On the constant in a transference inequality for the vector-valued Fourier transform
OON THE CONSTANT IN A TRANSFERENCE INEQUALITY FOR THEVECTOR-VALUED FOURIER TRANSFORM
DION GIJSWIJT & JAN VAN NEERVEN
Abstract.
The standard proof of the equivalence of Fourier type on R d and on the torus T d is usually stated in terms of an implicit constant, defined as the minimum of a sum of powersof sinc functions. In this note we compute this minimum explicitly. Introduction
The motivation of this paper comes from a well-known transference result for the vector-valuedFourier transform. Let X be a complex Banach space. The Fourier transform of a function f ∈ L ( R d ; X ) is defined by F R d f ( ξ ) := (cid:90) R d e − πix · ξ f ( x ) dx, ξ ∈ R d . Likewise, the
Fourier transform of a function f ∈ L ( T d ; X ) is defined by F T d f ( k ) := (cid:90) T d e − πik · t f ( t ) dt, k ∈ Z d . Proposition 1.
Let X be a complex Banach space, fix d ≥ and p ∈ (1 , , and let p + q = 1 .The following assertions are equivalent: (i) F R d extends to a bounded operator from L p ( R d ; X ) into L q ( R d ; X ) ; (ii) F T d extends to a bounded operator from L p ( T d ; X ) into (cid:96) q ( Z d ; X ) .In this situation, denoting the norms of these extensions by ϕ p,X ( R d ) and ϕ p,X ( T d ) , we have ϕ p,X ( R d ) ≤ ϕ p,X ( T d ) ≤ C − d/qq ϕ p,X ( R d ) , where C q is the global minimum of the periodic function x (cid:55)→ (cid:88) m ∈ Z (cid:12)(cid:12)(cid:12) sin( π ( x + m )) π ( x + m ) (cid:12)(cid:12)(cid:12) q , x ∈ R . This function, as well as several others considered below, have removable singularities. It isunderstood that we will always be working with their unique continuous extensions.A complex Banach space X which has the equivalent properties (i) and (ii) is said to have Fourier type p ; this notion has been introduced in [5]. Proposition 1 goes back to [4]; in its statedform the result can be found in [2, 3]. Related results may be found in [1]. These references do notcomment on the location of the global minimum. A quick computer plot (see Figure 1) suggeststhat the minimum is taken in the points + Z . To actually prove this turns out to be surprisinglydifficult. This is the modest objective of the present note: Proposition 2.
For every real number r ≥ , the function f r : [0 , → R defined by f r ( x ) := (cid:88) m ∈ Z (cid:12)(cid:12)(cid:12) sin( π ( x + m )) π ( x + m ) (cid:12)(cid:12)(cid:12) r , x ∈ [0 , , has a global minimum at x = . Date : November 5, 2018.2000
Mathematics Subject Classification.
Key words and phrases.
Hausdorff-Young inequality, Fourier type, transference. a r X i v : . [ m a t h . C A ] A p r DION GIJSWIJT & JAN VAN NEERVEN
Figure 1.
A plot of f r , where r = 1 . k for k = 1 , , , . . . , ϕ p,X ( R d ) ≤ ϕ p,X ( T d ) ≤ π d (cid:0) q − ζ ( q ) (cid:1) d/q · ϕ p,X ( R d ) , noting that (cid:88) m ∈ Z | + m | q = 2(2 q − ζ ( q ) . For even integers q = 2 n , the constant on the right-hand side may be evaluated explicitly in termsof the Bernoulli numbers. To further estimate this constant, recall that for any x ∈ (cid:96) ( Z ) thefunction q (cid:55)→ || x || q := ( (cid:80) m ∈ Z | x m | q ) /q is decreasing on [2 , ∞ ) and lim q →∞ || x || q = sup i ∈ Z | x i | .Taking x m := | + m | − we find ( (cid:80) m ∈ Z | + m | − q ) /q ≥ q ≥
2, and hence in particular ϕ p,X ( R d ) ≤ ϕ p,X ( T d ) ≤ ( π ) d ϕ p,X ( R d ) . The main result
The proof of the proposition is based on the following lemmas. The main idea is contained inthe first lemma.
Lemma 3.
