℓ ) iscomplemented in Lip ( B ℓ ) , then Lip ( B ℓ ) would fail to have the approximation propertyand we would have a negative answer to the question from the previous paragraph. Weare thus led to the following problem. Problem 1.1.
Identify P ( ℓ ) with a subspace of Lip ( B ℓ ) by restricting a polynomial on ℓ to B ℓ . Is P ( ℓ ) a complemented subspace of Lip ( B ℓ ) ? This question should also be compared to a classical and very useful result by JoramLindenstrauss [Li2, Theorem 2], according to which X ∗ is -complemented in Lip ( X ) , forevery infinite-dimensional Banach space X . Since, by the very definition, the space P ( X ) of -homogeneous polynomials is nothing but X ∗ , we can restate Lindenstrauss’ theoremas saying that P ( X ) is always complemented in Lip ( X ) . From this perspective, Problem1.1 asks whether Lindenstrauss’ result admits a counterpart for higher order polynomials.The main result of our paper, Theorem B below, answers Problem 1.1 in the negative.Consequently, although we were not able to answer Godefroy’s question, we can at leastsay that our strategy to approach the problem, though natural, is not effective, perhapsindicating that Lip ( B ℓ ) might have the approximation property. Moreover, our resultalso shows that Lindenstrauss’ theorem [Li2] admits no polynomial extension. As it isperhaps to be expected, our main theorem will be derived from its finite-dimensional andquantitative counterpart, that reads as follows. (The precise definitions of the spaces P ( E n ) , Lip ( B E n ) , and C ( B E n ) will be given in § 2.) Theorem A.
Let E n denote the n -dimensional Euclidean space ( R n , k·k ) . If Q is anyprojection from C ( B E n ) onto P ( E n ) , then k Q k > C (cid:16) n − √ (cid:17) / , where C := (cid:16) √ − · π (cid:17) / .In particular, the same estimate holds for every projection from Lip ( B E n ) onto P ( E n ) . The main reason why we stated the theorem in the stronger form concerning C ( B E n ) isthat the averaging argument that we shall need (see § 2.3) is simpler to describe in C ( B E n ) rather than in the bigger space Lip ( B E n ) . In this context, it is perhaps worth observing ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 3 that C ( B F ) fails to be complemented in Lip ( B F ) , for every finite-dimensional Banachspace F . Indeed C ([ − , is not complemented in Lip ([ − , , since the former isisometric to C ([ − , and the latter to L ∞ ([ − , . By ‘bootstrap’ from this observation,the claim for the general finite-dimensional Banach space F readily follows.The study of projections onto a finite-dimensional subspace of a Banach space is a veryclassical and wide topic and Theorem A can be considered as one more result in the area.By the classical Kadets–Snobar theorem [KS], every n -dimensional subspace of a Banachspace is √ n -complemented and this is asymptotically sharp [Kö]. Given these results, it isquite conceivable that the order of growth ∼ n / in Theorem A is far from being optimal,but we don’t know how to obtain better estimates. As a non-exhaustive list of resultson projection constants, let us quote [Gr, Le, FS, Ba], or the monographs [T-J, Wo].The particular case of projections onto spaces of polynomials is also very well studied(particularly in the trigonometric case), for its connections with harmonic ([Wo, III.B],[GM, OL]) and numerical ([FL, TV, TV2]) analysis. Let us stress that, in this context,one typically fixes the dimension of the space and lets the degree of the polynomials grow,while in our paper we are interested in the opposite situation.Having Theorem A at our disposal, we can formulate the main result of our paper, thatin particular answers in the negative Problem 1.1. Prior to the statement, we need to recallone definition. A Banach space X is said to contain uniformly complemented ( ℓ n ) ∞ n =1 ifthere are a constant C and a sequence ( F n ) ∞ n =1 of subspaces of X such that, for each n , F n is C -isomorphic to ℓ n and C -complemented in X . Theorem B. If X contains uniformly complemented ( ℓ n ) ∞ n =1 , then P ( X ) is not comple-mented in Lip ( B X ) . In particular: (i) If X has non-trivial type, P ( X ) is not complemented in Lip ( B X ) , (ii) P ( ℓ ) is not complemented in Lip ( B ℓ ) . As we will see in Section 4, the first clause of Theorem B is a purely formal consequenceof Theorem A. We shall not define type and cotype for a Banach space here, we just recallthat X has non-trivial type if it has type p , for some p > . For basic notions on typeand cotype we refer to [AK]; more advanced properties can be found in [DJT, MS]. (ii)is, of course, consequence of (i), since Hilbert spaces have type . More generally, everysuper-reflexive space has non-trivial type, [Pi]; on the other hand, there are non-reflexivespaces of type , [Ja].That (i) is a particular case of the first part of the theorem is a deep result due to Figieland Tomczak-Jaegermann, [FT-J]. Indeed, it is proved in [FT-J] that if the Banach space X has non-trivial type, there is a constant C such that, for every ε > and n ∈ N , X contains a C -complemented subspace that is (1 + ε ) -isomorphic to ℓ n . The result can alsobe found in the said monographs, [DJT, Theorem 19.3], [MS, Theorem 15.10]; a differentproof was given in [BG].At this stage, one might even be led to conjecture that P ( X ) is complemented in Lip ( B X ) , for no infinite-dimensional Banach space X (plainly, P ( X ) is complementedin Lip ( B X ) for finite-dimensional X ). However this has long been known to be false. P. HÁJEK AND T. RUSSO
Indeed, back in 1976, Aron and Schottenloher [AS] proved that P ( k ℓ ) is isomorphic to ℓ ∞ , hence even injective (for each k > ). More generally, Arias and Farmer [AF] provedthat P ( k X ) is isomorphic to ℓ ∞ whenever X is a separable L -space; in particular, P ( k L ) is complemented in Lip ( B L ) . Since the result in [AF] is stated in a somewhat differentform for L p -spaces, < p < ∞ , we decided to sketch the proof of the said result inProposition 4.1.Our paper is organised as follows: Section 2 collects basic definitions and ancillary resultsthat we need, while the main part of the paper is Section 3, where Theorem A is proved.Finally, in Section 4 we prove Theorem B and Proposition 4.1; we also mention extensionsof our results to higher order polynomials.2. Preliminary material
Our notation is standard, e.g. , as in [AK]. We denote by B X the closed unit ball of aBanach space X . ( E n , |·| ) indicates the n -dimensional Euclidean space ( R n , k·k ) ; whenthe dimension is not important, we just indicate by ( E, |·| ) , or E , a finite-dimensionalEuclidean space. Below we gather the definitions and some basic properties of the objectswe shall consider in our paper.2.1. Spaces of Lipschitz functions.
