Let $1 \leq p < \infty$ and $u \in W^{1,p}(\mathbb{R}$). Set
$$
D\_{h}u(x) = \frac{1}{h}(u(x+h) - u(x)), \ \ x \in \mathbb{R}, h> 0
$$
Show that $D\_{h}u \to u'$ in $L^{p}(\mathbb{R}$) as $h \to 0$.


**I'm trying to use the fact that $C\_{c}^{1}(\mathbb{R}$) is dense in $W^{1,p}(\mathbb{R}$)**



Exercise 8.13 - Brezis
The proof is not short, because it is done from first principles, without using any theorems about Sobolev space except its definition.


By the definition of $W^{1,p}$, there exist $v\_n \in C\_{c}^{1}(\mathbb{R})$
and $w \in L^p(\mathbb{R})$ such that $v\_n \to u$ in $L^p(\mathbb{R})$ and $v\_n' \to w$ in $L^p(\mathbb{R})$. In this case we write $u'=w$.
Note that $v\_n'$ is a classical derivative, so
$$D\_h v\_n(x)=\frac{v\_n(x+h)-v\_n}{h}= I\_h(v\_n')(x) \,, \tag{1}
$$
where for $f\in L^p(\mathbb{R})$, we write
$$I\_h(f)(x):=\int\_0^h \frac{f(x+t)}{h} \,dx \,.$$
By Jensen's inequality [1], for all $n \ge 1 $ and $x \in \mathbb{R}$, we have
$$|I\_h(v\_n')(x)-I\_h(u')(x)|^p \le \frac{1}{h} \int\_0^h |v\_n'(x+t)-u'(x+t)|^p \,dt \,.
$$
Integrating both sides $\,dx$ and using Fubini on the right-hand side, we obtain
$$\|I\_h(v\_n') -I\_h(u') \|^p \le \frac{1}{h}\int\_0^h \|v\_n'(\cdot+t)-u'(\cdot+t)\|\_p^p \,dt= \|v\_n' -u' \|\_p^p \,.\tag{2}
$$


Given $\epsilon>0$, find $k$ such that
$$ \|v\_k'-u'\|\_p<\epsilon \,. \tag{3}
$$
Let $M$ denote the Lebesgue measure of the support of $v\_k$.
Since $v\_k'$ is uniformly continuous, there exists $h\_0\in(0,1)$ such that
$$\forall t\in [0, h\_0], \quad \sup\_{x \in \mathbb{R}} |v\_k'(x+t)-v\_k'(x)|<\epsilon/(M+1) \,,$$
so for $h\in [0, h\_0]$ and all $x$, we have
$|I\_h (v\_k')(x)-v\_k'(x)|<\epsilon/(M+1)$, whence
$$\|I\_h (v\_k') -v\_k'\|\_p^p \le (M+h) (\epsilon/(M+1))^p <\epsilon^p \,.$$
In conjunction with $(2)$ and $(3)$, this gives
$$\|I\_h(u')-u' \|\_p \le \|I\_h(u')-I\_h(v\_k') \|\_p + \|I\_h(v\_k')-v\_k' \|\_p + \| v\_k' -u'\|\_p <3\epsilon \,. \tag{4}$$


Next, fix $h\in [0, h\_0]$, and choose $m=m(h,\epsilon)$ such that
$$\|u-v\_m \|\_p<\epsilon h \quad \text{and} \quad \|u'-v\_m'\|\_p<\epsilon \,.
$$
The first inequality implies that $\|D\_h(u) -D\_h(v\_m) \|\_p<2\epsilon$. Therefore, by $(1),\, (2)$ and $(4)$,
\begin{eqnarray}
\|D\_h(u)-u'\|\_p &\le& 
\|D\_h(u) -D\_h(v\_m) \|\_p+\|I\_h(v\_m')-I\_h(u')\|\_p+\|I\_h(u')-u' \|\_p \\
&<& 2\epsilon+\epsilon+3\epsilon=6\epsilon \,.
\end{eqnarray}
This completes the proof.


[1] <https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form>


