Positive definite functions, completely monotone functions, the Bernstein-Widder theorem, and Schoenberg s theorem Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto June 6, 15 1 Linear operators For a comple Hilbert space H let L H) be the bounded linear operators H H. It is a fact that A L H) is self-adjoint if and only if Ah, h R for all h H. 1 For a bounded self-adjoint operator A it is a fact that A sup Ah, h. h 1 A L H) is called positive if it is self-adjoint and Ah, h, h H; because we have taken H to be a comple Hilbert space, for A to be positive it suffices that the inequality is satisfied. For A, B L C n ), we define their Hadamard product A B L H) by So, A B)e i Ae i, e j Be i, e j e j. j1 A B)e i, e j Ae i, e j Be i, e j. The Schur product theorem states that if A, B L C n ) are positive then their Hadamard product A B is positive. 3 1 John B. Conway, A Course in Functional Analysis, second ed., p. 33, Proposition.1. John B. Conway, A Course in Functional Analysis, second ed., p. 34, Proposition.13. 3 Ward Cheney and Will Light, A Course in Approimation Theory, p. 81, chapter 1. 1
Positive definite functions Let X be a real or comple linear space, let f : X C be a function, and for 1,..., n X, define F f;1,..., n L C n ) by F f;1,..., n e i f i j )e j, j1 where {e 1,..., e n } is the standard basis for C n. Thus for u n i1 u ie i C n, n n F f;1,..., n u, u f i j )e j, u k e k u i i1 j1 n u i i1 j1 i1 j1 f i j ) u i u j f i j ). e j, k1 u k e k k1 We call f positive definite if for all 1,..., n X, F f;1,..., n operator, i.e. for u C n, is a positive F f;1,..., n u, u. We call f strictly positive definite for all distinct 1,..., n X and nonzero u C n, F f;1,..., n u, u >. 3 Completely monotone functions A function f : [, ) R is called completely monotone if 1. f C[, ). f C, ) 3. 1) k f k) ) for k and, ) Because a completely monotone function is continuous, f) tends to f) as. Because a completely monotone function is nonincreasing and conve, f) has a limit, which we call f ), as. The Bernstein-Widder theorem states that a function f satisfying f) 1 is completely monotone if and only if it is the Laplace transform of a Borel probability measure on [, ). 4 4 Peter D. La, Functional Analysis, p. 138, chapter 14, Theorem 3; http://djalil.chafai.net/blog/13/3/3/ the-bernstein-theorem-on-completely-monotone-functions/
Theorem 1 Bernstein-Widder theorem). A function f : [, ) R satisfies f) 1 and is completely monotone if and only if there is a Borel probability measure µ on [, ) such that f) e t dµt), [, ). Proof. If f is the Laplace transform of some probability measure µ on B [, ), then using the dominated convergence theorem yields that f is continuous and by induction that f C, ). For k and for, ), f k) ) t) k e t dµt), as t k e t dµt) so 1) k f k) ). Hence f is completely monotone, and f) dµt) 1. If f satisfies f) 1 and is completely monotone, then for each k, the function 1) k f k) :, ) R is nonnegative and is nonincreasing, so for k 1 and t, ), using that 1) k f k) is nondecreasing and that 1) k 1 f k 1) is nonnegative, t 1) k f k) t) 1) k f k) u)du t t/ ) t 1)k f k 1) t) f k 1) t/) Doing induction, for any k 1, 1) k f k) t) t 1)k 1 f k 1) t/). k 1 j1 k 1 j1 Because f) f) as, t f k 1 j t j t ) ) t f k 1 t k kk 1)/ f ) f and because f) f ) as, ) ) t t f k 1 f k ) k t f t k 1 t k 1 ) t f ) f ) t k, t, 3, t. t k k )) )).
