Best constants in the local Hausdorff Young inequality on compact Lie groups

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1 Best constants in the local Hausdorff Young inequality on compact Lie groups Michael G Cowling University of New South Wales Gwangju, August 7, / 24

2 Thanks It is a pleasure to be here! This is work in progress, joint with Alessio Martini, Detlef Müller and Javier Parcet. 2 / 24

3 Introduction background: Babenko and Beckner, Brascamp and Lieb a local Hausdorff-Young inequality for T n a Hausdorff-Young inequality for a compact Lie group remarks. 3 / 24

4 Background For f L 1 (R n ), define ˆf(ξ) = f(x)e 2πiξ x dx. Then ˆf f 1 and ˆf 2 = f 2. 4 / 24

5 Background For f L 1 (R n ), define ˆf(ξ) = f(x)e 2πiξ x dx. Then ˆf f 1 and ˆf 2 = f 2. Interpolation implies the Hausdorff Young inequality: when 1 p 2 and 1/p + 1/q = 1, ˆf q C f p, (HY) where C = 1. What is the best constant for this inequality? 4 / 24

6 Young s inequality Suppose that 1 p,q,r and 1 r = 1 p + 1 q 1. Then for all f L p (R n ) and g L q (R n ), f g L r (R n ) and f g r C f p g q, (Y) where C = 1. What is the best constant for this inequality? 5 / 24

7 The sharp Hausdorff Young inequality Define B p by ( p 1/p) 1/2 B p =, q 1/q where q is the conjugate index to p. Then B p < 1 when 1 < p < 2. 6 / 24

8 The sharp Hausdorff Young inequality Define B p by ( p 1/p) 1/2 B p =, q 1/q where q is the conjugate index to p. Then B p < 1 when 1 < p < 2. Theorem (Babenko (1961), Beckner (1975)) The best constant for the inequality (HY) on R n is (B p ) n. Babenko treated the case where q 2Z +, and Beckner did the general case. The extremal functions are Gaussians. 6 / 24

9 The sharp form of Young s inequality Theorem (Beckner (1975), Brascamp Lieb (1976)) The best constant in Young s inequality on R n is (B p B q B r ) n. Beckner s methods are more difficult than those of Brascamp and Lieb. Again, the extremal functions are Gaussians. 7 / 24

10 Generalisations One can generalise these results to locally compact abelian groups connected Lie groups. 8 / 24

11 Generalisations One can generalise these results to locally compact abelian groups connected Lie groups. On R a T b Z c, the best constant for inequality (HY) is B a p. The extremal functions are of the form γ χ δ, where γ is a Gaussian on R a χ is a character of T b δ is the characteristic function of a point in Z c 8 / 24

12 Generalisations One can generalise these results to locally compact abelian groups connected Lie groups. On R a T b Z c, the best constant for inequality (HY) is B a p. The extremal functions are of the form γ χ δ, where γ is a Gaussian on R a χ is a character of T b δ is the characteristic function of a point in Z c For semisimple Lie groups, e.g., SL(2,R), the picture is unclear. Since SL(2,R) is topologically the same as R T 2, perhaps we should expect B p rather than B 3 p. 8 / 24

13 Lie groups Many n-dimensional solvable Lie groups admit a chain of subgroups {e} = G 0 < G 1 < < G n 1 < G n where G j is normal in G j+1 and G j+1 /G j is isomorphic to R. Klein and Russo (1979) treated these groups. The best constant for Young s inequality is at most (B p B q B r ) n, for the full range of indices. The best constant for the Hausdorff Young inequality, is at most (B p ) n, at least when p 2Z +. For the Heisenberg group, there are no extremizers. 9 / 24