Let g : R + → R + be a non-decreasing convex function, and let x , . . . , x n ∈ R + and y , . . . , y n ∈ R + be such that (i) x + · · · + x n ≥ y + · · · + y n ; (ii) there exists t ∈ R + such that • x i ≤ y i if y i < t ; • x i ≥ y i if y i ≥ t .Then g ( x ) + · · · + g ( x n ) ≥ g ( y ) + · · · + g ( y n ) .Proof. We will prove the lemma by induction on n . The case n = 1 is clear: x ≥ y impliesthat g ( x ) ≥ g ( y ) since g is non-decreasing. Suppose now that the lemma has been proved for n = 1 , . . . , m − x i = y i for some index 1 ≤ i ≤ m , then we may remove x i and y i and apply the inductionhypothesis. N THE CONSTANT IN A TRANSFERENCE INEQUALITY 3 If x i ≥ y i for every index 1 ≤ i ≤ m , then again the result is immediate since g is non-decreasing.Therefore, we may assume that x i < y i for some index 1 ≤ i ≤ m . Then, by the first condition inthe lemma, there is also an index j for which x j > y j . By the second condition in the lemma wethen have x i < y i < t ≤ y j < x j .Let (cid:15) := min( y i − x i , x j − y j ) and define x (cid:48) i := x i + (cid:15) , x (cid:48) j := x j − (cid:15) , and x (cid:48) k := x k for all otherindices. Then x (cid:48) , . . . , x (cid:48) m , y , . . . , y m satisfy the conditions in the lemma (with the same t ) and x (cid:48) i = y i or x (cid:48) j = y j . Hence, by the induction hypothesis, we have g ( x (cid:48) ) + · · · + g ( x (cid:48) m ) ≥ g ( y ) + · · · + g ( y m ) . (1)Since x i ≤ x (cid:48) i ≤ x (cid:48) j ≤ x j , we can write x (cid:48) i = λx i + (1 − λ ) x j for some λ ∈ [0 , x (cid:48) j = x i + x j − x (cid:48) i , we have x (cid:48) j = (1 − λ ) x i + λx j . By the convexity of g it follows that g ( x (cid:48) i ) + g ( x (cid:48) j ) ≤ ( λg ( x i ) + (1 − λ ) g ( x j )) + ((1 − λ ) g ( x i ) + λg ( x j )) = g ( x i ) + g ( x j ) . (2)Combining inequalities (1) and (2) we obtain the lemma for n = m , thus completing the inductionstep. (cid:3) In order to apply this lemma we need a number of technical facts. The first (cf. [2, (6.14)]) iselementary and is left as an exercise.
Lemma 4. f ( x ) = 1 for all x ∈ [0 , . Let h : R → R be defined by h ( x ) := sinc ( πx ) = (cid:16) sin( πx ) πx (cid:17) , x ∈ R . Lemma 5.
Let r ≥ . The following assertions hold on the interval [0 , : (i) the function h ( x ) + h ( x − has a global minimum at x = ; (ii) for all m = 1 , , , . . . , h ( x + m ) + h ( x − ( m + 1)) has a global maximum at x = ; (iii) the function h ( x ) + h ( x − − ( h ( x ) r + h ( x − r ) /r has a global maximum at x = ; (iv) for all m = 1 , , , . . . and r ≥ , ( h ( x + m ) + h ( x − ( m + 1))) r − h ( x + m ) r − h ( x − ( m + 1)) r has a global maximum at x = . Assuming the lemmas for the moment, let us first show how the proposition can be deducedfrom them.
Proof of Proposition 2.
