A note on analytic continuation of characteristic functions
aa r X i v : . [ m a t h . C A ] S e p A note on analytic continuation of characteristic functions
Saulius Norvidas
Institute of Data Science and Digital Technologies, Vilnius University,Akademijos str. 4, Vilnius LT-04812, Lithuania(e-mail: [email protected])
Abstract
We derive necessary and sufficient conditions for a continuous bounded function f : R → C to be acharacteristic function of a probability measure. The Cauchy transform K f of f is used as analytic continuation of f to the upper and lower half-planes in C . The conditions depend on the behavior of K f ( z ) and its derivatives onthe imaginary axis in C . The main results are given in terms of completely monotonic and absolutely monotonicfunctions. Keywords : Characteristic function; complex-valued harmonic function; Cauchy transform; analytic function; com-pletely monotonic function; absolutely monotonic function.
Mathematics Subject Classification : 30E20 - 60E10
Suppose that σ is a probability measure on R . Let b σ ( x ) = Z ∞ − ∞ e − ixt d σ ( t ) (1.1)be the Fourier transform of σ . In the language of probability theory, f ( x ) : = b σ ( − x ) is called the characteristicfunction of σ (see [4, p. 10]). Any characteristic function f is continuous on R and satisfies f ( − x ) = f ( x ) for all x ∈ R . In particular, such an f is real-valued if and only if it is the Fourier transform of a symmetric distribution σ [4, p. 30], i.e. if σ satisfies σ ( − A ) = σ ( A ) for any measurable A ⊂ R .A function u = u + iu in a domain D ⊂ C is called complex-valued harmonic if both u and u are real harmonicfunctions in D . This means that u and u are twice continuously differentiable on D and satisfy there ∆ u = ∆ u =
0, where ∆ is the Laplace operator ∆ = ∂ ∂ x + ∂ ∂ y . For a function ϕ : R → C , the Dirichlet problem on the complex upper half-plane C + = { z = x + iy ∈ C : y > } is to extend ϕ on C + to a complex-valued harmonic function u ϕ , so that u ϕ ( z ) tends to ϕ ( x ) as z ∈ C + tends to x for each x ∈ R . Let us remember one of the most important case of this problem. If ϕ is a bounded continuousfunction on R , then u ϕ ( x , y ) = π Z ∞ − ∞ y ( x − t ) + y ϕ ( t ) dt = (cid:16) P y ∗ ϕ (cid:17) ( x ) , (1.2)is the unique solution to the Dirichlet problem on C + that is bounded in C + . Here P y ( x ) = π yx + y , y > , (1.3)1 ∈ R , is called the Poisson kernel for C + .In this paper, we will consider the problem of characterizing of conditions for ϕ to be a characteristic functionin terms of its analytic continuation in C \ R . If such a ϕ corresponds to a symmetric distribution on R , then asimilar problem for the harmonic continuation (1.2) was solved in [5]. Here and for later use we need the notion ofcompletely monotonic function (see [7, p.p. 144-145]). A function ω : ( a , b ) → R , − ∞ ≤ a < b ≤ ∞ , is said to becompletely monotonic if it is infinitely differentiable and ( − ) n ω ( n ) ( x ) ≥ x ∈ ( a , b ) and all n = , , , . . . . A function ω : [ a , b ] → R is called completely monotonic on [ a , b ] if it isthere continuous and completely monotonic on ( a , b ) . Theorem 1 [5, Theorem 2].
