So I was trying to solve the following implication in the hausdorff moment problem
Let $\{s_n\}_{n \geq 0}$ be a sequence of real numbers such that:
$\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}s_{n+i} \geq 0, \forall n,m \geq 0$ (1)
Prove that there exists a Borel measure $\mu$ on $[0,1]$ such that
$s_n = \int_{[0,1]}x^nd\mu$
My Attempt:
Let $P$ be the space of all polynomial on $[0,1]$ equiped with the supremum norm. Define a linear functional $\Phi$ on this space by defining $\Phi(x^k) = s_k$ Now assuming $\Phi$ is bounded on $P$ ,this gives that $\Phi$ is uniformly continous and as $P$ is dense in $C([0,1])$ $\Phi$ can be extended uniquely to a continous functional on $C([0,1])$.
$\Phi$ is linear on $C([0,1])$ follows as it is linear on $P$.If $f \in C([0,1])$ is non negative then consider the sequence of bernstein polynomials $B_Nf$
$B_Nf(x) = \sum_{k=0}^{N}f(\frac{k}{N})\binom{N}{k}x^k(1-x)^{N-k}$
So $\Phi(B_Nf) = \sum_{k=0}^{N}f(\frac{k}{N})\binom{N}{k}\Phi(x^k(1-x)^{N-k})$
By (1) it follows that $\Phi(B_Nf) \geq 0$.As $B_Nf$ converges to $f$ in $C([0,1])$, by continuity of $\Phi$ it follows that $\Phi(f) \geq 0$.So $\Phi$ is a positive linear function on $C[0,1] = C_c([0,1])$ and the conclusion follows by Reisz Representation theorem.
The problem:
How does one shows that $\Phi$ is bounded on $P$ ?