Fix r ≥ x ∈ [0 , s m ( x ) := h ( x + m ) + h ( x − ( m + 1)) ( m = 0 , , , . . . )and (cid:101) s ( x ) := (( h ( x )) r + ( h ( x − r ) /r . In view of part (iv) of Lemma 5 it suffices to prove that (cid:101) s r + s r + s r + · · · has a global minimum at x = .Fix an arbitrary x ∈ [0 ,
1] and set x m := s m ( x ) , y m := s m ( ) ( m = 0 , , , . . . )and (cid:101) x := (( h ( x )) r + h ( x − r ) /r , (cid:101) y := (( h ( )) r + h ( − ) r ) /r . In view of parts (i) and (ii) of Lemma 5 we have x ≥ y , x i ≤ y i ( i = 1 , , . . . ) (3) DION GIJSWIJT & JAN VAN NEERVEN
Lemma 4 implies x + x + x + · · · = y + y + y + · · · (4)By (3) and (4), x + x + · · · + x n ≥ y + y + · · · + y n ( n = 0 , , , . . . ) (5)Part (iii) of Lemma 5 implies (cid:101) x − x ≥ (cid:101) y − y . (6)By (5) and (6), (cid:101) x + x + · · · + x n ≥ (cid:101) y + y + · · · + y n ( n = 0 , , , . . . ) (7)Finally, by (3) and (6), (cid:101) x ≥ (cid:101) y . (8)A simple calculation shows that (cid:101) y > π and y i < π for i = 1 , , . . . . Taking t = π inLemma 3 and g ( x ) := x r now implies, by virtue of (3), (7), and (8), that (cid:101) x r + x r + · · · + x rn ≥ (cid:101) y r + y r + · · · + y rn holds for every n . Taking limits for n → ∞ completes the proof. (cid:3) Proof of Lemma 5
This section is devoted to the proof of Lemma 5, which is based on the following observations:
Lemma 6.
On the interval [0 , : (i) cos( πx )1 − x takes a global maximum at x = 0 ; (ii) ( x + 1) cos ( πx )(1 − x ) takes a global minimum at x = 0 .Proof. We start by showing that √ πx ) ≥ x for all x ∈ [0 , . (9)To this end, consider the function f ( x ) := √ πx ) − x . Observe that f (cid:48) ( x ) = π √ cos( πx ) − , f is concave. Since f (0) = f (1) = 0 this implies that f ( x ) ≥ x ∈ [0 , x = 0 of the given function equals 1, so it suffices to show that cos( πx ) ≤ − x for all x ∈ [0 , πx ) = 1 − ( πx ) ≤ − x . (ii): The given function has value 1 at x = 0, hence it suffices to show that for all x ∈ [0 , x + 1) cos ( πx ) ≥ (1 − x ) . On the interval [ ,
1] we substitute x = 1 − y . We then must prove that for y ∈ [0 , ],(2 − y + y ) sin ( πy ) ≥ (2 y − y ) . Since 2 y ∈ [0 , √ π · y ) ≥ y , and hence sin ( πy ) ≥ y .This implies that(2 − y + y ) sin ( πy ) ≥ (2 − y + y )(2 y ) = ( y + (2 − y ) ) y ≥ (2 − y ) y = (2 y − y ) , which concludes the proof on the interval [ , N THE CONSTANT IN A TRANSFERENCE INEQUALITY 5
For x ∈ [0 , ] we have( x + 1) cos ( πx ) ≥ ( x + 1) (cid:16) − π x (cid:17) = ( x + 1)(1 − π x + π x ) ≥ − π x + ( π − π x = 1 + (1 − π x + ( π − π − x + x ≥ (cid:104) (1 − π π − π − (cid:105) x + x ≥ − x + x = (1 − x ) , noting that π − π − < − π ) + ( π − π − ≈ − . · · · > − (cid:3) Proof of Lemma 5. (i): We have h ( x ) + h ( x −
