# 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 $$F(v):= \int_{\Omega} \|Du\|^2 dx$$ 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.
• Thank you, but how to prove triangle inequality for $|v|_0$? – user29999 Aug 14 '16 at 23:39
• @user29999 : The triangle inequality for $|\cdot|_0$ follows from the triangle inequality for the $L^2$ norm. $|v+w|_0 = \|Dv+Dw\|_{L^2}\le \|Dv\|_{L^2}+\|Dw\|_{L^2}=|v|_0+|w|_0$. – DisintegratingByParts Aug 15 '16 at 1:36
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.
• If $$\left |\|Du\|^2 -\|Dv\|^2\right | = \|Du\|^2 -\|Dv\|^2,$$ then $\|Du\|\ge \|Dv\|$. However $$\begin{array}{rccl} &\|Du-Dv\|^2 & =&\|Du\|^2 + 2 \langle Du, Dv \rangle + \|Dv\|^2\\ & &\ge & \|Du\|^2 -\|Dv\|^2 \\ \end{array}$$ Equivalentemente, $$\|Dv\|^2 \ge \langle Du, Dv \rangle\\$$ – user29999 Aug 6 '16 at 21:09
• Thanks again. However, whenever $\|u-v\|_{H}$ can become small, on the other hand, $\|u\|_{H^1}+\|u\|_{H^1}$ or not? I thought this too. – user29999 Aug 14 '16 at 23:25
• Not sure what your point is? The norms are finite and when e.g. $v$ converges to $u$ (in $H^1$) then $F(v)$ converges to $F(u)$. – H. H. Rugh Aug 14 '16 at 23:47