Weak derivatives in $L^2$ are derivatives in norm(!)

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I noticed something that strikes me as very curious regarding "weak derivatives" in $L^2$. The proof is so simple that the result is really just an "observation". But it's relevant to the definition of Sobolev spaces, so I'm wondering why I've never seen it. Of course I know nothing about Sobolev spaces, hence the question: is the result below actually out there somewhere?

$\newcommand\D{\mathcal D}$

If $f$ is a function on the line and $h\in\Bbb R$ define $$\tau_hf(t)=f(t+h).$$ Here on MSE we see people write $u(t+h)$ for distributions $u$, which is irritating; of course the right way to define a translate of $u\in\D'$ is $$\langle\tau_hu,\phi\rangle=\langle u,\tau_{-h}\phi\rangle\quad(\phi\in\D(\Bbb R)).$$

For $h\ne0$ define $$\Delta_hf=\frac{\tau_hf-f}{h},$$and similarly for $u$.

If $u\in \D'$ write $D_wu$ for the distribution-wise derivative. If $f,g\in L^2$ say $g=D_{L^2}f$ if $$||g-\Delta_hf||_2\to0\quad(h\to0).$$

Easy Lemma. If $u\in\D'$ then $\Delta_hu\to D_wu$ in $\D'$ as $h\to0$.

I believe this is just an exercise in Rudin Functional Analysis.

Proof. If $u\in\D'$ the definition of $\tau_hu$ shows that $$\langle\Delta_hu,\phi\rangle=-\langle u,\Delta_{-h}\phi\rangle,$$and it's easy to show that $\Delta_{-h}\phi\to\phi'$ in $\D$ as $h\to0$.

Curious Fact. Suppose $f,g\in L^2(\Bbb R)$. Then $g=D_wf$ if and only if $g=D_{L^2}f$.

Strikes me as very curious; given $f,g\in L^2$ we have $\Delta_hf\to g$ in $\D'$ if and only if $\Delta_hf\to g$ in norm.

But regardless it really is just an "observation", in that the proof is sort of evident: It doesn't take a great deal of wit to conjecture that the Fourier transform might be useful, and when you look at the Fourier transform the proof just falls out. For newbies:

Proof of the Curious Fact. One direction is trivial from the Easy Lemma: If $g=D_{L^2}f$ then $\Delta_hf\to g$ in norm; hence $\Delta_hf\to g$ in $\D'$, so $g=D_wf$.

Suppose then that $g=D_wf$. Then of course $$\hat g(\xi)=i\xi\hat f(\xi),$$so in particular $$\int|\xi|^2|\hat f(\xi)|^2\,d\xi<\infty.$$ Now $$||g-\Delta_hf||_2^2=\int\left|i\xi-\frac{e^{ih\xi}-1}{h}\right|^2|\hat f(\xi)|^2\,d\xi\to 0\quad(h\to0)$$by dominated convergence (for the "dominated" recall that $|1-e^{it}|\le|t|$ for $t\in\Bbb R$).

(Similarly, Fatou's lemma shows that

Suppose $f\in L^2(\Bbb R)$. Then $D_wf\in L^2$ if and only if $||\Delta_hf||_2$ is bounded.)