# Proof for Fourier transform in $L^2$

This question makes me really confused:

Let $f$ and $g$ two functions in $L^2$. Show that: $$\int \widehat f\cdot gdx= \int f\cdot\widehat gdx,$$ where $\widehat f$ is the Fourier transform for the function $f$.

Notice that I need to prove it in $L^2$. Can anyone please help me?

Thanks,

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Please write in Latex your question. – Davide Giraudo Oct 30 '12 at 22:47
sorry, I`ll try to learn how to do that ASAP. – Danny Oct 30 '12 at 23:36

In this thread, it's shown when $f$ and $g$ are $L^1$ functions. Then $f_n:=f\chi_{\{-n\leq f\leq n\}}$ and $g_n=g\chi_{\{-n\leq g\leq n\}}$ approximate respectively $f$ and $g$ in $L^2$, and are functions of $L^1$. So $$\int_{\Bbb R}f_n\widehat{g_n}dx=\int_{\Bbb R}\widehat{f_n}g_ndx.$$
Yes, and since it's a homework question, I don't give the full answer. The idea is the following: it's not hard to do it when the functions are integrable. Now we try to jump to this case for functions in $L^2$ approximating them by integrable one. – Davide Giraudo Oct 30 '12 at 23:37