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In Introduction to Fourier Analysis on Euclidean Spaces, the authors explore the $L^2$ extension of the Fourier transform and argue that it is onto $L^2(\mathbb{R})$ but I can't follow their reasoning. Here is their proof from page 17:

Theorem 2.3. The Fourier transform is a unitary operator on $L^2(\mathbb{R}^n)$.

Proof: Since $\mathcal{F}$ is an isometry its range is a closed subspace of $L^2(\mathbb{R}^n)$. If this subspace were not all of $L^2(\mathbb{R}^n)$, we could find a function $g$ such that $\int_{\mathbb{R}^n}\hat{f}gdx = 0$ for all $f\in L^2(\mathbb{R}^n)$ and $||g||\neq 0$. The multiplication formula [Parseval's theorem] obviously extends to all of $L^2$ and so we have $\int_{\mathbb{R}^n} f\hat{g}dx = \int_{\mathbb{R}^n} \hat{f}gdx = 0$. But this implies that $\hat{g} = 0$ almost everywhere, contradicting the fact that $||g|| = ||\hat{g}|| \neq 0$.

I understand that the range is closed from the closed range theorem but I fail to understand how such a $g$ exists. I'm sure it's something simple that I'm missing but I don't see it. Everything else follows quite logically from there, but I'm stuck on this one fact. Would anyone mind clearing this up for me?

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Cameron Williams
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    Any point perpendicular to the closed subspace will do. – copper.hat May 01 '13 at 18:43
  • Hi Williams, I have a question. The Parseval's theorem(or alternatively, the Plancherel Theorem), only tells us that $\int_{\mathbb{R}^n} f gdx = \int_{\mathbb{R}^n} \hat{f} \hat{g}dx $. Why we can have that $\int_{\mathbb{R}^n} f\hat{g}dx = \int_{\mathbb{R}^n} \hat{f}gdx$ as in your original post? Thanks. – Sam Wong Mar 01 '20 at 10:26

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It's standard Hilbert space theory. More generally, let $H$ be a Hilbert space and $K$ a closed subspace. If $K\ne H$ then the orthogonal complement $K^\perp$ is non-trivial.

Harald Hanche-Olsen
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