14 Lie groups Many n-dimensional solvable Lie groups admit a chain of subgroups {e} = G 0 < G 1 < < G n 1 < G n where G j is normal in G j+1 and G j+1 /G j is isomorphic to R. Klein and Russo (1979) treated these groups. The best constant for Young s inequality is at most (B p B q B r ) n, for the full range of indices. The best constant for the Hausdorff Young inequality, is at most (B p ) n, at least when p 2Z +. For the Heisenberg group, there are no extremizers. We can show that the constant in (HY) cannot be any smaller than (B p ) n for all p for Heisenberg groups. 9 / 24

15 A local version of Young s inequality Identify the torus T with {z C : z = 1}, and for δ (0,1/2), define U δ = {e 2πiθ : θ < δ} n T n. Let C(δ) be the best constant in the inequality f g r C f p g q supp(f),supp(g) U δ. Clearly C(δ) / 24

16 A local version of Young s inequality Identify the torus T with {z C : z = 1}, and for δ (0,1/2), define U δ = {e 2πiθ : θ < δ} n T n. Let C(δ) be the best constant in the inequality f g r C f p g q supp(f),supp(g) U δ. Clearly C(δ) 1. Theorem (obvious) With the above notation, { C(δ) = (Bp B q B r ) n if 0 < δ 1/4 C(δ) 1 as δ / 24

17 A local Hausdorff Young theorem Take δ < 1/2. Let C p (δ) be the best constant in the inequality Clearly C p (δ) 1. ˆf q C f p supp(f) U δ. 11 / 24

18 A local Hausdorff Young theorem Take δ < 1/2. Let C p (δ) be the best constant in the inequality Clearly C p (δ) 1. ˆf q C f p supp(f) U δ. Theorem (Andersson (1993), Sjölin (1995), Kamaly (1996)) With the above notation, lim C p(δ) = (B p ) n. δ / 24

19 Proof of theorem There is no loss of generality in supposing functions smooth enough for all the sums and integrals that occur to converge. Given f on T n, we define F on R n by { f(e 2πix ) when x ( 1 F(x) = 2, 1 2 ]n 0 otherwise. We say that F corresponds to f. When supp(f) U 1/2, it is clear that F is as smooth as f. The aim of the proof is to use the Hausdorff Young theorem for F and transfer it to f. 12 / 24

20 Proof (2) Suppose that supp(f) U 1/2. Since ˆf = ˆF Z n, the Poisson summation formula implies that ˆf(m) = ˆF(m) = F(m) = F(0) = ˆF(ξ)dξ, k Z n k Z n k Z n R n 13 / 24

21 Proof (2) Suppose that supp(f) U 1/2. Since ˆf = ˆF Z n, the Poisson summation formula implies that ˆf(m) = ˆF(m) = F(m) = F(0) = ˆF(ξ)dξ, k Z n k Z n k Z n R n that is, nˆf(m) = ˆF(ξ)dξ. k Z R n (PS) 13 / 24

22 Proof (3) Take k Z + and g such that supp(g) U δ1, where δ 1 = 1 4(k+1) ; set g (x) = ḡ( x). Then the 2k-fold convolution g g g g has support in U 1/2, and if g corresponds to G, then the convolution corresponds to G G G G. Further, the Fourier transforms of g g g g and G G G G are ĝ 2k and Ĝ 2k. 14 / 24

23 Proof (4) Take q = 2k and p dual to q. Applying (PS), the consequence of the Poisson summation formula, gives ĝ(k) 2k = Ĝ(ξ) 2k dξ k Z n R n = Ĝ q q (B p ) nq G q p = (B p ) nq g q p, so ĝ q (B p ) n g p. 15 / 24

24 Proof (5) We used the Poisson summation formula to show that when q 2Z +, Ĝ Z n q Ĝ q (actually, we proved equality). Now think of Ĝ on Rn as the function of interest. 16 / 24

