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.

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.