# The Fourier transform $\hat{f} \in L^1$ if both $f\in L^2$ and $f'\in L^2$

There is a problem in Folland's Fourier analysis and its applications(page 224, exercise 7)

Suppose that $f$ is continuous and piecewise smooth, then $\hat{f}\in L^1$ if both $f\in L^2$ and $f'\in L^2$

The hint he gives is that

first show $\int (1+\xi^2)|\hat{f}(\xi)|^2d\xi$ is finite and then use the Cauchy-Schwarz inequality.

However, when I try to prove that $\int (1+\xi^2)|\hat{f}(\xi)|^2d\xi$ is finite, there is a problem concerning about the derivative of Fourier transform. We know $\widehat{f'\;}(\xi)= i\xi\hat{f}(\xi)$ holds when $f'$ is integrable but he didn't assume that $f' \in L^1$.

So do we really need this extra condition or there is another way to approach this problem without assuming that $f'\in L^1$?

• It may help to think in terms of distribution theory: in this sense $\hat{f'}=i \xi \hat{f}$ holds in the sense of distribution theory as long as $f'$ is just a tempered distribution. – Ian Feb 9 '17 at 23:40
• @lan Thank you! But is there an approach without invoking distribution theory for this problem? I actually know nothing about distribution theory, and folland didn't talk about distribution theory as far as in this chapter – No One Feb 9 '17 at 23:44
• I dunno, I always have trouble dealing with the FT of functions not assumed to be $L^1$ a priori without using distribution theory. I know there are some tricks that can be pulled in the $L^2$ case, though (as the FT also acts nicely on $L^2$). The gist of these tricks is $L^2$-norm approximation of an $L^2$ function by a sequence of $L^1 \cap L^2$ functions. – Ian Feb 9 '17 at 23:44

Assume that $f$ is locally absolutely continuous and $f,f'\in L^2(\mathbb{R})$. Then $ff' \in L^1$, which gives the existence of the following limits $$\lim_{x\rightarrow\pm\infty}\int_{0}^{x}f(t)f'(t)dt=\lim_{x\rightarrow\infty}f(x)^2-f(0)^2.$$ The limits $\lim_{x\rightarrow\pm\infty}f(x)^2$ must be $0$ because $f$ is absolutely integrable.
The Plancherel Theorem gives \begin{align} \widehat{f} & = L^2\mbox{-}\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(x)e^{-i\xi x}dt \\ \widehat{f'} & = L^2\mbox{-}\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f'(x)e^{-i\xi x}dx. \end{align} An $L^2$ convergent sequence has a subsequence that converges pointwise a.e.. So there is a sequence $\{ R_n \}$ converging to $\infty$ such that the following holds for a.e. $\xi\in\mathbb{R}$: \begin{align} \widehat{f'}(\xi)&=\lim_{n}\frac{1}{\sqrt{2\pi}}\int_{-R_n}^{R_n}f'(x)e^{-i\xi x}dx \\ &= \lim_n \frac{1}{\sqrt{2\pi}}\left[f(x)e^{-i\xi x}|_{x=-R_n}^{R_n}+i\xi\frac{1}{\sqrt{2\pi}}\int_{-R_n}^{R_n}f(x)e^{-i\xi x}dx\right]. \end{align} Therefore, the following holds a.e., and the limit exist pointwise a.e. : $$\widehat{f'}(\xi)=\lim_n\frac{i\xi}{\sqrt{2\pi}}\int_{-R_n}^{R_n}f(t)e^{-i\xi t}dt = i\xi\hat{f}(\xi).$$ And that's what you need to do what you want.
Once you know that $\widehat{f'}(\xi)=i\xi\hat{f}(\xi)$, then $\xi\hat{f}(\xi)\in L^2$ by Plancerel's Theorem, and \begin{align} \int_{-\infty}^{\infty} |\hat{f}(\xi)|d\xi &= \int_{-\infty}^{\infty}\frac{1}{\sqrt{1+\xi^2}}(\sqrt{1+\xi^2}|\hat{f}(\xi)|)d\xi \\ &\le \left( \int_{-\infty}^{\infty}\frac{1}{1+\xi^2}d\xi\right)^{1/2} \left(\int_{-\infty}^{\infty}(1+\xi^2)|\hat{f}(\xi)|^2d\xi\right)^{1/2} \\ & = \sqrt{\pi}\left(\int_{-\infty}^{\infty}|f(x)|^2+|f'(x)|^2 dx\right)^{1/2} < \infty. \end{align}