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

Solution 1:

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 $\xi \widehat{f}(\xi) \in L^2(\mathbb{R})$.

Solution 2:

Transcribing comment to answer:

This doesn't address the original question, but the case you raise concerns about and trying an approximation method.

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.