Uniform convexity and polynomials question

Question

I wonder if someone can help me with the following question:

"Let $f \in C([0,1])$ and assume that there is a positive constant $C$ such that $\left|\int_0^1 p'(t)f(t) dt\right|\leq C \lVert p \lVert_2$ for all polynomials $p$ where $\lVert p \lVert_2^2 = \int_0^1 |p(t)|^2dt$. Show that there exists a unique $g \in L^2([0,1])$ such that $f(x)=\int_0^x g(t)dt, \int_0^1 g(t)dt=0$."

I know there's a theorem saying that if $X$ is a uniformly convex Banach space, then if $K$ is a closed convex subset for every $x \in X$ there is a unique point $z \in K$ that minimizes the distance between $x$ and $K$.

I can see that the set $K=\left\{ g \in L^2([0,1]), f(x)=\int_0^x g(t)dt, \int_0^1 g(t)dt=0\right\}$ is a convex set, but I'm not sure when it comes to closed?

I was thinking that if $\{g_k\} \in K$ and $\{\alpha_k\}$ is a sequence s.t $\sum_{k=1}^{\infty}\alpha_k =1$ then $\sum_{k=1}^{\infty}\alpha_k g_k(t)$ satisfies all the conditions for belonging to $K$ apart from being in $L^2([0,1])$ (which I havent succeeded to verify).

Then I'm not sure how to use the fact that $\left|\int_0^1 p'(t)f(t) dt\right|\leq C \lVert p \lVert_2$.

If anyone could help me I'd be very grateful!

Answer 1

Here is the idea.

Consider the functional $F(p)=-\int_0^1p'(t)f(t)\,dt$ on the space of all polynomials.

1. Since $|F(p)|\le C\|p\|_2$ it can be extended uniquely from polynomials to the whole $L_2[0,1]$ by continuity as polynomials are dense there. Call the extended functional $\hat F$.
2. By Riesz representation theorem there exists a unique function $g\in L_2[0,1]$ such that $$ \hat F(h)=\int_0^1h(t)g(t)\,dt,\qquad \forall h\in L_2. $$
3. Restricting $\hat F$ to $C_0^\infty[0,1]$ and using the fact that any smooth $h$ can be uniformly approximated with its first derivatives by polynomials, we will get that $$ \hat F(h)=-\int_0^1 h'(t)f(t)\,dt=\int_0^1 h(t)g(t)\,dt,\quad\forall h\in C_0^\infty. $$ Now the variant of Du Bois-Reymond lemma gives that $f'=g$.
4. Finally integration by parts for $h\in C^\infty$ $$ f(1)h(1)-f(0)h(0)=\int_0^1h'(t)f(t)\,dt+\int_0^1h(t)f'(t)\,dt=0. $$ Taking $h(0)=0$, $h(1)=1$ and then $h(0)=1$, $h(1)=0$ gives that $f(0)=f(1)=0$.