Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
See here for a nice argument. – t.b. May 6 '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 7 '12 at 21:21
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 7 '12 at 21:29
@t.b. Am I right in saying that $f(g-h)=0$ so that $g=h$? – Kuku May 7 '12 at 21:44
Yes, but ... why exactly? :) – t.b. May 7 '12 at 21:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.