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.

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.