1) = sin ( πx ) π x + sin ( πx ) π ( x − = (2 x − x + 1) sin ( πx ) π x ( x − =: g ( x ) . We must show that f ( x ) := g ( x + 12 ) = 8 π x + 1(4 x − cos ( πx )has a global minimum in x = 0 on the interval [ − , ]. But this follows from Lemma 6 and thefact that f is even.(ii): For m = 1 , , , . . . we have h ( x + m ) + h ( x − ( m + 1)) = [2 x − x + ( m + 1) + m ] sin ( πx ) π [( x + m ) ( x − ( m + 1)) ] =: g m ( x ) . We must show that f m ( x ) := g m ( x + 12 ) = 8 π x + 4 m + 4 m + 1[(4 x − (2 m + 1) ] cos ( πx )has a global maximum in x = 0 on the interval [ − , ]. For this, it suffices to check that thefunctions 4 x + 1(4 x − M ) cos ( πx ) and 1(4 x − M ) cos ( πx )are decreasing on [0 , ] for each M ≥
3, or equivalently, that √ x + 1 M − x cos( πx ) and 1 M − x cos( πx )are decreasing on [0 ,
1] for each M ≥
3. It suffices to prove this for the first function, since thiswill immediately imply the result for the second function.Straightforward algebra shows that the derivative of the function ψ M ( x ) := √ x + 1 M − x cos( πx )has a zero at x if and only if2 x ( x + 2 + M ) cos( πx ) = π ( M − x + ( M − x ) sin( πx ) . But, 2 x ( M + 2 + x ) cos( πx ) ≤ x ( M + 2 + x )and, since 0 ≤ x ≤ π ( M − x + ( M − x ) x ≤ π ( M − x + ( M − x ) sin( πx ) , DION GIJSWIJT & JAN VAN NEERVEN while also, using that M ≥ ≤ x ≤ M + 2 + x ) ≤ M + 2 + ( M − x ) < π ( M − M − x ) ≤ π ( M − x + ( M − x )since 2( M + 2) < π ( M −
1) for M ≥
3. It follows that the derivative of ψ M has no zeros on(0 , ψ M (0) = 1 M > ψ M (1)it follows that ψ M is decreasing on [0 , h ( x ) + h ( x − − (( h ( x )) r + ( h ( x − r ) /r = 1 π (cid:104) x + 1(1 − x ) − (cid:16) x r + 1(1 − x ) r (cid:17) /r (cid:105) sin ( πx ) =: g ( x ) . We must show that f ( x ) := g ( + x ) = 1 π (cid:104) ( − x ) + ( + x ) − (( − x ) r + ( + x ) r ) /r (cid:105) cos ( πx )( − x ) has a global maximum in x = 0 on the interval [ − , ]. The function f is even, and by Lemma 6,cos ( πx ) / ( − x ) takes its maximum at x = 0. It thus remains to show that on the interval[0 , ] the function φ r ( x ) := ( − x ) + ( + x ) − (( − x ) r + ( + x ) r ) /r is decreasing on [0 , ]. The derivative of this function equals φ (cid:48) r ( x ) = 4 x − (cid:0) ( − x ) r + ( + x ) r (cid:1) /r − (cid:0) ( + x ) r − − ( − x ) r − (cid:1) . To show that φ (cid:48) r ( x ) ≤ (cid:0) ( + x ) r + ( − x ) r (cid:1) /r − (cid:0) ( + x ) r − − ( − x ) r − (cid:1) ≥ x for x ∈ [0 , ], or, after substituting a = + x and b = − x , that a r − − b r − ≥ ( a − b ) (cid:0) a r + b r (cid:1) − /r for all a ∈ [ , (cid:0) a r + b r (cid:1) − /r = (cid:104)(cid:0) a r + b r (cid:1) / (2 r ) (cid:105) r − ≤ (cid:104)(cid:0) a r − + b r − (cid:1) / (2 r − (cid:105) r − = (cid:0) a r − + b r − (cid:1) − / (2 r − , with p := 2 r − a p − b p ≥ ( a − b )( a p + b p ) − /p for all a ≥ b ≥
0. We can further simplify this upon dividing both sides by b p . In the new variable x = a/b we then have to prove that x p − ≥ ( x − x p + 1) − /p for all x ≥ y ) α ≤ αy for y ≥ ≤ α ≤
1, we have( x − x p + 1) − /p = ( x p − x p − )(1 + x − p ) − /p ≤ ( x p − x p − )[1 + (1 − p ) x − p ] . Therefore it remains to prove that for x ≥ p ≥ x p − ≥ ( x p − x p − )[1 + (1 − p ) x − p ] , or, multiplying both sides with x , that x p +1 − x ≥ x p +1 − x p + (1 − p ) ( x − . N THE CONSTANT IN A TRANSFERENCE INEQUALITY 7 that is, we must show that f p ( x ) := x p ≥ x + (1 − p )( x −