Let ( M, d, M ) be a pointed metric space, namelya metric space ( M, d ) with a distinguished point M ∈ M . The vector space Lip ( M ) comprising all Lipschitz functions f : M → R such that f (0 M ) = 0 becomes a Banachspace when endowed with the norm given by the best Lipschitz constant k f k Lip := Lip ( f ) := sup (cid:26) | f ( x ) − f ( y ) | d ( x, y ) : x = y ∈ M (cid:27) . In our paper we will only consider the cases when M is a Banach space X or its closedunit ball B X . In either case, the distinguished point will be the origin of the Banach spaceand it will be denoted simply by . Let us recall also here that, for every Banach space X ,the spaces Lip ( X ) and Lip ( B X ) are isomorphic, [Ka, Corollary 3.3].The open unit ball of the Euclidean space E will be denoted B o E . We denote ∇ f thegradient of a differentiable function f : B o E → R . Let C ( B E ) := (cid:26) f : B E → R : f ∈ C ( B E ) ∩ C ( B o E ) , f (0) = 0 , ∇ f is uniformly continuous on B o E (cid:27) . By uniform continuity, the function ∇ f is bounded on B o E and it admits a (unique) exten-sion, that we also denote ∇ f , to B E . Plainly, k f k Lip = k∇ f k ∞ , for every f ∈ C ( B E ) ; inparticular, C ( B E ) ⊆ Lip ( B E ) . Moreover, it is a standard calculus result that C ( B E ) isa closed subspace of Lip ( B E ) . ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 5
Polynomials.
In this section we shall briefly revise basic results on polynomials; werefer to [HJ, § 1.1] for further details. We denote by L ( n X ) the vector space of all n -linearforms M : X × · · · × X → R such that k M k L := sup x j ∈ B X j =1 ,...,n | M ( x , . . . , x n ) | < ∞ . L ( n X ) is a Banach space when endowed with the norm k·k L . L s ( n X ) denotes the closedsubspace of L ( n X ) comprising all symmetric n -linear forms.A function P : X → R is an n -homogeneous polynomial if there is M ∈ L ( n X ) such that P ( x ) = M ( x, . . . , x ) ( x ∈ X ). The vector space P ( n X ) of all n -homogeneous polynomialsis turned into a Banach space when furnished with the norm k P k P := sup x ∈ B X | P ( x ) | . By a standard symmetrisation argument, for every P ∈ P ( n X ) there is a unique symmetric n -linear form q P ∈ L s ( n X ) such that P ( x ) = q P ( x, . . . , x ) . Moreover, the form q P satisfies k P k P (cid:13)(cid:13)(cid:13) q P (cid:13)(cid:13)(cid:13) L n n n ! k P k P . (2.1)Quite importantly, however, in case of a Hilbert space H one has k P k P = k q P k L forevery P ∈ P ( n H ) , [HJ, Theorem 1.16]. These results imply that P ( n X ) and L s ( n X ) areisomorphic Banach spaces for every Banach space X and that they are indeed isometricwhen X = H is a Hilbert space.We shall next discuss the crucial fact mentioned in the Introduction that P ( n X ) naturallyembeds into Lip ( B X ) , for every Banach space X . Fact 2.1.
Let X be any Banach space and P ∈ P ( n X ) . Then n k P k P (cid:13)(cid:13) P ↾ B X (cid:13)(cid:13) Lip n (cid:13)(cid:13)(cid:13) q P (cid:13)(cid:13)(cid:13) L . Proof.
For fixed x ∈ B X , the function [ − , ∋ t P ( tx ) = t n P ( x ) has Lipschitz constantequal to n · | P ( x ) | . Therefore, Lip ( P ↾ B X ) > n · | P ( x ) | and the former inequality follows.For the latter, fix x, y ∈ B X and write P ( x ) − P ( y ) = q P ( x, . . . , x ) − q P ( y, . . . , y )= n X k =1 q P ( x, . . . , x | {z } k -many , y, . . . , y ) − q P ( x, . . . , x | {z } ( k − -many , y, . . . , y )= n X k =1 q P ( x, . . . , x, x − y | {z } k -th place , y, . . . , y ) . Thus, | P ( x ) − P ( y ) | P nk =1 (cid:13)(cid:13)(cid:13) q P (cid:13)(cid:13)(cid:13) L k x k k − k y k n − k k x − y k n (cid:13)(cid:13)(cid:13) q P (cid:13)(cid:13)(cid:13) L k x − y k . (cid:4) P. HÁJEK AND T. RUSSO
Therefore, it follows from Fact 2.1 and (2.1) that the map P P ↾ B X defines an iso-morphic embedding of P ( n X ) into Lip ( B X ) . In the case of a Hilbert space, we have n k P k P = k P ↾ B X k Lip ; hence, such an embedding is additionally multiple of an isometricone. Throughout the paper, when we consider P ( n X ) as a subspace of Lip ( B X ) it will bethis embedding the one under consideration.2.3. Averaging and invariant projections.