25 Proof (5) We used the Poisson summation formula to show that when q 2Z +, Ĝ Z n q Ĝ q (actually, we proved equality). Now think of Ĝ on Rn as the function of interest. If we could interpolate, it would follow that Ĝ Z n q Ĝ q, for all q [2, ), and we could deduce that ĝ q Ĝ q (B p ) n G p = (B p ) n g p But we cannot interpolate because of the support restriction. 16 / 24

26 Proof (6) Take ϕ : R n [0,1] such that supp(ϕ) U δ1 and ϕ(x) = 1 on 1 U δ2, where δ 2 = 4(R+1)(k+1) for some R > 0, 17 / 24

27 Proof (6) Take ϕ : R n [0,1] such that supp(ϕ) U δ1 and ϕ(x) = 1 on 1 U δ2, where δ 2 = 4(R+1)(k+1) for some R > 0, and define TH = ˆϕ H, for H on R n, then TH is the Fourier transform of a function supported in U δ1, whence (TH) Z n 2k = TH2k ˆϕ 1H2k and the same holds when k is replaced by k + 1. Now we can interpolate and deduce that (TH) Z n q ˆϕ 1 H q for q [2k,2(k + 1)]. 17 / 24

28 Proof (7) If supp(ĥ) U δ 2, then TH = H. Thus, if supp(g) U δ2, then ĝ q = (T Ĝ) Z n q ˆϕ 1Ĝq 18 / 24

29 Proof (7) If supp(ĥ) U δ 2, then TH = H. Thus, if supp(g) U δ2, then ĝ q = (T Ĝ) Z n q ˆϕ 1Ĝq This now gives ĝ q ˆϕ 1 Ĝ q ˆϕ 1 (B p ) n G p = ˆϕ 1 (B p ) n g p, which is our sharp local Hausdorff-Young inequality. 18 / 24

30 Compact Lie groups García-Cuerva, Marco, and Parcet (2003) proved a central local Hausdorff Young inequality for a compact Lie group G, and applied this to functional analytic questions about Banach spaces. They showed that the constant was less than 1, but they did not find the best constant. 19 / 24

31 Compact Lie groups García-Cuerva, Marco, and Parcet (2003) proved a central local Hausdorff Young inequality for a compact Lie group G, and applied this to functional analytic questions about Banach spaces. They showed that the constant was less than 1, but they did not find the best constant. A function f is central if f(xy) = f(yx), that is, if f is constant on conjugacy classes. 19 / 24

32 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2, / 24

33 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2,.... They parametrise the collection of irreducible unitary representations π λ of G. 20 / 24

34 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2,.... They parametrise the collection of irreducible unitary representations π λ of G. Each such representation π λ is of finite dimension d λ, and has a character χ λ, given by tr(π λ ( )). 20 / 24

35 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2,.... They parametrise the collection of irreducible unitary representations π λ of G. Each such representation π λ is of finite dimension d λ, and has a character χ λ, given by tr(π λ ( )). The characters are central. 20 / 24

36 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2,.... They parametrise the collection of irreducible unitary representations π λ of G. Each such representation π λ is of finite dimension d λ, and has a character χ λ, given by tr(π λ ( )). The characters are central. For a central function f, we define ˆf(λ) by ˆf(λ) = f(x) χ λ (x)dx. Then f = λ Λ G ˆf(λ)χ λ and f 2 = λ Λ ˆf(λ) / 24

37 Representations and characters in a nutshell A compact Lie group comes with a set Λ + of dominant integral weights. For SU(2), these are 0, 1 2,1, 3 2,2,.... They parametrise the collection of irreducible unitary representations π λ of G. Each such representation π λ is of finite dimension d λ, and has a character χ λ, given by tr(π λ ( )). The characters are central. For a central function f, we define ˆf(λ) by ˆf(λ) = f(x) χ λ (x)dx. Then f = λ Λ G ˆf(λ)χ λ and f 2 = λ Λ ˆf(λ) 2. Further ˆf(λ) dλ f / 24

