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Suppose that $c_k$ is an decreasing sequence of non-negative real numbers, such that $c_0=1$ and $c_{k}\leq \frac{1}{2}(c_{k-1}+c_{k+1})$.

Is it true that the generated function of $c_k$ admit an integral representation as below $$ \sum_{k=0}^{\infty}c_kz^k=\int_{\partial\mathbb D} \frac{1}{1-\zeta z}d\mu(\zeta), $$ where $\mu$ is a Borel Probability Measure in $\partial\mathbb D$ ?

Motivation: This question is related a possible slight different solution for the question asked in https://math.stackexchange.com/questions/2188/complex-analysis-question whose the answer, as pointed out by damiano, can be found at the IMC website.

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Leandro
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1 Answers1

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I think this is not possible. We need a stronger condition on the $c_k$.

Assume such a measure exists, then $L:C(S^1) \to \mathbb{C}$ given by $$L(f) = \int_{S^1} f \, d\mu$$ is a linear functional so we can extend it to $L^2(S^1)$. To see it is continuous note that

$$|L(f)| \leq \|f\|_1 \leq \|f\|_2$$

Also write this extension as $L$.

By Riesz representation theorem we know that $L(f) = \langle f, x \rangle$ for some $x \in L^2(S^1)$.

So $L(\sum \zeta^n z^n) = \sum b_n z^n$ where $b_n = \langle \zeta^n, x \rangle$ and we know that this is equal to $\sum c_n z^n$. So that $c_n = \langle \zeta^n, x \rangle$. But we also know that $\sum |c_n|^2 < \infty$. This is a stronger condition than convexity as George pointed out below.

I hope it is correct this time.

The problem looks a lot like the Hausdorff moment problem and the standard work on moment problems is probably the book by Akhiezer: The classical moment problem and some related questions in analysis. Hafner Publishing Co., New York 1965.

JT_NL
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  • Hi Jonas, I think you argument is fine to find finite borel measures in $\partial \mathbb D$. But for probability borel measures (that it is the case here) I am not sure. Unfortunately using your integral representation we are not able to solve the problem I linked in the question, because in the last inequality (in the argument I gave there) we will get the total variation of the measure and in this case it is not clear to me how to compute it. Anyway thanks to think about the problem. – Leandro Aug 23 '10 at 00:07
  • Okay, I see. By the way, I still didn't give an argument why $L$ is bounded, but the convexity does that (I think). I will think about it a little more, probably it is fixable one way or the other. Do you know of the existence of the Hausdorff Moment Problem? Your problem made me think of that. – JT_NL Aug 23 '10 at 11:32
  • Jonas, $c_n=1/\sqrt{n}$ is convex, but $\sum_nc_n^2=\infty$. – George Lowther Aug 23 '10 at 22:36
  • Also, you need to show that L is positive, otherwise the Riesz representation theorem only gives a signed measure. – George Lowther Aug 23 '10 at 22:38
  • Oops, you're right about the convexity, I hope it can fix that it is still bounded. You're also right on the positivity. Now I only get a complex valued measure. I hope it is fixable.

    Sorry for all the corrections, I thought I give it a shot since it was open for several days already, I will think about it some more and if it is not fixable I'll delete it.

     – JT_NL Aug 23 '10 at 23:38
  • Should have said, $c_n=1/sqrt{1+n}$, which is convex and $c_0=1$, $\sum_nc_n^2=\infty$. – George Lowther Aug 24 '10 at 00:16
  • No I don't knew the Hausdorff Moment Problem. Thank you for the link. In Mathoverflow, I made similar question and I get there a link for the trigonometric moment problem. But the criterion you are pointing out, it seems more computable than I found in the link I told you. There to solve the problem is required to shown that some Toeplitz matrix is positive semi-definite ( perhaps they are similar, I did not make the calculations. ) – Leandro Aug 24 '10 at 02:24
  • Hey guys, I made a new attempt, this time I think it is not possible. I hope you're not tired yet of checking. – JT_NL Aug 24 '10 at 18:48