It is a standard and useful fact that pro-jections of minimal norm tend to respect the symmetries of the Banach spaces they act on(see, e.g. , [Po, Ri, Ru]). In this part, we shall recall the version of this result that we need;we also sketch a proof, for the sake of completeness.Let us denote by ( E, |·| ) the Euclidean space ( R n , k·k ) and by O n its orthogonal group[Ch, § 1.1]. Since O n is a compact topological group, we can select a normalised Haarmeasure, which we denote µ , on it [HR, Chapter 4].Given a projection Q : C ( B E ) → P ( E ) , we shall consider the projection ˜ Q : C ( B E ) →P ( E ) defined by ˜ Q ( f ) := Z ω ∈O n Q ( f ◦ ω ) ◦ ω − d µ ( ω ) ( f ∈ C ( B E )) . We first observe that ˜ Q ( f ) is well defined as a Bochner integral of the function ω Q ( f ◦ ω ) ◦ ω − , from O n to P ( E ) . Indeed, it suffices to check that the said function iscontinuous. Fact 2.2.
The map ω Q ( f ◦ ω ) ◦ ω − is continuous.Proof. We only need to show that the map ω f ◦ ω , from O n to C ( B E ) , is continuousat the identity matrix . For every x ∈ B E we have |∇ ( f ◦ ω )( x ) − ∇ f ( x ) | = |∇ f ( ωx ) ω − ∇ f ( x ) | |∇ f ( ωx ) ω − ∇ f ( ωx ) | + |∇ f ( ωx ) − ∇ f ( x ) | k∇ f k ∞ · k ω − k + ̟ ∇ f ( k ω − k ) , where ̟ ∇ f denotes the modulus of continuity of ∇ f . Thus, as ω → , k∇ ( f ◦ ω − f ) k ∞ k∇ f k ∞ · k ω − k + ̟ ∇ f ( k ω − k ) → . (cid:4) Once we know that ˜ Q is well defined, it is straightforward to verify the following prop-erties (see [Wo, Theorem III.B.13]). Property (i) claims that ˜ Q is invariant under O n . Lemma 2.3 ([Ru]) . ˜ Q is a projection from C ( B E ) onto P ( E ) . Moreover, (i) ˜ Q ( f ◦ ω ) = ˜ Q ( f ) ◦ ω for each f ∈ C ( B E ) and ω ∈ O n ; (ii) k ˜ Q k k Q k . ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 7
Moreover, recall that Bochner integrals commute with bounded linear operators. Thusfor every bounded linear operator T : P ( E ) → Y one has T (cid:16) ˜ Qf (cid:17) = Z ω ∈O n T (cid:0) Q ( f ◦ ω ) ◦ ω − (cid:1) d µ ( ω ) ( f ∈ C ( B E )) . In particular, this also yields the pointwise formula ˜ Q ( f )( x ) = Z ω ∈O n Q ( f ◦ ω )( ω − x ) d µ ( ω ) ( x ∈ B E ) . Similarly as the average of a projection, we can also define the average of a function f ∈ C ( B E ) to be the function ˜ f := Z ω ∈O n f ◦ ω d µ ( ω ) . As before, we see that ˜ f is well defined as a Bochner integral. Hence ˜ f ∈ C ( B E ) and ∇ ˜ f = Z ω ∈O n ∇ ( f ◦ ω ) d µ ( ω ) . Moreover, we easily see that
Lip ( ˜ f ) Lip ( f ) and that ˜ f is invariant under O n , in thesense that ˜ f ◦ ω = ˜ f for every ω ∈ O n . Finally, ˜ f = f , if f is invariant under O n .3. Proof of Theorem A
This section is dedicated to the proof of the finite-dimensional result Theorem A. Beforeentering the details of the argument, we shall sketch here the the main steps of the proofand the very idea behind it. We will construct a C function ψ : B R → R (hence Lipschitz)which is equal to zero on a neighbourhood of the coordinate axes and whose average overall rotations of the plane is close to a ‘large’ multiple of the polynomial N := ( k·k ) . Sucha function can be considered as the -dimensional analogue of the function ̺ : [ − , → R defined by ̺ ( x ) := ( | x | − ε ) [2 ε, ( | x | ) ( x ∈ [ − , ); actually, the radial behaviour of ψ will be determined by ̺ .We will then define a C function Ψ : B R n → R by means of ψ . We shall show thatthe Lipschitz constant of Ψ is bounded uniformly in n (but depending on ε ), while theprojection of Ψ has a Lipschitz constant that grows with n . Proof of Theorem A.
For the sake of simplicity, we shall just denote by E the n -dimensionalEuclidean space E n = ( R n , k·k ) and |·| its norm. Let K n := inf (cid:8) k Q k : Q is a projection from C ( B E ) onto P ( E ) (cid:9) . Since P ( E ) is a finite-dimensional Banach space, a standard compactness argumentshows that K n is attained, [IS, Theorem 3] (see, e.g. , the argument in [FH e , Lemma5.17.(iii)], or [Li1, p. 12]). Therefore, we can pick a projection Q from C ( B E ) onto P ( E ) P. HÁJEK AND T. RUSSO such that k Q k = K n . Moreover, Lemma 2.3 allows us to assume additionally that Q isinvariant under O n .We are now ready to start the first step of the proof. Step 1.
