I'm working on a project and for it I needed to read and understand a paper by Berestycki-Lions about the existence of Nonlinear Scalar Field equations. (If someone has it or is further interested in the context its page $319$ in Section $2$ Necessary Conditions.) Anyway, the thing is at some point it says:
We define fhe functional $T(u)$: $$T(u) = \int_R^N |\nabla u|^2\, dx $$ one readily checks that it has the property that for a scale change in $R^N$: for $\sigma > 0$ we define $u_\sigma (x) = u(x/ \sigma)$: $$T(u_\sigma) = \sigma^{N-2} T(u).$$
My problem is, I don't really see how this works, so any help will be very much appreciated. I don't know if it's necessary for the proof but u is furthermore spherically symmetric.
I presume your integral is over all $R^N$ and $\nabla u=\partial u/\partial x$ is a vector in $R^N$. Then: $$ \begin{aligned} T(u_\sigma) &= \int_{R^N} |\nabla u_\sigma(x)|^2\, dx = \int_{R^N} |\nabla u(x/\sigma)|^2\, dx = \int_{R^N} \left|{\partial u(x/\sigma)\over\partial x}\right|^2\, \sigma^N d(x/\sigma)\\ &= \int_{R^N} \left|{\partial u(x/\sigma)\over\sigma \partial (x/\sigma)}\right|^2\, \sigma^N d(x/\sigma) = \sigma^{N-2} \int_{R^N} \left|{\partial u(x/\sigma)\over\partial (x/\sigma)}\right|^2\, d(x/\sigma)=\sigma^{N-2}T(u). \end{aligned} $$ For the case you mention in the comment below: $$ \begin{aligned} T(u_\sigma)(x) &= \int_{R^N} \left|{u_\sigma(x)-u_\sigma(y)\over|x-y|^{N+s}}\right|^2\, dy = \int_{R^N} \left|{u(x/\sigma)-u(y/\sigma)\over\sigma^{N+s}|x/\sigma-y/\sigma|^{N+s}}\right|^2\, \sigma^N d(y/\sigma) \\ &=\sigma^{-N-2s}\int_{R^N} \left|{u(x/\sigma)-u(y/\sigma)\over|x/\sigma-y/\sigma|^{N+s}}\right|^2\, d(y/\sigma)=\sigma^{-N-2s}T(u)(x/\sigma). \end{aligned} $$