Convex sequences and Integral representation for the generating function 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.
 A: 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.
