# A function is $L^2$-differentiable if and only if $\xi\widehat{f}(\xi) \in L^2$.

In this previous question, I defined $L^p$ derivatives of functions in $L^p(\mathbb{R}^n)$. I've been struggling for a while now to prove the following:

If $f \in L^2(\mathbb{R})$, then $f$ is $L^2$-differentiable if and only if $\xi\widehat{f} \in L^2(\mathbb{R})$, where $\widehat{f}$ is the Fourier transform of $f$. When this happens, then $\widehat{f^{\prime}}(\xi) = 2\pi{}i\xi{}\widehat{f}(\xi)$.

Here's my problem. Say that we're first trying to show that if $f'$ exists, then $\xi{}\widehat{f} \in L^2(\mathbb{R})$. Under this assumption, we know that $\widehat{f'}$ exists and is in $L^2$, so it would suffice to show that it equals $2\pi{}i\xi{}\widehat{f}$. The naive approach would be to compute

$$\widehat{f'}(\xi) = \int_{\mathbb{R}}f'(x)e^{\large -2\pi{}ix\xi}dx = ({2\pi{}ix\xi})\int_{\mathbb{R}}f(x)e^{\large -2\pi{}ix\xi}dx = ({2\pi{}ix\xi})\widehat{f}(\xi),$$ and we're done.

But I believe (and please correct me if I'm wrong!) that there are two fundamental mistakes here:

$1$) The Fourier transform of an $L^2$ function $f$ is not given by the formula above, but rather by a limit of Fourier transforms (in the above formula) of (say) $L^1 \cap L^2$ functions which converge (in $L^2$) to $f$.

$2$) The integration by parts above only holds when the other function, namely $e^{\large -2\pi{}ix\xi}$, has compact support.

Then I tried approximating $f'$ or $f$ by convolutions and the like, but I always arrive at a double integral which does not converge absolutely, and I'd have to interchange the order of integration.

• Hint: Convolution of $f$ with $\frac{1}{\sqrt{2\pi \epsilon}^n}e^{-|x|^2/\epsilon}$ might help.
– Sam
Aug 23 '13 at 19:36
• I don't know why we'd need a compact support to integrate by parts, but the thing that bothers me is referring to the $L^2$ derivative as if it were the pointwise derivative of $f$, for use in the integration-by-parts schema. Aug 23 '13 at 19:38
• @JonathanY. It's not that you need compact support to integrate by parts, it's that that particular formula only holds when the other function has compact support. You get a boundary term that doesn't go to zero if the other function doesn't have compact support. Also, I'm not referring to the $L^2$ derivative as if it were the pointwise derivative. The prime is just a convenient notation.
– L..
Aug 23 '13 at 20:01
• @Sam could you elaborate a little bit further? I did try convolution with very rapidly decaying functions, even ones with compact support, but that really doesn't seem to help since I arrive at a double integral for which Fubini doesn't apply.
– L..
Aug 23 '13 at 20:03
• Another common way to attack these is to multiply the integral by $\exp\{-\epsilon |x|^2/2\}$ and let $\epsilon \to 0$, which may be significantly easier? Because now you have regularized the Fourier frequency you are integrating against. Don't use convolution I guess. This is one way to define the Fourier transform for an $L^2$ function. This way you also don't have to choose whether you are approximating $f$ or $f'$.
– Evan
Aug 23 '13 at 20:15

Let's start with the easier direction. If $\xi \mapsto \xi\cdot\widehat{f}(\xi) \in L^2(\mathbb{R})$, then $f$ is $L^2$-differentiable, and $\widehat{f'}(\xi) = 2\pi i \xi\widehat{f}(\xi)$.

Let $\tau_hf(x) = f(x+h)$. Then $\widehat{\tau_hf}(\xi) = e^{2\pi ih\xi}\widehat{f}(\xi)$, and

$$\frac{\widehat{\tau_hf}-\widehat{f}}{h}(\xi) = \frac{e^{2\pi ih\xi}-1}{h}\widehat{f}(\xi) \to 2\pi i \xi\widehat{f}(\xi)$$

pointwise, and since

$$\left\lvert\frac{e^{2\pi ih\xi}-1}{h} \right\rvert \leqslant 2\pi \lvert \xi\rvert,$$

by the dominated convergence theorem also in $L^2$.

Applying the inverse Fourier transform - which is an isometry of $L^2$ by Plancherel's theorem - we find that

$$\frac{\tau_hf - f}{h} \xrightarrow{L^2} \mathcal{F}^{-1}(2\pi i \xi\widehat{f}),$$

i.e. $f$ is $L^2$-differentiable with $f' = \mathcal{F}^{-1}(2\pi i \xi\widehat{f})$.

Conversely, if $f$ is $L^2$-differentiable, then by the continuity of the Fourier transform we have

$$\frac{\widehat{\tau_hf}-\widehat{f}}{h}(\xi) = \frac{e^{2\pi ih\xi}-1}{h}\widehat{f}(\xi) \xrightarrow{L^2} \widehat{f'},$$

and since the left hand side converges to $2\pi i \xi \widehat{f}(\xi)$ pointwise, we conclude that $\xi \widehat{f}(\xi) \in L^2(\mathbb{R})$.

• Let me see if I understand the last paragraph: convergence in $L^2$ implies the existence of a subsequence that converges pointwise (almost everywhere) to $\widehat{f'}(\xi)$. But since it converges everywhere to $2\pi{}i\xi\widehat{f}(\xi)$, we conclude they are equal a.e., therefore equal as elements of $L^2$?
– L..
Aug 26 '13 at 3:36
• Yep, exactly that. Aug 26 '13 at 8:34

A common way to attack these is to multiply the integral by $g_\epsilon(x) = \exp\{-\epsilon |x|^2/2\}$ and let $\epsilon \to 0$. Now you have regularized the Fourier frequency you are integrating against. This is one way to define the Fourier transform for an $L^2$ function, as now $fg_\epsilon \in L_1 \cap L_2$ with $fg_\epsilon \to f$ in $L^2$. This way you also don't have to choose whether you are approximating $f$ or $f'$.
Integration by parts is now clearly valid, and taking $\epsilon \to 0$ gives you the result.