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.
Please help me!
 A: 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 $\xi \widehat{f}(\xi) \in L^2(\mathbb{R})$.
A: 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.
