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On the Convolution Inequality F ⩾ On the convolution inequality f > f ? f Eric Carlen Department of Mathematics, Rutgers University [email protected] Ian Jauslin Department of Physics, Princeton University [email protected] Elliott H. Lieb Departments of Mathematics and Physics, Princeton University [email protected] Michael Loss School of Mathematics, Georgia Institute of Technology [email protected] Abstract 1 d We consider the inequality f > f ? f for real functions in L (R ) where f ? f denotes the convolution of f with itself. We show that all such functions f are non-negative, which is not p the case for the same inequality in L for any 1 < p 6 2, for which the convolution is defined. 1 d R 1 R 1 We also show that all solutions in L (R ) satisfy d f(x)dx 6 . Moreover, if d f(x)dx = , R R 2 R 2 then f must decay fairly slowly: d jxjf(x)dx = 1, and this is sharp since for all r < 1, there R 1 R R r R 1 are solutions with d f(x)dx = 2 and d jxj f(x)dx < 1. However, if d f(x)dx =: a < 2 , the R R 1 R decay at infinity can be much more rapid: we show that for all a < 2 , there are solutions such that for some > 0, R ejxjf(x)dx < 1. Rd © 2020 by the authors. This paper may be reproduced, in its entirety, for non-commercial purposes. Our subject is the set of real, integrable solutions of the inequality d f(x) > f ? f(x); 8x 2 R ; (1) R where f ? f(x) denotes the convolution f ? f(x) = d f(x − y)f(y)dy. By Young's inequality p R d [LL96, Theorem 4.2], for all 1 6 p 6 2 and all f 2 L (R ), f ? f is well defined as an element of p=(2−p) d p d L (R ). Thus, one may consider this inequality in L (R ) for all 1 6 p 6 2, but the case p = 1 is special: the solution set of (1) is restricted in a number of surprising ways. Integrating both sides of (1), one sees immediately that R f(x)dx 1. We prove that, in fact, all integrable Rd 6 solutions satisfy R f(x)dx 1 , and this upper bound is sharp. Rd 6 2 Perhaps even more surprising, we prove that all integrable solutions of (1) are non-negative. p d p d This is not true for solutions in L (R ), 1 < p 6 2. For f 2 L (R ), 1 6 p 6 2, the Fourier transform f(k) = R e−i2πk·xf(x)dx is well defined as an element of Lp=(p−1)( d). If f solves the b Rd R equation f = f ? f, then fb = fb2, and hence fb is the indicator function of a measurable set. By 1 d the Riemann-Lebesgue Theorem, if f 2 L (R ), then fb is continuous and vanishes at infinity, and the only such indicator function is the indicator function of the empty set. Hence the only integrable solution of f = f ? f is the trivial solution f = 0. However, for 1 < p 6 2, solutions abound: take d = 1 and define g to be the indicator function of the interval [−a; a]. Define Z sin 2πxa f(x) = e−i2πkxg(k)dk = ; (2) R πx p which is not integrable, but which belongs to L (R) for all p > 1. By the Fourier Inversion Theorem fb = g. Taking products, one gets examples in any dimension. −2πjkjt To construct a family of solutions to (1), fix a; t > 0, and define ga;t(k) = ae . By [SW71, Theorem 1.14], Z t f (x) = e−i2πkxg (k)dk = aΓ((d + 1)=2)π−(d+1)=2 : a;t a;t 2 2 (d+1)=2 Rd (t + x ) 2 Since ga;t(k) = ga2;2t, fa;t ? fa;t = fa2;2t, Thus, fa;t > fa;t ? fa;t reduces to t 2at (t2 + x2)(d+1)=2 > (4t2 + x2)(d+1)=2 R which is satisfied for all a 1=2. Since d fa;t(x)dx = a, this provides a class of solutions of (1) 6 R that are non-negative and satisfy Z 1 f(x)dx 6 ; (3) Rd 2 all of which have fairly slow decay at infinity, so that in every case, Z jxjf(x)dx = 1 : (4) Rd Our results show that this class of examples of integrable solutions of (1) is surprisingly typical of all integrable solutions: every real integrable solution f of (1) is positive, satisfies (3), and if 1 d there is equality in (3), f also satisfies (4). The positivity of all real solutions of (1) in L (R ) p d may be considered surprising