Suppose that ϕ : R → R is a bounded continuous even function and ϕ ( ) = . Then ϕ is a characteristic function if and only if the functiony → u ϕ ( , y ) is completely monotonic on [ , ∞ ) . Other form of this theorem (in other terms) has been shown by Egorov [1]. Moreover, [1] deals with absolutelyintegrable and infinitely differentiable ϕ satisfying more other conditions.If ϕ is a complex-valued function, then it is easy to see that Theorem 1 fails in general. Indeed, suppose that ϕ ( x ) = ( − α ) e − ix + α e ix , (1.5)where α ∈ R . Since b P y ( ξ ) = e − y | ξ | , ξ ∈ R , for all y >
0, we obtain by straightforward calculation in (1.2) that u ϕ ( , y ) = e − y . Therefore, u ϕ ( , y ) is completelymonotonic on [ , ∞ ) for all α ∈ R . On the other hand, by the Bochner theorem for characteristic function, (1.5) ischaracteristic if and only if α ∈ [ , ] .Here we will extend Theorem 1 for complex-valued characteristic functions. To this end, let us introduce somenotions and basic facts.If u is a complex-valued harmonic function in a domain D ⊂ C , then another complex-valued harmonic function v in D is harmonic conjugate of u if u + iv is analytic in D . Recall that if D = /0, then the harmonic conjugate v of u isunique, up to adding a constant.Let f : R → C be a continuous bounded function. The integral v f ( x , y ) = π Z ∞ − ∞ x − t ( x − t ) + y f ( t ) dt can be chosen as the harmonic conjugate of u f in C + (see, for example, [3, p. 108]). In that case u f + iv f coincideswith the usual Cauchy transform (the Cauchy integral) of fk f ( z ) = i π Z ∞ − ∞ f ( t ) z − t dt . (1.6)Here the integral is absolutely convergent as long as Z ∞ − ∞ | f ( t ) | + | t | dt < ∞ . (1.7)2n particular, this condition is satisfied for any f ∈ L p ( R ) , 1 ≤ p < ∞ , but not in the case of an arbitrary f ∈ L ∞ ( R ) .In general, if f ∈ L p ( R ) , 1 ≤ p ≤ ∞ , then we shall use the followings harmonic conjugate of (1.2) V f ( x , y ) = π Z ∞ − ∞ (cid:16) x − t ( x − t ) + y + tt + (cid:17) f ( t ) dt (see [3, p.p. 108-109]). Denote K f ( z ) = i π Z ∞ − ∞ (cid:16) z − t + tt + (cid:17) f ( t ) dt , (1.8) z ∈ C \ R . We call K f the modified Cauchy transform of f . Both integrals in (1.6) and in (1.8) define analyticfunctions in C \ R = C + ∪ C − (e.g., see [6, p.p. 144-145]), where C − denotes the open lower half-plane in C . Letus write k ( ± ) f ( z ) = k f ( z ) for z ∈ C ± , and K ( ± ) f ( z ) = K f ( z ) for z ∈ C ± .A function ω ( x ) is said to be absolutely monotonic on ( a , b ) if and only if ω ( − x ) is completely monotonic on ( − b , − a ) (see [7, p.p. 144-145]). It is obvious that such a function ω ( x ) can be characterized by the inequalities ω ( n ) ( x ) ≥ , (1.9) n = , , , . . . .The main results are the following theorems: Theorem 2 . Suppose that f : R → C is a bounded continuous function and f ( ) = . Then f is a characteristicfunction if and only if there is a constant a f ∈ R such that:( i ) a f + K (+) f ( iy ) is completely monotonic for y ∈ ( , ∞ ) , and( ii ) − (cid:16) a f + K ( − ) f ( iy ) (cid:17) is absolutely monotonic for y ∈ ( − ∞ , ) . Theorem 3 . Suppose that f is as in Theorem 2 and satisfies (1.7). Then f is a characteristic function if and onlyif:( i ) k (+) f ( iy ) is completely monotonic for y ∈ ( , ∞ ) , and( ii ) − k ( − ) f ( iy ) is absolutely monotonic for y ∈ ( − ∞ , ) . A function ϕ on R is said to be positive definite if n ∑ i , j = ϕ ( x i − x j ) c i c j ≥ x , . . . , x n ∈ R , for every choice of c , . . . , c n ∈ C , and all n ∈ N . By the Bochner theorem (e.g., see[2, Theorem 33.3]), we have that a continuous function ϕ is positive definite if and only if there exists a non-negativefinite measure µ on R such that ϕ ( x ) = b µ ( − x ) .We will need to use later the following lemma: Lemma 1 [2, Theorem 33.10].
