Prove that $W^{1,p}$-functions are "$L^{p}$-Lipschitz". Let $U$ be open in $\mathbb{R}^{n}$ and $V\subset\subset U$. Let
$u\in W^{1,p}(U)$ for $p\in[1,\infty)$.
Want to show for $x\in V$ and $h\in\mathbb{R}$, 
$$
\sup_{h}\int\left(\frac{u(x+he_{1})-u(x)}{h}\right)^{p}dx\leq C\|Du\|_{L^{p}(U)} ^p
$$
for some $C$ and $0<|h|<\frac{1}{2}$dist$(V,\partial U)$ where $Du$ : weak derivative of $u$.
I have no idea with this. Anyone gives me a hint for this? I've learnt Sobolev inequalities. Since $U$ is not surely bounded, I can't use partition of unity argument.
 A: I suppose $e_1$ is some unital vector, i.e. $|e_1| = 1$. Note that it is sufficient to show the inequality for $u \in C_{c}^{\infty}(\mathbb{R})$, because this implies the inequality for general $u \in W^{1,p}(U)$ by a standard density argument. For a classically differentiable function $u$, we have
\begin{align}
 u(x+he_1) - u(x) &= \int_{0}^{1} \frac{d}{dt}[u(x+the_1 )] dt \\
                  &= \int_{0}^{1} \nabla u(x+the_1)\cdot he_1 dt
\end{align}
As the function $x \mapsto |x|^p$ is convex for $p \geq 1$, this implies by Jensen's inequality
\begin{align}
\left|\frac{ u(x+he_1) - u(x)}{h}\right|^p \leq \int_{0}^{1} \left| \nabla u(x+the_1)\cdot e_1 \right|^p  \leq \int_{0}^{1} \left| \nabla u(x+the_1)\right|^pdt
\end{align} 
Integrating this over $V$ yields
\begin{align}
\int_{V} \left|\frac{ u(x+he_1) - u(x)}{h}\right|^p dx &\leq \int_{V} \int_{0}^{1} \left| \nabla u(x+the_1)\right|^p dt \;dx \\
&=\int_{0}^{1} \int_{V} \left| \nabla u(x+the_1)\right|^p dx \;dt \\
&\leq \int_{0}^{1} \int_{U} \left| \nabla u(y)\right|^p dy \;dt \\
&= \|Du\|_{L^p (U)}^p
\end{align}
where the last inequality holds because we have chosen $h$ small enough such that $x+the_1 \in U$ for $0 < t < 1$ and $x \in V$.
EDIT: For the density argument, let $u_n \in C_{c}^{\infty}(\mathbb{R})$ converge against $u$ in $W^{1,p}(U)$. From what we've shown so far, we know that 
$$\int_{V}\left|\frac{u_n(x+he_{1})-u_n(x)}{h}\right|^{p}dx\leq \|Du_n\|_{L^{p}(U)} ^p $$
holds for any $n \in \mathbb{N}$. Due to the continuity of the $W^{1,p}$-Norm, the right hand side clearly converges against $\|Du\|_{L^{p}(U)}^{p}$, so it suffices to show that the left hand side converges against $\left\|\frac{u(\cdot+he_{1})-u(\cdot)}{h}\right\|_{L^{p}(V)}^{p}$. A simple change of variable shows that $u_n (\cdot + he_1) \to u(\cdot + he_1)$ in $L^{p}(V)$:
$$\int_{V} \left|u_n(x+he_1) - u(x+he_1)\right|^p dx = \int_{V+he_1} \left|u_n(y) - u(y)\right|^p dy \leq \|u_n - u\|_{L^{p}(U)} \to 0 $$
where the last inequality follows again because $V + he_1 \subseteq U$. Therefore, we may conclude that $$ \frac{u_n(x+he_{1})-u_n(x)}{h} \to \frac{u(x+he_{1})-u(x)}{h} \quad \text{in}\quad L^{p}(V)$$ 
holds. This implies convergence of the respective $L^{p}(V)$-norms, which proves 
$$\int_{V}\left|\frac{u(x+he_{1})-u(x)}{h}\right|^{p}dx\leq \|Du\|_{L^{p}(U)} ^p $$
for all $u \in W^{1,p}(U)$ and $h$ satisfying $0 < |h| < \text{dist}(V,\partial U)$. Therefore, we may take the supremum over all such $h$ on the left hand side and the inequality will remain true.
