Is $F: H^{1}(\Omega) \longrightarrow \mathbb{R}$ defined by $F(v):= \int_{\Omega} | Du|^2 dx$ continuous? 
I'm wondering about the folowing question motiveted by existence of weak solutions to the Dirichlet Problem. The function $F: H^{1}(\Omega) \longrightarrow \mathbb{R}$ defined by
  \begin{equation}
F(v):= \int_{\Omega} \|Du\|^2 dx
\end{equation}
  is continous? 

My idea:
\begin{eqnarray}
\left | \int_{\Omega} \| Du\|^2 dx - \int_{\Omega} \| Dv\|^2 dx \right | &\le &  \int_{\Omega}  \left | \| Du\|^2  -  \| Dv\|^2 \right | dx  \\
&\le &  \int_{\Omega} \|D(u)-D(v)\|^2dx?\\
&\le &  \int_{\Omega} \|D(u-v)\|^2dx\\
&\le &  \|u-v\|_{H^1(\Omega)}
\end{eqnarray}
How to foud out an inequality like the second one? Maybe nust exist a constant before.
 A: The function $|v|_0=\left(\int_{\Omega}|Dv|^2dx\right)^{1/2}$ has all of the properties of a norm except possibly strict positivity. That's enough to give you the triangle inequality and reverse triangle inequality:
$$
              |\,|v|_0-|u|_0| \le |v-u|_0.
$$
And that's enough to establish the continuity of $G: H^1\rightarrow\mathbb{R}$ defined as $G(v)=|v|_0$ because
$$
    |G(u)-G(v)| \le |u-v|_0 \le \|u-v\|_{H^1}.
$$
Your function is $F(v)=G(v)^2$, which is the composition of the square function on $\mathbb{R}$ with $G$. So $F$ is continuous.
A: Hint: (your end result is not correct, e.g. wrong homogeneity). Use instead $$ \| Du \| = \|Dv + Du-Dv\| \leq \|Dv\| + \|Du-Dv\|$$
(and the same with $u$, $v$ interchanged) to arrive at:
 $$ \left| \; \| Du \| -  \|Dv\| \right| \leq \|Du-Dv\|.$$
Whence
$$ \left| \; \| Du \|^2 -  \|Dv\|^2 \right|
 \leq \|D(u-v)\|(\| Du \| +  \|Dv\|) .$$
After a little calculation (using Cauchy-Schwarz for $L^2$ functions):
 $$ |F(u)-F(v)|\leq \int \left| \; \| Du \|^2 -  \|Dv\|^2 \right|\; dx
 \leq \|u-v\|_{H^1} (\|u\|_H^1 + \|v\|_{H^1}) .$$
Thus implying continuity.