Hence for each k 1, and ft) o k t k ), t, 1) ft) o k t k ), t. ) Furthermore, for any, ), f k) t) f k) ) as t, so it is immediate that t ) k f k) t), t. 3) For and k 1, integrating by parts, using ) and 1) or 3) respectively as or >, f) f ) f t)dt t )f t) Hence for and n, Define + f t)t )dt f t)t )dt t ) f t) f t ) t) dt f t ) t) dt 1) k f k) t )k 1 t) dt. k 1)! f) f ) 1)n+1 n! For n 1, by change of variables, For t, define f) f ) 1)n+1 n! φ n y) 1 y/n) n 1 [,n] y). 1)n+1 n 1)! 1)n+1 n 1)! 1)n+1 n 1)! s n t) 1)n+1 n 1)! /n /n f n+1) t)t ) n dt. f n+1) nu)nu ) n ndu f n+1) nu)nu) n 1 ) n du nu nu) n φ n /u)f n+1) nu)du n/t) n φ n t)f n+1) n/t)t dt. 1/t 4 nu) n f n+1) nu)du,
where s n ), and for t < let s n t). s n is continuous and because f is completely monotone, s n is nondecreasing, so there is a unique positive measure σ n on B R such that 5 σna, b]) snb) sna), a b. On the other hand, s n is absolutely continuous, so σ n is absolutely continuous with respect to Lebesgue measure λ 1, and for λ 1 -almost all t R, 6 dσ n dλ 1 t) s nt). Now for t >, by the fundamental theorem of calculus and the chain rule, and therefore s nt) 1)n+1 n 1)! n/t)n f n+1) n/t) t, f) f ) φ n t)s nt)dλ 1 t) φ n t) dσ n dλ 1 t)dλ 1 t) φ n t)dσ n t). The total variation of σ n is equal to the total variation of s n, and because s n is nondecreasing, which is σ n s nt) dt σ n 1)n+1 n 1)! s nt)dt s n ) s n ) s n ), nu) n f n+1) nu)du f) f ), showing that {σ n : n 1} is bounded for the total variation norm. We claim that {σ n : n 1} is tight: for each ɛ > there is a compact subset K ɛ of R such that σ n Kɛ c ) < ɛ for all n. Taking this for granted, Prokhorov s theorem 7 states that there is a subsequence σ kn of σ n that converges narrowly to some positive measure σ on B R. Finally, the sequence t φ n t) tends in C b [, )) to t e t, and it thus follows that 8 φ n t)dσ n t) e t dσt), 5 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker s Guide, third ed., p. 393, Theorem 1.48. 6 H. L. Royden, Real Analysis, third ed., p. 33, Eercise 16. 7 V. I. Bogachev, Measure Theory, volume II, p., Theorem 8.6.. 8 cf. Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker s Guide, third ed., p. 511, Corollary 15.7. 5
so Let with which hence f) f ) e t dµt) µ σ + f )δ, f) e t dσt). e t dσt) + f ), e t dµt). Because f) 1, dµt) 1, showing that µ is a probability measure. 4 Fourier transforms For a topological space X and a positive Borel measure µ on X, F X is called a support of µ if i) F is closed, ii) µf c ), and iii) if G is open and G F then µg F ) >. If F 1 and F are supports of µ, it is straightforward that F 1 F. It is a fact that if X is second-countable then µ has a support, which we denote by supp µ. 9 Lemma. If µ is a Borel measure on a topological space X and µ has a support supp µ, if f : X [, ) is continuous and fdµ then f) for all X supp µ. Proof. Let F supp µ and let E { X : f) }. E is an open subset of X. Suppose by contradiction that there is some E F, i.e. that E F. Because f is continuous and f) >, there is some open neighborhood G of for which fy) > f)/ for y U. Then G F, so G F and because F is the support of µ, µg F ) > and a fortiori µg) >. Then X fdµ G fy)dµy) G f) f) dµy) µg) >, a contradiction. Therefore E F, i.e. for all F, f). The following lemma asserts that a certain function is nonzero λ d -almost everywhere, where λ d is Lebesgue measure on R d. 1 9 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker s Guide, third ed., p. 44, Theorem 1.14. 1 Ward Cheney and Will Light, A Course in Approimation Theory, p. 91, chapter 13, Lemma 6. 6
Lemma 3. Let 1,..., n be distinct points in R d, let u C n not be the zero vector, and define gy) u j e πij y, y R d. j1 For λ d -almost all y R d, gy). The following theorem gives conditions under which the Fourier transform of a Borel measure on R d is strictly positive definite. 11 Theorem 4. If µ is a finite Borel measure on R d and λ d supp µ) >, then ˆµ : R d C is strictly positive definite. Proof. For distinct 1,..., n R d and for nonzero u C n, u j u k ˆµ j k ) R d u j u k e R πij k) y dµy) d n ) u j e πij y u k e πi k y dµy) j1 k1 u j e πij y dµy) R d j1 gy) dµy). R d It is apparent that this is nonnegative. If it is equal to then because g is continuous we obtain from Lemma that gy) for all y supp µ, i.e. gy) for all y supp µ. In other words, supp µ {y R d : gy) }. But by Lemma 3, λ d {y R d : gy) }), so λ d supp µ), contradicting the hypothesis λ d supp µ) >. Therefore R d gy) dµy) >, which shows that ˆµ is strictly positive definite. 5 Schoenberg s theorem Let X,, ) be a real inner product space. We call a function F : X R radial when y implies that F ) F y). 11 Ward Cheney and Will Light, A Course in Approimation Theory, p. 9, chapter 13, Theorem 3. 7