Let ϕ be a continuous positive definite function on R such that ϕ ∈ L ( R ) . Then b ϕ is nonnegative, b ϕ is in L ( R ) , and ( ˇ ϕ ) b ( x ) = ϕ ( x ) for all x ∈ R . Here ˇ ϕ ( t ) = π Z ∞ − ∞ e it ξ ϕ ( ξ ) d ξ is the inverse Fourier transform of ϕ . We first prove our Theorem 3.3 roof of Theorem 3. Suppose that f is a characteristic function. Let z = x + iy ∈ C with y =
0. According to (1.7),the integral in (1.6) converges absolutely. Moreover, it is easily checked [6, p.p. 144-145] that d n dy n k f ( iy ) = i π Z ∞ − ∞ ∂ n ∂ y n (cid:16) iy − t (cid:17) f ( t ) dt (2.1)for all m = , , , . . . .Let now z ∈ C + . Then by direct calculation we obtain that Z ∞ x n e − yx e − ixt dx = ( − ) n ∂ n ∂ y n (cid:16) iiy − t (cid:17) . This, together with Bochner’s theorem shows that for any y > t ∈ R . Set ϕ n ( t ) = ( − ) n ∂ n ∂ y n (cid:16) iiy − t (cid:17) f ( t ) . Under the condition (1.7), we have that ϕ n satisfies the hypotheses of Lemma 1. Hence ( − ) n Z ∞ − ∞ ∂ n ∂ y n (cid:16) iiy − t (cid:17) f ( t ) dt = Z ∞ − ∞ ϕ n ( t ) dt = b ϕ n ( ) ≥ m = , , , . . . . Now, by (2.1), we get that the function y → k (+) f ( iy ) satisfies (1.4), i.e., it is completelymonotonic on ( , ∞ ) .In the case where z ∈ C − , we have Z − ∞ | x | n e − yx e − ixt dx = − ∂ n ∂ y n (cid:16) iiy − t (cid:17) for n = , , , . . . . Again, we see that the right side of this equality is continuous and positive definite as a functionof t ∈ R . Applying now Lemma 1 to ϕ n ( t ) = − ∂ n ∂ y n (cid:16) iiy − t (cid:17) f ( t ) , we get as in the previous case that the function y → − k ( − ) f ( iy ) satisfies (1.9). Thus, it is absolutely monotonic on ( − ∞ , ) .Suppose that k (+) f ( iy ) is completely monotonic for y ∈ ( , ∞ ) . By the Bernstein-Widder theorem (see [7, p. 116]),there exists a nonnegative (not necessarily finite) measure µ supported on [ , ∞ ) such that k (+) f ( iy ) = Z ∞ e − yt d µ ( t ) , (2.2)where the integral converges for all y ∈ ( , ∞ ) . This means that for any given τ >
0, the function u τ ( iy ) : = k (+) f ( i ( y + τ )) (2.3)is completely monotonic for y ∈ [ , ∞ ) . Now, there is a finite nonnegative measure µ τ on [ , ∞ ) such that u τ ( iy ) = Z ∞ e − yt d µ τ ( t ) , y ∈ [ , ∞ ) . Any such u τ can be continued analytically to C + as the Laplace transform of finite µ τ . According to(2.3), we have that the Laplace transform of µ L µ ( z ) = Z ∞ e izt d µ ( t ) (2.4)4s well-defined and is also analytic in C + . With equation (2.2) in mind, the applications of the uniqueness theoremfor analytic functions (1.6) and (2.4) in C + yields i π Z ∞ − ∞ f ( t ) z − t dt = k f ( z ) = k (+) f ( z ) = Z ∞ e izt d µ ( t ) , (2.5) z ∈ C + .Let y → − k ( − ) f ( iy ) be absolutely monotonic on ( − ∞ , ) . By definition, the function − k ( − ) f ( − iy ) is completelymonotonic for y ∈ ( , ∞ ) . Using the same argument as before, we have that there exists a nonnegative measure η on ( − ∞ , ] such that k f ( z ) = k ( − ) f ( z ) = − Z − ∞ e izt d η ( t ) , (2.6)where the integral is absolutely convergent for each z ∈ C − .Fix y >
