Convergence in $L^2$ of difference quotients to derivative of function in $H^1$

Question

Is it true that if $u\in H^1({\mathbb R})$, then $(u(x+h)-u(x))/h$ converges to $u'(x)$ in $L^2({\mathbb R})$, as $h\to 0$? It's hard for me to get a handle on this, since $u'$ doesn't have to be continuous (so there's no uniform convergence, even on compacts, since $u$ does have to be continuous), but I can't seem to construct a counterexample either.

Answer 1

I randomly found this unaswered question asked about 6 months ago... Shall I..? Yeaahhhhh.

First a comment about "since $u$ does have to be continuous": the Sobolev embedding theorem for this case ($N = 1$, $p = 2$, $k = 1$) gives $H^1(\mathbb{R}) \hookrightarrow C^{0,1/2}(\mathbb{R})$. (Holder continuous functions with Holder exponent $1/2$). Hence $u$ needs to be continuous.

Moving on, the result follows immediately from the "Difference quotients theorem" for Sobolev spaces:

Theorem:

Let $\Omega \subset \mathbb{R}^N$ be an open set, $1 \le p < \infty$, $u \in W^{1,p}(\Omega)$. Then for every $i = 1, \dots , N$, $h > 0$ $$\int_{\Omega_{h,i}}\Big|\frac{u(x + he_i) - u(x)}{h}\Big|^p\,dx \le \int_{\Omega}\Big|\frac{\partial u}{\partial x_i}(x)\Big|^p\,dx$$ and $$\lim_{h \to 0^+}\int_{\Omega_{h,i}}\Big|\frac{u(x + he_i) - u(x)}{h}\Big|^p\,dx = \int_{\Omega}\Big|\frac{\partial u}{\partial x_i}(x)\Big|^p\,dx.$$

Here $\Omega_{h,i} = \{x \in \Omega : x + he_i \in \Omega\}$ (you don't care about this part since you are in $\mathbb{R}$, hence this condition is always satisfied.)

The proof is a little bit technical, but most definitely comprehensible: you need to play with mollifications a lot and you need few facts about the relation between sobolev functions and absolutely continuous functions. A proof is outlined in Evan's book, chapter 5 if I remember correctly.

By the way, for $p > 1$ the theorem has a converse which turns out to be a blessing when proving regularity for elliptic problems! :)