A signed version of the Hausdorff moment problem This is a follow-up to this question. Let $C([0, 1])$ denote the space of real-valued continuous functions, let $\mathbb R^{\infty}$ denote the space of all real sequences, and consider the mapping of $f\in C([0, 1])$ to $(\mu_n)\in \mathbb{R}^\infty$, where 
$$\tag{1}
\mu_n(f):=\int_0^1 s^n f(s)\, ds.$$
(The letter $\mu$ stands for "moment"). As we discovered in the linked question, this mapping is not surjective, because $\mu_n(f)$ must satisfy the bound 
$$\tag{2}
|\mu_n|\le \frac{C}{n+1},$$
where $C=\sup\left\{ |f(s)|\ :\ s\in[0,1]\right\}$. 

Question. Suppose that $(\mu_n)\in \mathbb R^{\infty}$ is a sequence that satisfies the bound (2), for some constant $C>0$. Does there exist $f\in C([0, 1])$ such that $\mu_n=\mu_n(f)$, as in (1)?

This is similar to the Hausdorff moment problem, but with signed continuous functions, not just positive ones, in place of positive probability measures.

EDIT. The above is not the right question to ask, as mathworker21 clearly shows in his answer. The right question involves the mapping 
$$\tag{3}
\mu_n(m):=\int_0^1 s^n dm(s), $$
where $m$ is a signed Borel measure on $[0,1]$. This sequence needs not satisfy (2); take, for example, $m(x) =\delta(x-1)$. Instead, it satisfies $$\tag{4} \lvert \mu_n \rvert \le C, $$
where $C$ equals the total variation of $m$. 

Refined Question. Suppose that $(\mu_n)\in\mathbb R^\infty$ is a sequence that satisfies (4) for some constant $C>0$. Does there exist a signed Borel measure $m$ such that $\mu_n=\mu_n(m)$, as in (3)?

I have noticed that the same question has been already asked here, without response. 
My feeling is that the answer to the "Refined question" is affirmative, but that the proof is not very easy. I think that the method of proof of the Hausdorff moment problem can be adapted to the present case, but I don't know enough to do this adaptation myself. 
In case I do not get any feedback, I will accept mathworker21 answer, which fully solves the original question.
 A: I think the answer is "no". The point is that the map $f \mapsto (\mu_n(f))_n$ should usually be injective. Let $g(x) = 1_{[0,\frac{1}{2}]}(x)$. Then since $||g||_\infty \le 1$, we still have $\mu_n(g) \le \frac{1}{n+1}$ for each $n$. Now suppose there is some continuous function $f$ so that $\int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$. Then $\int_0^1 p(x)[f(x)-g(x)]dx = 0$ for any polynomial $p$, so $\int_0^1 \phi(x)[f(x)-g(x)]dx = 0$ for any $\phi \in C^\infty([0,1])$, meaning $f(x) = g(x)$ a.e.. But then $f$ can't be continuous, contradiction.
A: No
The two Banach spaces $l^\infty(\mathbb Z)$ and $M[0,1]$ are not isomorphic.  [reference needed]
Here,
$l^\infty(\mathbb Z)$ is the space of all bounded sequences $\big(\mu_n\big)_{n \in \mathbb Z}$ with norm $\sup\{|\mu_n| : n \in \mathbb Z\}$ 
$M[0,1]$ is the space of all finite signed measures on $[0,1]$ with variation norm.
Now the moment map $M[0,1] \to l^\infty(\mathbb Z)$ is linear and (by the closed graph theorem) continuous.  But since those two spaces are not isomorphic, either the moment map is not injective or it is not surjective.