An identity that is worth memorizing is that for y R, e π e πiy d e πy. R Using this and Fubini s theorem yields, y R d, e π,y e π y. R d e π Lemma 5. For α > and y R d, π ) ) d/ ep π R α α e πi,y d e α y. d Proof. Define T : R d R d by T ) π α, Rd. T ) π α I L Rd ) and J T ) det T ) π d/. α) Let u R d and define f) e π e πi,u. By the change of variables formula, 1 f T ) J T dλ d fdλ d, R d T R d ) and because T is self-adjoint this is π T ) e πi,t u π ) d/ d α R d e and therefore R d π α e π R d e πi,u d, ) ) d/ ep π α e πi,t u d e π u. For u T 1 y) α π y this is π ) ) d/ ep π R α α e πi,y d e α y, d proving the claim. We now prove that on a real inner product space, e α positive definite whenever α >. 13 is strictly 1 Charalambos D. Aliprantis and Owen Burkinshaw, Principles of Real Analysis, third ed., p. 393, Theorem 4.7. 13 Ward Cheney and Will Light, A Course in Approimation Theory, p. 14, chapter 15, Theorem. 8
Theorem 6. Let X,, ) be a real inner product space. If α >, then is radial and strictly positive definite. e α, X, Proof. Let 1,..., n be distinct points in X. There is an n-dimensional linear subspace V of X that contains 1,..., n. By the Gram-Schmidt process, V has an orthonormal basis {v 1,..., v n }. Define T : V R n by T v j e j, where {e 1,..., e n } is the standard basis for R n, which is an orthogonal transformation, and define fu) e α u, u R d. For u C n, u, u j u k e α j k u j u k ep α T j k ) ) u j u k ft j T k ). Now, let µ be the Borel measure on R d whose density with respect to λ d is ) π d/ y ep α) π α y. Because µ is absolutely continuous with respect to λ d, λ d supp µ) >, so Theorem 4 states that the Fourier transform ˆµ : R d C is strictly positive definite. Applying Lemma 5, the Fourier transform of µ is π ) ) d/ ˆµu) ep π R α α y e πi y,u dy e α u fu), d so f is strictly positive definite. Because T is an orthogonal transformation it is in particular one-to-one, so T 1,..., T n are distinct points in R d. Thus the fact that f is strictly positive definite means that u j u k e α j k u j u k ft j T k ) >, which establishes that e α is strictly positive definite. The following is Schoenberg s theorem. 14 14 Ward Cheney and Will Light, A Course in Approimation Theory, p. 11, chapter 15, Theorem 1; René L. Schilling, Renming Song, and Zoran Vondraček, Bernstein Functions: Theory and Applications, p. 14, Theorem 1.14; William F. Donoghue Jr., Distributions and Fourier Transforms, p. 5, 41. 9
Theorem 7 Schoenberg s theorem). Let X,, ) be a real inner product space. If f : [, ) R is completely monotone, f) 1, and f is not constant, then f ), X [, ), is radial and strictly positive definite. Proof. Because f is completely monotone, the Bernstein-Widder theorem Theorem 1) tells us that there is a Borel probability measure µ on [, ) such that ft) e st dµs), t [, ), that is, f is the Laplace transform of µ. Now, the Laplace transform of δ is t 1, and because f is not constant, the Laplace transform of µ is not equal to the Laplace transform of δ, which implies that µ δ. 15 Therefore µ, )) >. Let 1,..., n be distinct points in X and let u C n, u. Then, because n n u ju k, u j u k f j k ) + u j u k u j u k µ{}) 1, ) s) 1, ) s) gs)dµs). ep s j k ) dµs) u j u k ep s j k ) dµs) u j u k ep s j k ) dµs) u j u k ep s j k ) dµs) Assume by contradiction that gs)dµs). Because g, this implies that µ{s [, ) : gs) > }). 16 By Theorem 6, for each s >, u j u k ep s j k ) >, 15 Bert Fristedt and Lawrence Gray, A Modern Approach to Probability Theory, p. 18, 13.5, Theorem 6. 16 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker s Guide, third ed., p. 411, Theorem 11.16. 1
so gs) > when s >. Thus µ, )), a contradiction. Therefore, u j u k f j k ) gs)dµs) >, which shows that f ) is strictly positive definite. 11