A function in the plane.
We shall define a C function ψ : B R → R with some symmetries and with the propertythat, if ψ ( x, y ) = 0 , then | x | , | y | > ε . The function ψ will be defined in polar coordinates,as follows.Fix a parameter ε > (whose value will be chosen at the end of the argument in Step5 ). Consider the functions ̺ : [0 , → R and τ : [0 , π/ → R defined by ̺ ( r ) := ( ( r − ε ) r > ε r < ε ( r ∈ [0 , τ ( θ ) := max n π − (cid:12)(cid:12)(cid:12) θ − π (cid:12)(cid:12)(cid:12) , o ( θ ∈ [0 , π/ . Since τ is not C , we approximate it by a smooth function that shares the same proper-ties. More precisely, we fix δ > with δ < π/ and we pick a C function τ : [0 , π/ → R such that(i) τ > and τ − δ τ τ ,(ii) τ is symmetric with respect to θ = π/ ,(iii) Lip ( τ ) .Such a function τ can be easily obtained by a standard convolution argument. The function ψ is the function whose expression in polar coordinates in the first quadrant is given by ̺ · τ . More precisely, we set ψ ( x, y ) := ̺ (cid:16)p x + y (cid:17) · τ (cid:18) arctg (cid:12)(cid:12)(cid:12)(cid:12) xy (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (( x, y ) ∈ B R ) . Fact 3.1.
The function ψ : B R → R has the following properties: (i) ψ ( x, y ) = ψ ( | x | , | y | ) = ψ ( y, x ) , (ii) ψ ( x, y ) = 0 only when | x | , | y | > ε , (iii) ψ ∈ C ( B R ) and Lip ( ψ ) .Proof of Fact 3.1. (i) The fact that ψ ( x, y ) = ψ ( y, x ) is consequence of the symmetry of τ with respect to θ = π/ . The equality ψ ( x, y ) = ψ ( | x | , | y | ) is obvious by definition.(ii) If ψ ( x, y ) = 0 , then x + y > ε and √ | x | | y | √ | x | . Substituting | y | √ | x | into the first inequality yields | x | > ε ; similarly, one gets | y | > ε .(iii) The fact that ψ is C on B R is clear (the smoothness in the points of the two axesfollows, e.g. , from (ii)). Thus, it suffices to show that |∇ ψ ( x, y ) | for each ( x, y ) ∈ B R .Via the expression for the gradient in polar coordinates, we get ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 9 |∇ ψ ( x, y ) | = (cid:16) ̺ ′ (cid:16)p x + y (cid:17) · τ (cid:16) arctg (cid:12)(cid:12)(cid:12) yx (cid:12)(cid:12)(cid:12)(cid:17)(cid:17) + 1 x + y (cid:16) ̺ (cid:16)p x + y (cid:17) · τ ′ (cid:16) arctg (cid:12)(cid:12)(cid:12) yx (cid:12)(cid:12)(cid:12)(cid:17)(cid:17) (2 · π/ + 1 x + y (cid:16)p x + y (cid:17) π /
36 + 1 . (cid:3) Step 2.
The functions Ψ and Ψ d . We shall now pass to defining functions on B E . Fix indices i < j n and define ψ ij : B E → R by ψ ij ( x ) := ψ ( x i , x j ) ( x = ( x , . . . , x n ) ∈ B E ) . Next, we set d := ⌊ n/ √ ⌋ and we define the C functions Ψ := X i Lip (Ψ) , Lip (Ψ d ) /ε .Proof of Fact 3.2. Fix any x ∈ B E , any ξ > and let I := (cid:26) i = i, . . . , n : | x i | > ε ξ (cid:27) . Observe that | I | (cid:18) ξε (cid:19) . Indeed, > n X i =1 x i > X i ∈ I (cid:18) ε ξ (cid:19) = (cid:18) ε ξ (cid:19) · | I | . Now pick any y ∈ B E with k x − y k εξ/ ξ and assume that ψ ij ( y ) = 0 . Fact 3.1yields | y i | , | y j | > ε , whence | x i | , | x j | > ε/ ξ . Therefore, i, j ∈ I ; in other words, in theball B ( x, εξ/ ξ ) only the functions ψ ij with i, j ∈ I are different from . Consequently, Ψ is locally sum of at most · (cid:0) ξε (cid:1) functions whose Lipschitz constant is at most . Itfollows that Ψ is (cid:0) ξε (cid:1) -Lipschitz and letting ξ → proves the claim for Ψ .The argument for Ψ d is identical. (cid:3) Step 3. Computation of Q (Ψ) and Q (Ψ d ) and estimate of max { Lip ( Q Ψ) , Lip ( Q Ψ d ) } . We start giving formulas for the -homogeneous polynomials Q Ψ and Q Ψ d exploitingthe symmetries of Ψ and Ψ d and the invariance of Q . Since ψ ij is a rotation of ψ , itsuffices to compute Qψ .Let ( a ij ) ni,j =1 be scalars with a ij = a ji and such that Qψ ( x ) = n X i,j =1 a ij · x i x j . Fix k = 1 , . . . , n and let ω k be the reflection ( x , . . . , x n ) ( x , . . . , x k − , − x k , x k +1 , . . . , x n ) .Since ψ is invariant under ω k , the same is true for Qψ , namely n X i,j =1 a ij · x i x j = n X i,j =1 i,j = k a ij · x i x j + a kk · x k − n X i =1 i = k a ik · x i x k . Therefore, a ik = 0 , whenever i = k . Since k = 1 , . . . , n was arbitrary, it follows that a ij = 0 for distinct i, j = 1 , . . . , n ; hence Qψ ( x ) = P ni =1 a ii · x i .Next, the invariance of ψ under the reflection ( x , . . . , x n ) ( x , x , x , . . . , x n ) impliesthat a = a . Similarly, the invariance of ψ under the reflection that permutes thecoordinates i and j ( i, j = 3 , . . . , n ) yields that a = · · · = a nn .In conclusion, there are scalars α and β such that Qψ ( x ) = α (cid:0) x + x (cid:1) + β (cid:0) x + · · · + x n (cid:1) . (3.1)This also implies Qψ ij ( x ) = α (cid:0) x i + x j (cid:1) + β n X ℓ =1 ℓ = i,j x ℓ . Consequently, we can now compute Q Ψ( x ) = X i 1) + β ( n − n − (cid:21) (cid:0) x + · · · + x n (cid:1) . ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 11 Letting N ∈ P ( E ) be the square of the norm, i.e. , the polynomial N ( x ) := x + · · · + x n ,we rewrite the above as Q Ψ = (cid:20) α ( n − 1) + β ( n − n − (cid:21) · N. (3.2)Similarly, we compute Q Ψ d . Q Ψ d ( x ) = X i 1) + β ( d − d − (cid:21) (cid:0) x + · · · + x d (cid:1) + β d ( d − (cid:0) x d +1 + · · · + x n (cid:1) . As before, we can set N d ( x ) := x + · · · + x d and obtain Q Ψ d = (cid:20) α ( d − 1) + β ( d − d − (cid:21) · N d + β d ( d − N − N d ) . (3.3)Since the polynomials N and N d are -Lipschitz on B E , the equations (3.2) and (3.3)give Lip ( Q Ψ) = 2 (cid:12)(cid:12)(cid:12)(cid:12) α ( n − 1) + β ( n − n − (cid:12)(cid:12)(cid:12)(cid:12) (3.4) Lip ( Q Ψ d ) > (cid:12)(cid:12)(cid:12)(cid:12) α ( d − 1) + β ( d − d − (cid:12)(cid:12)(cid:12)(cid:12) . (3.5)Note that in the second estimate we used the fact that Lip ( Q Ψ d ) > Lip (( Q Ψ d ) ↾ R d ) .We are now in a position to give a better estimate that only depends on α (and not on β ). Claim 3.3. The polynomials Q Ψ and Q Ψ d satisfy max { Lip ( Q Ψ) , Lip ( Q Ψ d ) } > (cid:16) √ − (cid:17) | α | (cid:16) n − √ (cid:17) . (3.6) Proof of Claim 3.3. We fix a parameter λ > , whose value will be determined later, andwe distinguish two cases. • In case | β | > λ | α | n − , from (3.4) we have Lip ( Q Ψ) = 2 (cid:12)(cid:12)(cid:12)(cid:12) α ( n − 1) + β ( n − n − (cid:12)(cid:12)(cid:12)(cid:12) > | β | ( n − n − − | α | ( n − > λ | α | ( n − − | α | ( n − 1) = ( λ − | α | ( n − . • In case | β | λ | α | n − , on the other hand, we use (3.5) and we obtain Lip ( Q Ψ d ) > (cid:12)(cid:12)(cid:12)(cid:12) α ( d − 1) + β ( d − d − (cid:12)(cid:12)(cid:12)(cid:12) > | α | ( d − − | β | ( d − d − > | α | ( d − − λ | α | ( d − d − n − > | α | ( d − − λ | α | ( d − 1) 1 √ > | α | ( d − − λ/ √ > (cid:16) √ − λ/ (cid:17) | α | (cid:16) n − √ (cid:17) . Here, we used the inequalities d − n − > / √ and d − > n − √ that follow readilyfrom our previous choice d := ⌊ n/ √ ⌋ .The optimal choice for λ , namely λ := (cid:0) √ (cid:1) yields λ − √ − λ/ (cid:0) √ − (cid:1) .Thus, we have: • In case | β | > λ | α | n − , Lip ( Q Ψ) > (cid:16) √ − (cid:17) | α | ( n − > (cid:16) √ − (cid:17) | α | (cid:16) n − √ (cid:17) . • In case | β | λ | α | n − , Lip ( Q Ψ d ) > (cid:16) √ − (cid:17) | α | (cid:16) n − √ (cid:17) . Therefore, the estimate (3.6) is proved. (cid:3) Step 4. Estimate of α . The estimate (3.6) that we obtained in Claim 3.3 in the previous step is still not sufficient,since we have no information on the parameter α . It is thus the purpose of this step tohandle this and to give a lower bound for α . To wit, we prove here the following claim. Claim 3.4. Assume that εK n . Then, the parameter α that appears in (3.6) satisfies α > (cid:16) π − δ (cid:17) · (cid:0) − εK n (cid:1) . (3.7)The proof is based on the fact, mentioned at the beginning of the present section, thatthe average of ψ under rotations of the plane is close to a multiple of the square of thenorm of R . Therefore, its projection onto the set of -homogeneous polynomials sharesthe same property. ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 13 Proof of Claim 3.4. We identify the group S O with the subgroup of O n of rotations of span { e , e } . Accordingly, for ω ∈ S O , we also write ω to denote the rotation ω ⊕ n − ∈O n defined by ω ⊕ n − ( x , . . . , x n ) = ( ω ( x , x ) , x , . . . , x n ) . We also denote µ the Haarmeasure of S O .Let ˜ ψ be the average of ψ under the group S O (see § 2.3), i.e. , let ˜ ψ := Z ω ∈ S O ψ ◦ ω d µ ( ω ) . Then for every x ∈ B E , we have ˜ ψ ( x ) = Z ω ∈ S O ψ ( ω ( x , x ) , x , . . . , x n ) d µ ( ω )= Z ω ∈ S O ψ ( ω ( x , x )) d µ ( ω )= Z ω ∈ [0 , π ] ̺ (cid:18)q x + x (cid:19) · τ ( ω ) d ω π =: ̺ (cid:18)q x + x (cid:19) · η, where we defined η := R [0 , π ] τ ( ω ) d ω π .Letting, as before, N ∈ P ( E ) be the polynomial N ( x ) = x + x and recalling that ̺ ( r ) := ( r − ε ) · [2 ε, ( r ) , the above yields (cid:16) ηN − ˜ ψ (cid:17) ( x ) = η · ( x + x if x + x ε ε p x + x − ε elsewhere . Consequently, Lip (cid:16) ηN − ˜ ψ (cid:17) εη. (3.8)Moreover, since τ − δ τ τ and R [0 , π ] τ ( ω ) d ω π = π , we have the estimate < π − δ η π . (3.9)On the other hand, using the facts that Bochner integrals commute with linear operators,that Q is invariant under O n (hence, under its subgroup S O ) and that Qψ is invariantunder S O by (3.1), we derive Q ˜ ψ = Z ω ∈ S O Q ( ψ ◦ ω ) d µ ( ω )= Z ω ∈ S O Q ( ψ ) ◦ ω d µ ( ω )= Z ω ∈ S O Q ( ψ ) d µ ( ω ) = Q ( ψ ) . Hence, using again Qψ = αN + β ( N − N ) from (3.1), we obtain Q (cid:16) ηN − ˜ ψ (cid:17) = ηN − Qψ = ( η − α ) N + β ( N − N ) , whence Lip (cid:16) Q (cid:16) ηN − ˜ ψ (cid:17)(cid:17) > Lip (cid:16) Q (cid:16) ηN − ˜ ψ (cid:17) ↾ R (cid:17) = Lip (( η − α ) N ) = 2 | α − η | . (3.10)Finally, combining (3.8) with (3.10) yields | α − η | Lip (cid:16) Q (cid:16) ηN − ˜ ψ (cid:17)(cid:17) K n · Lip (cid:16) ηN − ˜ ψ (cid:17) εη · K n . Therefore, α > η (1 − εK n ) whence, using (3.9), the conclusion follows. (cid:3) Step 5. Plugging estimates together. In this step, we shall combine the estimates obtained in Fact 3.2 and Claim 3.3 with thelower bound on α from Claim 3.4 and we shall conclude the proof.From Fact 3.2 and Claim 3.3 we deduce (cid:16) √ − (cid:17) | α | (cid:16) n − √ (cid:17) max (cid:8) k Q Ψ k Lip , k Q Ψ d k Lip (cid:9) K n · max (cid:8) k Ψ k Lip , k Ψ d k Lip (cid:9) K n · ε . (3.11)We now set (the reason behind such a choice will be clear in a few lines) c := 23 (cid:16) √ − (cid:17) π and ε := 2 / c / (cid:0) n − √ (cid:1) / and we distinguish two cases, depending on whether − εK n > or not. ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 15 • If εK n , we see from Claim 3.4 that α > (recall that δ < π/ ). Therefore,we can plug the estimate (3.7) for α in (3.11) and we obtain ε · K n > (cid:16) √ − (cid:17) · (cid:16) π − δ (cid:17) · (1 − εK n ) (cid:16) n − √ (cid:17) . (3.12)We can now let δ → and obtain ε · K n > (cid:16) √ − (cid:17) · π 72 (1 − εK n ) (cid:16) n − √ (cid:17) =: c · (1 − εK n ) (cid:16) n − √ (cid:17) , which we rewrite as follows: c · (cid:16) n − √ (cid:17) (cid:20) ε + 2 εc (cid:16) n − √ (cid:17)(cid:21) K n . The above choice of ε minimises the value of the square parenthesis above and givesthe value (cid:20) ε + 2 εc (cid:16) n − √ (cid:17)(cid:21) = c / / (cid:16) n − √ (cid:17) / . Therefore, we have K n > / c / (cid:16) n − √ (cid:17) / = C (cid:16) n − √ (cid:17) / . Here, C := 2 / c / = 25 √ − · π ! / , hence we obtained the inequality claimed in Theorem A. • In the case εK n > , then K n > ε = 12 / c / (cid:16) n − √ (cid:17) / > C (cid:16) n − √ (cid:17) / , since / > / . Consequently, also in this case, the desired estimate is valid. (cid:4) Infinite-dimensional results The present section is dedicated to our infinite-dimensional results. We shall start withthe proof of our main result, Theorem B. We then briefly mention the situation for separable L -spaces. Finally we conclude the section discussing how to extend our results to higherorder polynomials. Proof of Theorem B. As we already observed in the Introduction, (i) follows from [FT-J]and the first part of the theorem, while (ii) is a particular case of (i). We now prove thefirst clause.Let X be a Banach space that contains uniformly complemented ( ℓ n ) ∞ n =1 ; towards acontradiction, assume that there exists a projection Q from Lip ( B X ) onto P ( X ) . Bydefinition, we may pick a sequence ( F n ) ∞ n =1 of C -complemented subspaces of X such that F n is C -isomorphic to ℓ n ( n ∈ N ). Let P n be projections from X onto F n with k P n k C and let T n : ℓ n → F n be isomorphisms with k T n k and k T − n k C , for each n ∈ N .We define a projection Q n from Lip ( C · B ℓ n ) onto P ( ℓ n ) by the following formula. Q n ( f ) := (cid:0) Q ( f ◦ T − n ◦ P n ) (cid:1) ↾ F n ◦ T n ( f ∈ Lip ( C · B ℓ n )) . Lip ( C · B ℓ n ) Q n (cid:15) (cid:15) ( T − n ◦ P n ) ♯ / / Lip ( B X ) Q (cid:15) (cid:15) P ( ℓ n ) P ( F n ) T ♯n o o P ( X ) · ↾ Fn o o Here, T ♯n is defined by T ♯n g := g ◦ T n (and similarly for ( T − n ◦ P n ) ♯ ). Notice that in thiscase we consider P ( ℓ n ) as a subspace of Lip ( C · B ℓ n ) via the embedding P P ↾ C · B ℓn .Since k T − n ◦ P n k C , f ◦ T − n ◦ P n is indeed