38 The central local Hausdorff Young inequality ( d p 2 λ λ Λ + The sharp constant is 1. ˆf(λ) q ) 1/q Cd f p f central. For a neighbourhood U of the identity in G, let C p (U) be the best constant in the same inequality with the additional support restriction that supp(f) U. 21 / 24

39 The central local Hausdorff Young inequality ( d p 2 λ λ Λ + The sharp constant is 1. ˆf(λ) q ) 1/q Cd f p f central. For a neighbourhood U of the identity in G, let C p (U) be the best constant in the same inequality with the additional support restriction that supp(f) U. Theorem (C-M-M-P) Let G be a compact connected Lie group. Then lim C p(u) = (B p ) dim(g). U {e} 21 / 24

40 Proof The exponential map exp maps the Lie algebra g of G onto G, sends 0 to the identity e of G, and is a bijection near / 24

41 Proof The exponential map exp maps the Lie algebra g of G onto G, sends 0 to the identity e of G, and is a bijection near 0. To a function f on G supported near e, there is a corresponding function F on g supported near 0, namely, F = f exp. 22 / 24

42 Proof The exponential map exp maps the Lie algebra g of G onto G, sends 0 to the identity e of G, and is a bijection near 0. To a function f on G supported near e, there is a corresponding function F on g supported near 0, namely, F = f exp. The Fourier transform ˆF of F is defined on the dual g of g: ˆF(ξ) = F(X)e 2πiξ X dx. g 22 / 24

43 Proof The exponential map exp maps the Lie algebra g of G onto G, sends 0 to the identity e of G, and is a bijection near 0. To a function f on G supported near e, there is a corresponding function F on g supported near 0, namely, F = f exp. The Fourier transform ˆF of F is defined on the dual g of g: ˆF(ξ) = F(X)e 2πiξ X dx. g The function J on g is the square root of the Jacobian of the exponential map; it is close to 1 near / 24

44 Proof (2) The following analogue of (PS) holds for central functions with small support on G: 1 d λˆf(λ) = (J 1 F)ˆ(ξ)dξ. λ Λ + g 1 Check 23 / 24

45 Proof (2) The following analogue of (PS) holds for central functions with small support on G: 1 d λˆf(λ) = (J 1 F)ˆ(ξ)dξ. λ Λ + g For central functions f and g on G, the convolution h = f g is again central, and ĥ = ˆf ĝ; further, if F, G and H correspond to f, g and h, then J 1 H = (J 1 F) (J 1 G). Then multiple convolutions on G turn into multiple convolutions on g whose Fourier transforms on g multiply as in the classical case. 1 Check 23 / 24

46 Proof (2) The following analogue of (PS) holds for central functions with small support on G: 1 d λˆf(λ) = (J 1 F)ˆ(ξ)dξ. λ Λ + g For central functions f and g on G, the convolution h = f g is again central, and ĥ = ˆf ĝ; further, if F, G and H correspond to f, g and h, then J 1 H = (J 1 F) (J 1 G). Then multiple convolutions on G turn into multiple convolutions on g whose Fourier transforms on g multiply as in the classical case. We can therefore extend the proof of the sharp result for T n to central functions on a compact Lie group G. 1 Check 23 / 24

47 Remarks We are hoping to extend the result to all functions. For functions supported in a small neighbourhood of the identity on any connected Lie group G, the group convolution turns into a perturbed convolution on the Lie algebra: f g(exp X) = F(Y)G(X Y P(X,Y))J(Y)dY, where the perturbation P(X,Y) is a convergent power series in the variables X and Y, given by the higher order terms in the Baker Campbell Hausdorff formula. When X and Y are very small, convolution looks like euclidean convolution, and the Jacobians are close to 1. So there should be local versions of many classical results on Lie groups. 24 / 24

Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is. e ikx f(x) dx. (1.

Chapter 1 Fourier transforms 1.1 Introduction Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is ˆf(k) = e ikx f(x) dx. (1.1) It