Let $u \in H^1(\mathbb{R})$. Show that
$$\left\| \frac{u(x+h)-u(x)}{h} – u' \right\|_2 \to 0\quad \text{ as } h \to 0, $$
where $u' \in L^2(\mathbb{R})$ is the weak derivative of $u$. In other words, the weak derivative can be written as an $L^2$ limit of a difference quotient.
We know that functions in $H^1(\mathbb{R})$ are differentiable almost everywhere, so that the pointwise limit
$$ \lim_{h \to 0} \frac{u(x+h)-u(x)}{h} $$
exists and is equal to $u'(x)$ almost everywhere.
I would like to take the limit inside the integral (using dominated convergence), but I'm not positive that there is a dominating function… Does anybody have any hints, or an outright solution?
Note: This is the first exercise from Chapter 7 of Renardy & Roger's An Introduction to Partial Differential Equations. I am not doing it for homework. Just for "fun."
Best Answer
If $u\in H^1(\mathbb{R})$, then in particular it's absolutely continuous, so that the fundamental theorem of calculus gives $$ |u(x+h)-u(x)|\leq \int_0^1 |u'(x+th)||h|dt $$ So that squaring and integrating over $\mathbb{R}$ we get (using Jensen's inequality and Fubini's theorem) $$ \left\|\frac{u(\cdot +h)-u(\cdot)}{h}\right\|_2^2 \leq \| \int_0^1 u'(\cdot+th)dt\|_2^2 \leq \| u'\|_2^2, \quad \forall h>0. $$
Now, since for smooth compactly supported functions the limit is clear, we pick $u\in H^1$ and $u_k\in C_c^\infty$ with $u_k\to u$ in $H^1$. Then we get $$ \left\|\frac{u(\cdot +h)-u(\cdot)}{h}-u'(\cdot)\right\|_2 \leq \left\|\frac{u(\cdot +h)-u(\cdot)}{h} - \frac{u_k(\cdot +h)-u_k(\cdot)}{h}\right\|_2 + \left\|\frac{u_k(\cdot +h)-u_k(\cdot)}{h}-u_k'(\cdot)\right\|_2 + \| u'-u_k'\|_2 $$ The first term is bounded by $\| u'-u_k'\|_2$ by the previous inequality; the second term goes to zero since $u_k\in C_c^\infty$ so that we can make the LHS arbitrarily small for appropriate $k$ and $h$ small enough.