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I think the solution to this question somehow involves Riesz Representation Theorem, but I don't see how to apply it.

Suppose $g$ is a measurable function on $[0,1]$ such that $\int_0^1 fg~dx$ exists for all $f\in L^2[0,1]$. Then $g\in L^2[0,1]$.

How do I use the above mentioned theorem to show it.

Guy Fsone
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Kuku
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    See here for a nice argument. – t.b. May 06 '12 at 13:56
  • @t.b. Thanks for the link. Could you please help me with the conclusion. I don't quite get how showing that $| f g|_1 \leq M|f|_2$ shows that $g\in L^2[0,1]$. Thanks. – Kuku May 07 '12 at 21:21
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    the point is that this shows that $f \mapsto \int fg$ is a continuous linear functional on $L^2$. The Riesz representation theorem classifies the continuous linear functionals on $L^2$ and shows they can be uniquely represented by an $L^2$-function, so there's an $L^2$ function $h$ such that $\int fg = \int fh$ for all $f \in L^2$. Can you take it from here? – t.b. May 07 '12 at 21:29
  • @t.b. Am I right in saying that $f(g-h)=0$ so that $g=h$? – Kuku May 07 '12 at 21:44
  • Yes, but ... why exactly? :) – t.b. May 07 '12 at 21:51
  • @t.b. Because $\int f(g-h) =0$ – Kuku May 07 '12 at 22:06
  • @kuku $\int f(g-h)=0$ does not imply that $g=h$. For example take $f=\chi_{[0,1]},; g(x)=x, h\equiv 0.5$. Then $\int f(g-h)=0$ but $f\neq g$ as you can see. However $\int |g-h|=0$ indeed implies that $f=g$. Notice that you have $\int f(g-h)=0$ for each $f\in L^2[0,1]$. Can you find $f\in L^2[0,1]$ so that $f(g-h)=|g-h|$? – leo May 08 '12 at 15:02
  • @leo: I don't get your point. Do you mean $\int|g-h|=0$ implies $g=h$? – Kuku May 09 '12 at 03:40
  • @kuku You have that for all $f\in L^2[0,1]$, $\int f(g-h)=0$. It is true that $\int |g-h|=0$ implies $g=h$ almost everywhere. So, it is enough to find $f\in L^2[0,1]$ so that $f(g-h)=|g-h|$ in order to conclude that $g=h$ almost everywhere. – leo May 09 '12 at 06:17
  • @leo: then $f$ would have to 1. right? – Kuku May 09 '12 at 17:05
  • Not exactly. For $x\in\mathbb R$ define $$\operatorname{sign}(x)=\begin{cases} 1 &\text{if } x\gt 0\ 0 &\text{if } x=0\ -1 &\text{if } x\lt 0\end{cases}$$ You can prove that $\operatorname{sign}(x)\cdot x=|x|$. Now, I ask you, can you find $f\in L^2[0,1]$ so that $f(g-h)=|g-h|$? – leo May 09 '12 at 19:21
  • And yes would have to be $1$ but in the right set. You can argue as follows: let $f_1=\chi_{{x\in [0,1]:g(x)\gt h(x)}}$. Then $\int f_1^2\leq \int_0^1 dx=1$, so $f\in L^2$, then $\int f_1(g-h)=0$. Similarly if $f_2=-\chi_{{x\in [0,1]:g(x)\lt h(x)}}$ you can see that $f_2\in L^2$ etc. If you consider $f=f_1+f_2$, then $\int f(g-h)=0$ is equivalent to $\int |g-h|=0$, therefore $g=h$ almost everywhere. – leo May 09 '12 at 19:33

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