The functional $\int_{\Omega} |\nabla u|^p dx$ is invariant under dialation and translation in $\mathbb{R}^N$. How?

Question

The functional $\int_{\Omega} |\nabla u|^p dx$ is invariant under dialation and translation in $\mathbb{R}^N$. How?

Answer 1

It is easy to see that it is translation invariant since $$ \int_{\mathbb{R}^N} |\nabla u(x+t)|^p dx=\int_{\mathbb{R}^N} |\nabla u(x)|^p dx. $$ You can see this by performing the change of variable $x \mapsto x+t$ for any $t \in \mathbb{R}^N$.
However, it is in general not dilation invariant. Consider the dilation $x \mapsto \lambda x$ for some positive $\lambda$. Then we have $$ \int_{\mathbb{R}^N} |\nabla (u(\lambda x))|^p dx=\int_{\mathbb{R}^N} \lambda^p |\nabla u(\lambda x)|^p dx=\int_{\mathbb{R}^N} \frac{\lambda^p}{\lambda^N} |\nabla u( x)|^p dx $$ The last equality again follows by a change of variables. So the functional is dilation invariant iff $p=N$.