1) =: g p ( x ) . Now f (cid:48) p ( x ) = px p − , g (cid:48) p ( x ) = 2 − p . It follows that f (cid:48) p ( x ) ≥ g (cid:48) p ( x ) ≥ x ≥
1, since p ≥ − p (multiply both sides by p ). Togetherwith f p (1) = g p (1) it follows that f p ( x ) ≥ g p ( x ) for x ≥ p ≥
1. This concludes the proof of(iii).(iv): Fix m ≥
1. For x ∈ [ − , ] we have( h ( x + + m ) + h ( x + − ( m + 1))) r − h ( x + + m ) r − h ( x + − ( m + 1)) r = (cid:20)(cid:16) x + ( m + )) + 1( x − ( m + )) (cid:17) r − (cid:16) x + ( m + )) (cid:17) r − (cid:16) x − ( m + )) (cid:17) r (cid:21) × π − r (cid:0) cos ( πx ) (cid:1) r . We must show that this function has a global maximum on [ − , ] at x = 0. Since by Lemma 6cos( πx ) / (1 − x ) has a global maximum at x = 0, it suffices to prove that (cid:2) (( a + x ) − + ( a − x ) − ) r − ( a + x ) − r − ( a − x ) − r (cid:3) · (1 − x ) r has a global maximum at x = 0, where we have written a := m + ≥ . Since the function x (cid:55)→ x r is convex, we have ( a + x ) − r + ( a − x ) − r ≥ ( ( a + x ) − + ( a − x ) − ) r and hence2 − r (( a + x ) − + ( a − x ) − ) r − ( a + x ) − r − ( a − x ) − r ≤ x = 0. Therefore, it suffices to show that(1 − − r )(( a + x ) − + ( a − x ) − ) r (1 − x ) r has a global maximum at x = 0. It is enough to show that g ( x ) := (( a + x ) − +( a − x ) − )(1 − x ) is decreasing on [0 , ].Computing the derivative of g we find g (cid:48) ( x ) = − x (1 − x )(( a + x ) − + ( a − x ) − ) + (1 − x ) ( − a + x ) − + 2( a − x ) − )= (1 − x )( a + x ) − ( a − x ) − k ( x ) , where k ( x ) = − x ( a − x )(( a + x ) + ( a − x ) ) + (1 − x )(2( a + x ) − a − x ) )= − x · a − x ) + (1 − x ) · x · (3 a + x )= 4 x [ − a − x ) + (1 − x )(3 a + x )]= 4 x [4 x + (1 − a ) x + (3 a − a )] . Since a > (cid:113) , the function p ( y ) := 4 y + (1 − a ) y + (3 a − a ) has a positive and a negativeroot. The sum of the two roots equals a − and therefore the positive root is larger than 3 a − ≥ . It follows that p is negative on [0 , ] and hence g (cid:48) ( x ) = (1 − x )( a + x ) − ( a − x ) − · x · p ( x ) ≤ , ], which finishes the proof. (cid:3) Added in proof . After this paper had been accepted for publication, Tom Koornwinder sent usthe following interesting proof for the case that the parameter r in Proposition 2 is integral. Withhis kind permission we reproduce it here.We consider f r ( x ) on (0 , ζ ( s, q ) (see [6, Eq. 25.11.1])we have f r ( x ) = π − r sin r ( πx )( ζ (2 r, x ) + ζ (2 r, − x )) , r = 1 , , . . . DION GIJSWIJT & JAN VAN NEERVEN
In terms of the digamma function ψ ( z ) = Γ (cid:48) ( z ) / Γ( z ) (see [6, Eq. 25.11.12]) this can be rewrittenas f r ( x ) = π − r sin r ( πx )(2 r − (cid:18) ddx (cid:19) r − ( ψ ( x ) − ψ (1 − x )) . Applying the reflection formula ψ (1 − z ) − ψ ( z ) = π cot( πz ) (see [6, Eq. 5.5.4]) we obtain f r ( x ) = − π − r sin r ( πx )(2 r − (cid:18) ddx (cid:19) r − cot( πx ) . Substitution of t = πx simplifies this expression to(2 r − r t · f r ( t/π ) = − (cid:18) ddt (cid:19) r − cot t. Since ( d/dt ) cot t = − / sin t , we have f ( t/π ) = 1. Also, we obtain the following recursionrelation: (2 r + 1)!sin r +2 t · f r +1 ( t/π ) = − (cid:18) ddt (cid:19) r +1 cot t = (2 r − (cid:18) ddt (cid:19) f r ( t/π )sin r t . (10)A small computation shows that (cid:18) ddt (cid:19) f r ( t/π )sin r t = (cid:18) ddt (cid:19) (cid:20) (sin − r t ) (cid:18) ddt (cid:19) f r ( t/π ) − r (cos t )(sin − r − t ) f r ( t/π ) (cid:21) = (sin − r t ) (cid:18) ddt (cid:19) f r ( t/π ) − r (cos t )(sin − r − t ) (cid:18) ddt (cid:19) f r ( t/π )+ (cid:0) r (2 r + 1) cos t sin − r − t + 2 r sin − r t (cid:1) · f r ( t/π ) . Hence, (10) implies that(2 r + 1)!(2 r − f r +1 ( t/π ) = (sin t ) (cid:18) ddt (cid:19) f r ( t/π ) − r (cos t )(sin t ) (cid:18) ddt (cid:19) f r ( t/π )+(2 r (2 r + 1) cos t + 2 r sin t ) · f r ( t/π ) . (11)Set y := cos t and D := d/dy . So d/dt = − t cos t ) D and (cid:18) ddt (cid:19) = ddt ( − t cos t ) D = − t − sin t ) D − (2 sin t cos t ) (cid:18) ddt (cid:19) D = − t − sin t ) D + (4 sin t cos t ) D = ( − y + 2) D + 4 y (1 − y ) D . Equation (11) can therefore be rewritten as(2 r + 1)!(2 r − f r +1 ( t/π ) = (cid:2) y (1 − y ) D + ((8 r − y + 2)(1 − y ) D + 2 r (2 ry + 1) (cid:3) f r ( t/π )= (cid:2) y ( r − yD ) + 8( r − yD ) yD + 2 yD + 2 r + 4 yD + 2 D (cid:3) f r ( t/π ) . (12)Observe that ( r − yD ) y k = ( r − k ) y k . Hence, if p = p ( y ) is a polynomial of degree n < r withnonnegative coefficients, then the same holds for ( r − yD ) p . The recursion (12) and the factthat f ( t/π ) = 1 now imply that f r ( t/π ) is a polynomial in y of degree r − N THE CONSTANT IN A TRANSFERENCE INEQUALITY 9 coefficients. The first few are given explicitly by f ( t/π ) = 1 f ( t/π ) = 13 + 23 cos tf ( t/π ) = 215 + 1115 cos t + 215 cos tf ( t/π ) = 17315 + 47 cos t + 38105 cos t + 4315 cos tf ( t/π ) = 622835 + 10722835 cos t + 484945 cos t + 2472835 cos t + 22835 cos t For integers r , Proposition 2 is an immediate consequence.For half-integers r = n + one could observe that the identity ψ (2 n ) ( x ) = − (2 n )! ∞ (cid:88) m =0 x + m ) n +1 allows one to express the inequality of Proposition 2 in terms of the polygamma functions ψ (2 n ) .We have not been able, however, to use this fact to give a simpler proof in that case. References [1] J. Bourgain, Vector-valued Hausdorff-Young inequalities and applications. In: Geometric aspects of functionalanalysis (1986/87), 239–249, Lecture Notes in Math., Vol. 1317, Springer, Berlin, 1988.[2] J. Garcia-Cuerva, K.S. Kazarian, V.I. Kolyada, & J.L. Torrea, Vector-valued Hausdorff-Young inequality andapplications. Russian Mathematical Surveys, (3) (1998), 435–513.[3] H. K¨onig, On the Fourier-coefficients of vector-valued functions. Math. Nachr. (1991), 215–227.[4] S. Kwapie´n, Isomorphic characterizations of inner product spaces by orthogonal series with vector valued coef-ficients. Studia Math. (1972), 583–595.[5] J. Peetre, Sur la transformation de Fourier des fonctions `a valeurs vectorielles. Rend. Sem. Mat. Univ. Padova (1969), 15–26.[6] http://dlmf.nist.gov Delft University of Technology, Faculty EEMCS/DIAM, P.O. Box 5031, 2600 GA Delft, The Nether-lands
E-mail address ::