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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.

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up vote 1 down vote accepted

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 $f\equiv -G$
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Thanks alot I see I had a typo in my try. And shouldnt it be -$\int_0^1p(t)G'(t)dt = \int_0^1p(t)g(t) dt $ and $G(1) = 0$ is given from the beginning. –  Johan Dec 29 '12 at 8:34
    
@Johan No, $G(1)=0$ is not given. This is something you are told to prove in the sentence that begins with "Show that..". (note that the formula $\int_0^1 g=0$ is a part of that sentence). When you are asked "show that there exists a unicorn with five legs", it does not mean you can assume that unicorns have five legs. It means your duty is to not only to produce a unicorn, but one that has five legs. No four-legged unicorns will be accepted. –  user53153 Dec 29 '12 at 8:44
    
Sorry, I can not read what I'm typing, u are correct! –  Johan Dec 29 '12 at 8:54
    
So the last things following by putting $p(t) =1 $? –  Johan Dec 29 '12 at 9:49
    
@Johan Yes, that's right. –  user53153 Dec 29 '12 at 15:55
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