Can integration of polynomial equals the p'(0)? Im trying to solve the following problem (don't know whether is it true):
There exist a complex measure $\mu$ on $[0, 1]$ such that $p'(0) = \int p \, d\mu$ for all polynomial $p$.
In fact, the question is originally restricted to polynomial of degree $\leq n$. In this case, I can solve it. In fact, if $M:=$ { a polynomial of degree $\leq n$}, then the linear functional $f(p) = p'(0)$ is a bounded linear functional (domain is of finite dimension). And applying the Hahn-Banach, we obtain the bounded linear functional $F$ on $C([0, 1])$ which is an extension of $f$. Since the space of regular complex measures and the $C([0, 1])^*$ are the same, so such measure $\mu$ exists.
However in general case, the functional $f$ is not bounded, so we cannot extend it. And I guess that the statement is false, but I have no idea to show this formally.

There exists a polynomial $q_n$ such that $|q_n(x)-\frac 1  {\sqrt {x+\frac 1n}}| <\frac 1 n$ for all $x$. Integrating this we get $|2\sqrt {x+\frac  1 n} -p_n(x)|  <\frac 1 n$ for all $x$ where $p_n(x)=\int_0^{x} q_n(t)dt$. Note that $q_n(0)> \sqrt n -\frac  1n \to \infty $. Now $$|2\int \sqrt x d\mu(x)| =\lim |\int 2  {\sqrt {x+\frac 1n}} d\mu (x)|>|\int p_n(x)d\mu (x)| -\frac 1 n \|\mu\|$$ $$=|p_n'(0)|+\frac 1 n=q_n(0)+\frac  1n $$, a contradiction.
