Krein condition and the Hilbert transform

Krein condition have been used as a qualitative result to show the M-indeterminacy of some kind of densities. In this work we use results from the theory of the Hilbert transform to construct families of densities having all the same finite moment sequence as a density $f$ with finite logarithmic integral. Actually, our approach explicitly gives Stieltjes classes with center at $f$ and perturbations involving the Hilbert transform of $\ln f.$ We consider densities supported on the whole real line or the positive half line.


Introduction
Let F be a distribution supported on I = R + or R such that I x n dF (x) < ∞ for all n ≥ 1.
Under this assumption we say that F has a finite moment sequence on I. A distribution F with finite moment sequence on I is called M -indeterminate if there are other distributions supported on I having the same moments as F .
In 1945 Krein proved that if F is an absolutely continuous distribution on R with finite moment sequence whose density f has finite logarithmic integral, i.e. (1) then F is M -indeterminate. This is the so-called Krein criterion.
About the Krein criterion, in [6] the authors say that it "is a qualitative result; there is no indication of how to write other distributions with the same moments as F ". In [5,Theorem 1] the author used the theory of the Hardy space on the upper half plane H 1 to get a simple proof of the Krein criterion. In fact, if f is a density satisfying the Krein condition (1), the author proved the existence of a density g having the same moment sequence as f . In this work we go a step further, we combine the ideas in the proof of Theorem 1 in [5] with some results of the Hilbert transform and the space H 1 , to obtain an explicit description of the latter density g.
Actually, in this setting, we get a family of densities having the same moment sequence as f . To do this, we consider a construction introduced in [8] to exhibit some densities with the same moment sequence.
Let f be a density with finite moment sequence on I. Assume that there exists a bounded measurable function h with sup x∈I |h (x)| ≤ 1, such that and the function f h is not identically zero, then the Stieltjes class S I (f, h) with center at f and perturbation h is given by When I = R + we have a similar result.
Theorem 2. Let f be a density on R + with finite moment sequence. If f satisfies the condition This work is organized as follows. In the next section we give some facts about the Hilbert transform and compute the Hilbert transform of two important cases. In the last section we prove the results and analyze two examples to show the usefulness of our approach.

Preliminaries
The following results can be found in [4, pages 60-65]. Suppose that the function u : R → R satisfies Hence the following integral converges absolutely on H := {z ∈ C : ℑz > 0} and defines an analytic function on H. Notice that U is the Poisson integral of u and is the unique harmonic extension of u to H. Moreover, U is the unique conjugate harmonic function of U such that U (i) = 0.
It is known the existence of the non-tangential limits of U and U at almost t ∈ R; the non-tangential limit of U is u, and the non-tangential limit of U is called the Hilbert transform of u and is denoted by Hu. The Hilbert transform of u can be written as the principal value of a singular integral: Remark 3. a) If u is an even function satisfying (3) then In particular, Hu is an odd function on R. We also can see that (3) and Let µ ∈ (0, 1). From Remark 3 and the previous case we obtain Proof. From Remark 3 we have and it is sufficient to consider t > 0. Now, from the identity we get for ε > 0 small enough that then we use that arctanh(x) = 2 −1 ln 1+x 1−x , |x| < 1, and apply the Weierstrass M-test considering ε ∈ [0, ε 0 ) with ε 0 small enough, to obtain , t > 0.

Proof of the results
Proof of Theorem 1. Since ln f ≤ f , condition (1) is equivalent to ln f ∈ L 1 (dt/(1+ t 2 )), so we can set u = ln f and proceed as at the beginning of Section 2: consider the holomorphic function F (z) = U (z) + i U (z) on H, where U is the harmonic extension of u to H, and U is the unique conjugate harmonic function of U satisfying U (i) = 0. Now we introduce the function G ∈ hol(H) given by By using the Jensen's inequality we get that Thus G ∈ H 1 and Theorem 3.1 in [2, page 55] implies that there exists a function g ∈ L 1 (R) such that y) , a.e. x ∈ R. By the other hand, we have In particular, we notice that g has finite moments of all nonnegative orders.
By Lemma 3.7 in [2, page 59] we have We set t = 0 to get Since f is a density and |g| = f a.e. on R, it follows that at least one of the functions ℜg, ℑg is a nonzero function. From (5) and (6) we get that cos(H ln f ) and sin(H ln f ) are perturbations for Stieltjes classes with center at f . Example 6. Odd powers of the normal distribution. Let X be a random variable with X ∼ N (0, 1 2 ), then X 2n+1 , n ≥ 1, has the density f n (x) := 1 (2n + 1) √ π |x| −2n/(2n+1) exp(−|x| 2/(2n+1) ), x ∈ R.
Proof of Theorem 2. We set f * (x) = |x|f (x 2 ), x = 0. Clearly f * is a density on R and verifies The hypothesis about f imply that f * has a finite moment sequence. Actually, since f * is an even function, the moments of odd order of f * vanish.
Hence f * satisfies the hypothesis in Theorem 1 and we can proceed as in (6) to get for all k ≥ 0. Unfortunaly, we notice the function x 1/2 sin H ln f * (x 1/2 ) is not bounded on R + .

Remark 3 and Lemma 5 imply that
Finally, for t > 0 we have therefore cos H ln f * (t 1/2 ) = sin(H e ln f (t)) is a perturbation for the Stieltjes class with center at f . Example 7. Let X ∼ N (0, 1 2 ). For r > 0 the random variable |X| r has a density supported on R + given by f r (x) := 2 r √ π x 1/r−1 exp(−x 2/r ), x > 0.
This perturbation was also obtained in [1].
Conclusion Krein condition is no longer just a qualitative result to show the M -indeterminacy of a density f but provides families of densities having all the same moment sequence as f .