Uniqueness of for integration functional Let $f\in C([0,1])$ and assume that there exists a positive constant C such that
$\left| \int_0^1p'(t)f(t) dt \right| \leq C\|p\|_2 $
for all polynomials $p$, where $\|p\|^2_2 = \int_0^1 |p(t)|^2 dt$. Show that there exists a unique $g\in L^2([0,1])$ such that
\begin{equation}
f(x) = \int_0^x g(t) dt, \\
\int_0^1 g(t)dt =0
\end{equation}
My try: let $\ell(p) = \int_0^1 p'(t) f(t) ds$ then $$\ell(p) = \int_0^1 p'(t) \left(\int_0^tg(x) dx\right) dt$$
Integrate by parts:
$$\ell(p) = - \int_0^1 p(t)g(t) dt$$
and this unique by Riesz theorem.
Im pretty sure I need some kind of extension here but I cannot see where, please correct me and fill in some blanks.
 A: To define $\ell(p) = \int_0^1 p'(t) f(t)\, ds$ was the right move. I don't understand what happens after that line, though: the second formula for $\ell(p)$ does not seem to agree with the first. 
Uniqueness of $g$ is not so hard: If both $g_1$ and $g_2$ serve the purpose, then $\int_0^x (g_1-g_2)=f(x)-f(x)=0$ for all $x$, which implies [why?] that $g_1$ and $g_2$ are the same element of $L^2$. 
For existence, we certainly need the Riesz representation theorem. But the first step should probably be to extend $\ell$ to a bounded linear functional on all of $L^2$, by Hahn-Banach. Then the Riesz representation theorem gives us $g\in L^2$ such that $\ell(p)=\langle p,g\rangle $ for all $g$. 
You are also correct in that we need integration by parts. On which side of
$$\int_0^1 p'(t)f(t)\,dt = \int_0^1 p(t)g(t)\,dt$$
can we do it? Only on the right, because $f$ is not known to be differentiable. So, introduce $G(x)=\int_0^x g(t)\,dt$ and perform the magic: 
$$\int_0^1 p(t)g(t)\,dt = p(1)G(1) - \int_0^1 p'(t)G(t)\,dt$$
This is what we have so far: 
$$\int_0^1 p'(t)f(t)\,dt  = p(1)G(1) - \int_0^1 p'(t)G(t)\,dt$$
I leave it for you to finish the proof. To-do items: 


*

*show that $G(1)=0$

*show that $f\equiv -G$ 