a Lipschitz function on B X , when f ∈ Lip ( C · B ℓ n ) . Thus, Q ( f ◦ T − n ◦ P n ) is a polynomial on X whose Lipschitz constant on B X is at most C k Q k · Lip ( f ) . Hence, its Lipschitz constant on C · B X is bounded by C k Q k · Lip ( f ) . Consequently, we obtain that k Q n k C k Q k . Moreover, it is easy torealise that Q n is a projection from Lip ( C · B ℓ n ) onto P ( ℓ n ) .By scaling, we also obtain projections from Lip ( B ℓ n ) onto P ( ℓ n ) with norms at most C k Q k , for every n ∈ N . However, for large n , this contradicts Theorem A and concludesthe proof. (cid:4) We already mentioned in the Introduction that there is no hope to extend the conclusionof Theorem B to every infinite-dimensional Banach space. The first result in this directionwas that P ( k ℓ ) is isomorphic to ℓ ∞ , [AS]. Its proof goes as follows: one first showsthe rather straightforward fact that L ( k ℓ ) is isometric to ℓ ∞ . Then, by the Polarisationformula, L s ( k ℓ ) is complemented in L ( k ℓ ) (recall that L s ( k X ) is isomorphic to P ( k X ) forevery X ). The conclusion follows from Lindenstrauss’ result that complemented subspacesof ℓ ∞ are isomorphic to it, [Li3]. Alternatively, one could directly use that L s ( k X ) isisomorphic to L ( k X ) when X is isomorphic to X , [DD]. Proposition 4.1 ([AF]) . If X is a separable L -space, P ( k X ) is isomorphic to ℓ ∞ , forevery k > . Hence, P ( k X ) is complemented in Lip ( B X ) . Combining this with the comments below on general polynomials, it also follows that P k ( X ) is isomorphic to ℓ ∞ . We refer to [LT] for the definition of the L -spaces. ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 17 Proof. As before, it suffices to embed L ( k X ) as a complemented subspace of ℓ ∞ and appealto [Li3]. Since X is a separable L -space, there are two uniformly bounded sequences ( ϕ j ) ∞ j =1 and ( ψ j ) ∞ j =1 of linear maps ϕ j : X → ℓ j , ψ j : ℓ j → X such that ϕ j ◦ ψ j = id and ( ψ j ( ℓ j )) ∞ j =1 is an increasing sequence whose union is dense in X ([LT, Proposition II.5.9]).Define a bounded linear map Ψ : L ( k X ) → (cid:16)P ∞ j =1 L ( k ℓ j ) (cid:17) ∞ by Ψ( M ) := ( M ◦ ψ j ) ∞ j =1 ;notice that (cid:16)P ∞ j =1 L ( k ℓ j ) (cid:17) ∞ is isometric to ℓ ∞ .We also define Φ : (cid:16)P ∞ j =1 L ( k ℓ j ) (cid:17) ∞ → L ( k X ) by ( M j ) ∞ j =1 lim U M j ◦ ϕ j where U isa free ultrafilter on N and the limit is in the pointwise topology. Plainly, Φ is a boundedlinear map. Moreover, it is easy to see that Φ ◦ Ψ is the identity of L ( k X ) [AF, Lemma2.3]. Hence, L ( k X ) embeds as a complemented subspace of ℓ ∞ , as desired. (cid:4) Remark . Let X be a separable L p -space, < p < ∞ . The same argument as aboveshows that L ( k X ) embeds as a complemented subspace of (cid:16)P ∞ j =1 L ( k ℓ jp ) (cid:17) p , which is com-plemented in L ( k ℓ p ) . Then, one shows that L ( k ℓ p ) embeds complemented in L ( k X ) [AF,Lemma 2.4] and invokes Pełczyński’s decomposition method to conclude that L ( k X ) isisomorphic to L ( k ℓ p ) , [AF, Theorem 2.1].Let us also point out the similarity between this argument and some techniques from[CCD]. This is also true for the statement of some their results, e.g. , [CCD, Theorem 3.3].In conclusion to our article, we shall mention how to extend our results to k -homogeneouspolynomials ( k > ) and general polynomials. For the definition of k -homogeneous poly-nomials we refer to Section 2.2. A polynomial of degree at most k on X is a function P : X → R of the form P = P kj =0 P j , where P j ∈ P ( j X ) . The collection of all polynomialsof degree at most k is denoted by P k ( X ) and it is a Banach space under the same norm k P k P := sup x ∈ B X | P ( x ) | . We denote by P k ( X ) the closed subspace comprising polynomi-als that vanish at , namely those where P = 0 (of course, such restriction is necessary inorder to embed into Lip ( B X ) ).It is an important fact that P k ( X ) = P ( X ) ⊕ P ( X ) ⊕ · · · ⊕ P ( k X ) via the naturaldecomposition P = P kj =0 P j ( P j ) kj =0 . Moreover, the norm of the projection P P j doesnot depend on X . (This can be shown via Vandermonde matrices as in [HJ, Fact 1.1.42],or via the Polarisation formula, [HJ, Lemma 1.1.47].) In particular, the map P P ↾ B X also defines an isomorphic embedding of P k ( X ) into Lip ( B X ) .Since P ( X ) is complemented in P k ( X ) , a formal consequence of Theorem B