since it is false in L (R ) for all p > 1, as the example (2) shows. We also show that when strict inequality holds in (3) for a solution f of (1), it is possible for f to have rather fast decay; we construct examples such that R ejxjf(x)dx < 1 for some > 0. Rd The conjecture that integrable solutions of (1) are necessarily positive was motivated by recent work [CJL20, CJL20b] on a partial differential equation involving a quadratic nonlinearity of f ? f type, and the result proved here is the key to the proof of positivity for solutions of this partial differential equations; see [CJL20]. Autoconvolutions f ? f have been studied extensively; see [MV10] and the work quoted there. However, the questions investigated by these authors are quite different from those considered here. 1 Theorem 1 1 d Let f be a real valued function in L (R ) such that f(x) − f ? f(x) =: u(x) > 0 (5) for all x. Then R f(x) dx 1 , and f is given by the series Rd 6 2 1 1 X f(x) = c 4n(?nu)(x) (6) 2 n n=1 1 d pwhich converges in L (R ), and where the cn > 0 are the Taylor coefficients in the expansion of 1 − x 1 p X (2n − 3)!! 1 − x = 1 − c xn; c = ∼ n−3=2 (7) n n 2nn! n=1 In particular, f is positive. Moreover, if u 0 is any integrable function with R u(x)dx 1 , then > Rd 6 4 the sum on the right in (6) defines an integrable function f that satisfies (5), and R f(x)dx = 1 Rd 2 if and only if R u(x)dx = 1 . Rd 4 Proof: Note that u is integrable. Let a := R f(x)dx and b := R u(x)dx 0. Fourier Rd Rd > transforming, (5) becomes 2 fb(k) = fb(k) + ub(k) : (8) 2 1 2 1 1 At k = 0, a − a = −b, so that a − 2 = 4 − b. Thus 0 6 b 6 4 . Furthermore, since u > 0, 1 jub(k)j 6 ub(0) 6 4 (9) p and the first inequality is strict for k 6= 0. Hence for k 6= 0, 1 − 4ub(k) 6= 0. By the Riemann- Lebesgue Theorem, fb(k) and ub(k) are both continuous and vanish at infinity, and hence we must have that 1 1 p fb(k) = 2 − 2 1 − 4ub(k) (10) p for all sufficiently large k, and in any case fb(k) = 1 ± 1 1 − 4u(k). But by continuity and the p 2 2 b fact that 1 − 4ub(k) 6= 0 for any k 6= 0, the sign cannot switch. Hence (10) is valid for all k, 1 p including k = 0, again by continuity. At k = 0, a = 2 − 1 − 4b, which proves (3). The fact that −3=2 cn as specified in (7) satisfiesp cn ∼ n is a simple application of Stirling's formula, and it shows that the power series for 1 − z converges absolutely and uniformly everywhere on the closed 1 p X n unit disc. Since j4ub(k)j 6 1, 1 − 4ub(k) = 1 − cn(4ub(k)) . Inverting the Fourier transform, n=1 R n n 1 d yields (6), and since d 4 ? u(x)dx 6 1, the convergence of the sum in L (R ) follows from the P1 R convergence of n=1 cn. The final statement follows from the fact that if f is defined in terms of u in this manner, then (10) is valid, and then (8) and (5) are satisfied. Theorem 2 Let f 2 L1( d) satisfy (1) and R f(x) dx = 1 . Then R jxjf(x) dx = 1. R Rd 2 Rd R 1 R Proof: If d f(x) dx = 2 , d 4u(x) dx = 1, then w(x) = 4u(x) is a probability density, and R 1 P1 n R we can write f(x) = 2 n=0 ? w. Aiming for a contradiction, suppose that jxjf(x) is integrable. Then jxjw(x) is integrable. Let m := R xw(x)dx. Since first moments add under convolution, Rd the trivial inequality jmjjxj > m · x yields Z Z n n 2 jmj jxj ? w(x)dx > m · x ? w(x)dx = njmj : Rd Rd 2 R jmj P1 It follows that d jxjf(x)dx ncn = 1. Hence m = 0. R > 2 n=1 Suppose temporarily that in addition, jxj2w(x) is integrable. Let σ2 be the variance of w; i.e., σ2 = R jxj2w(x)dx Define the function '(x) = minf1; jxjg. Then Rd Z Z Z n 1=2 n 1=2 d=2 1=2 n 1=2 d=2 jxj ? w(x)dx = jn xj ? w(n x)n dx > n '(x) ? w(n x)n dx: Rd Rd Rd By the Central Limit Theorem, since ' is bounded and continuous, Z Z lim '(x) ?n w(n1=2x)nd=2dx = '(x)γ(x)dx =: C > 0 (11) n!1 Rd Rd where γ(x) is a centered Gaussian probability density with variance σ2.
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