If $f \in L^2(R)$, then $\hat{f}\in L^2(R)$?

Question

The Fourier transform of $f$ is as follows: $$\hat{f}(\omega)=\frac{1}{‎‎‎\sqrt{2‎\pi‎}}\int_{-\infty}^{+\infty} e^{-i\omega t}f(t)dt$$ I need to know that if $f \in L^2(R)$, then can we conclude that $\hat{f}\in L^2(R)$? It would be appreciated if someone could help me.

Answer 1

If $f\in L^2(\mathbb{R})$, then $f \in L^1[-R,R]\cap L^2[-R,R]$ for any $0 < R < \infty$. Consequently $$ \hat{f_{R}}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt \in L^2(\mathbb{R}) $$ is a continuous function of $s$, and, by Parseval's identity, $$ \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt\right\|_{L^2}^2=\int_{-R}^{R}|f(t)|^2dt. $$ As $R\rightarrow\infty$, the function $\hat{f_{R}}$ converges in $L^2(\mathbb{R})$ to the definition of $\hat{f}(s)$, and the Parseval identity $\|\hat{f}\|_{L^2}=\|f\|_{L^2}$ holds in the limit.

Answer 2

We need at least $f\in L^1(\mathbb{R})$ so that the Fourier integral $$\hat{f}(\omega):=\frac{1}{‎‎‎\sqrt{2‎\pi‎}}\int_{\mathbb{R}} e^{-i\omega t}f(t)\ dt \tag{*}$$ converges. Not every $L^2(\mathbb{R})$ function is integrable; for example, the function $$ g(x):=(1+x^2)^{-1/2}\in L^2(\mathbb{R}) $$ but $g\not\in L^1(\mathbb{R})$. Thus we cannot define the Fourier transform of a general $L^2$-function directly by means of its Fourier integral $(*)$. But since the Schwartz space $\mathcal{S}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$, we can extend the Fourier transform on $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$. Moreover, we have a theorem which says that the extension maps $L^2$ to itself (see for instance the section Fourier transform on $L^2$ in Stein and Shakarchi's Real Analysis).