is that P k ( X ) is not complemented in Lip ( B X ) , whenever X is a Banach space as in TheoremB. For P ( k X ) the argument is more complicated and the idea is to embed P ( X ) as acomplemented subspace of P ( k X ) and mimic the proof of Theorem A. This strategy ismotivated by the result that P ( X ) embeds as a complemented subspace of P ( k X ) , [AS,Proposition 5.3] (also see [Bl, Proposition 5] for a stronger result). However, for us it willbe more convenient to use the canonical isomorphism H between P k ( X ) and P ( k X ⊕ R ) ,given by homogenisation. More precisely, let ζ : X ⊕ R → R be the functional ζ ( x, t ) := t . The action of H on P = P kj =0 P j ∈ P k ( X ) is given by P h P := P kj =0 ζ k − j P j . Conversely, every P ∈ P ( k X ⊕ R ) has the form P = P kj =0 ζ k − j P j , with P j ∈ P ( j X ) , and H − ( P ) = P kj =0 P j (roughly speaking, this amounts to evaluating P on t = 1 ). Moreover, the isomorphismconstant between P k ( X ) and P ( k X ⊕ R ) depends on k , but not on X . Theorem 4.3. Let X be a Banach space that contains uniformly complemented ( ℓ n ) ∞ n =1 .Then, for every k > , P ( k X ) and P k ( X ) are not complemented in Lip ( B X ) .Proof. We only need to show the assertion concerning P ( k X ) .Assume that X is a Banach space that contains uniformly complemented ( ℓ n ) ∞ n =1 and,towards a contradiction, assume that P ( k X ) is complemented in Lip ( B X ) . The samediagram chasing argument as in the proof of Theorem B shows that there are projectionsfrom Lip ( B ℓ n ) onto P ( k ℓ n ) whose norms are bounded uniformly in n ∈ N . We let F n := ℓ n ⊕ ∞ R . Since F n is √ -isomorphic to ℓ n +12 , it also follows that P ( k F n ) is complementedin Lip ( B F n ) , uniformly in n .Next, we exploit the isomorphism H . P ( ℓ n ) is complemented in P k ( ℓ n ) , uniformly in n ,and the latter space is isomorphic to P ( k F n ) , uniformly in n as well. Therefore, the imageof P ( ℓ n ) under H is complemented in Lip ( B F n ) , uniformly in n . However, we shall showthat this is not the case.To simplify the notation, we let E := ℓ n and F := F n = E ⊕ ∞ R . Moreover, let P := H (cid:0) P ( E ) (cid:1) = (cid:8) P ∈ P ( k F ) : P = ζ k − · P , for some P ∈ P ( E ) (cid:9) . Recall that ζ was defined on F by ζ ( x, t ) := t ( x ∈ E , t ∈ R ). Heuristically, P consists ofpolynomials of the form t k − P ( x ) , for some -homogeneous polynomial P on E .Our goal will be to show that if Q is any projection from Lip ( B F ) onto P , then thenorm of Q diverges with n , which leads to the desired contradiction. As in Theorem A, weactually consider the restriction of Q to C ( B F ) and, by averaging, we assume that Q isinvariant under O n , the group of rotations of the Euclidean space E . We then follow theproof of Theorem A, but multiplying the Lipschitz functions there by ζ k − .Consider the functions ϕ ij := ψ ij · ζ k − , Φ := Ψ · ζ k − , and Φ d := Ψ d · ζ k − . Thepresence of the factor ζ k − now leads to an extra k/ factor in Fact 3.2, namely we have Lip (Φ) , Lip (Φ d ) k/ ε (indeed, each p ij is k -Lipschitz). The invariance argument in Step 3 proceeds identically to give that Q Φ = ( Q Ψ) · ζ k − and analogously for Φ d . Hence,Claim 3.3 remains true with the same estimate. In the estimate of α , the unique differenceis an extra ( k − factor; in other words, (3.7) becomes α > (cid:16) π − δ (cid:17) · (cid:0) − εK n (cid:1) · ( k − . (4.1)Finally, when putting estimates together in Step 5 , we obtain an inequality strongerthan (3.12), since k/ k − . Consequently, we actually even obtain the same lowerbound on k Q k as the one in Theorem A. We omit further details. (cid:4) ROJECTING LIPSCHITZ FUNCTIONS ONTO SPACES OF POLYNOMIALS 19 References [AACD] F. Albiac, J.L. Ansorena, M. Cúth, and M. Doucha, Lipschitz algebras and Lipschitz-free spacesover unbounded metric spaces , arXiv:2011.12993 .[AK] F. Albiac and N. 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Hájek) Department of Mathematics, Faculty of Electrical Engineering, Czech Tech-nical University in Prague, Technická 2, 166 27 Prague 6, Czech Republic Email address : [email protected] (T. Russo) Institute of Mathematics, Czech Academy of Sciences, Žitná 25, 115 67 Prague1, Czech Republic; Department of Mathematics, Faculty of Electrical Engineering, CzechTechnical University in Prague, Technická 2, 166 27 Prague 6, Czech Republic Email address ::