0. Using (2.5) and (2.6), we get (cid:16) P y ∗ f (cid:17) ( x ) = h k f ( x + iy ) − k f ( x − iy ) i = Z ∞ − ∞ e ixt d ϑ y ( t ) , (2.7)where ϑ y ( t ) = e −| y | t h µ ( t ) + η ( t ) i . Since the integrals (2.4) and (2.6) are absolutely convergent, it follows that ϑ y is a finite measure on R . Now,applying the Bochner theorem to (2.7), we see that ( P y ∗ f )( x ) is continuous positive definite function on R . Recallthat the family of Poisson kernels ( P y ) y > form an approximate unit in L ( R ) (e.g., see [3, p. 111]). Hencelim y → (cid:16) P y ∗ f (cid:17) ( x ) = f ( x ) for any point x = x of continuity of f . According to the fact that the pointwise limit of positive definite functionsalso is a positive definite, we have that f is positive definite on R . This proves Theorem 3. Proof of Theorem 2 . Fix z ∈ C \ R . Since the modified Cauchy kernel γ ( z ; t ) = i π (cid:16) z − t + tt + (cid:17) is an integrable function, it follows that its Fourier transform is well-defined and is also continuous function. Let ζ A denote the indicator function of a measurable subset A ⊂ R . If z = iy , y =
0, then by direct calculation we obtainthat b γ ( iy ; − x ) = Z ∞ − ∞ γ ( iy ; t ) e ixt dt = ( ζ [ , ∞ ) ( x ) e − yx − sign ( x ) e −| x | , if x = , , if x = , (2.8)for y >
0, and b γ ( iy ; − x ) = ( − ζ ( − ∞ , ] ( x ) e − yx − sign ( x ) e −| x | , if x = , − , if x = , (2.9)for y < f is a characteristic function. Then f ( t ) = Z ∞ − ∞ e ixt d σ ( x ) , (2.10)where σ is a probability measure on R . Since the function (2.8) is positive for small y and is negative for large y ,we have that γ ( iy ; t ) is not necessary positive definite as a function of x ∈ R in general. Therefore, we cannot applyLemma 1, as in the proof of Theorem 3. But, on the other hand, both (2.8) and (2.9) are integrable functions for x ∈ R . 5et y >
0. Substituting (2.10) into (1.8), applying (2.8) and using Fubini’s theorem, we get K f ( iy ) = Z ∞ − ∞ γ ( iy ; t ) f ( t ) dt = Z ∞ − ∞ b γ ( iy ; − x ) d σ ( x ) = Z ∞ e − yx d σ ( x ) − a f , where σ denotes the nonnegative finite measure 2 σ · ζ [ , ∞ ) − σ { } , while a f = Z ∞ − ∞ sign ( x ) e −| x | d σ ( x ) . (2.11)Here σ { } is the measure σ of the one-point set { } . So by the Bernstein-Widder theorem [7, p. 116], the function K f ( iy ) + a f is completely monotonic for y > y <
0, then combining (1.8), (2.9), and (2.10) we obtain − K f ( iy ) = Z − ∞ e − yx d σ ( x ) + a f with σ = σ · ζ ( − ∞ , ] − σ { } and a f defined in (2.11). Finally, it is easy to verify by straightforward calculation ofderivatives that the function − (cid:16) a f + K ( − ) f ( iy ) (cid:17) = Z − ∞ e − yx d σ ( x ) satisfies (1.9) for y ∈ ( − ∞ , ) .The sufficiency can be proved in a manner similar to the proof of the sufficiency of Theorem 3. References [1] A.V. Egorov, On the theory of characteristic functions, Russian Math. Surveys 59(3) (2004) 567-568.[2] E. Hewitt, K.A. Ross, Abstract Harmonic Analysis, vol. 2, Springer, Berlin-Heidelberg, 1997.[3] P. Koosis, Introduction to H pp