To show that $ J(u) = \frac{1}{2}\int_\Omega |\nabla u|^2 -\frac{1}{p+1}\int_\Omega |u|^{p+1} $ is not bounded above for $1 < p < 2^*-1$ For a bounded $ \Omega\subset\mathbb{R}^n $ with smooth boundary, and for $ 1 < p < \frac{n+2}{n-2} = 2^* -1 $ where $ \frac{1}{2^*} = \frac{1}{2}-\frac{1}{n}$, I have the functional $ J : H^1_0(\Omega) \rightarrow \mathbb{R} $ given by 
$$ J(u) = \frac{1}{2}\int_\Omega |\nabla u|^2 -\frac{1}{p+1}\int_\Omega |u|^{p+1} $$ I have to show that $J$ is not bounded above. Now I have been told that for $ u \in C^\infty_c(\Omega) $ if we define for every $\epsilon > 0$ and $ \alpha = n/(p+1) $
$$ u_\epsilon(x) = \frac{1}{\epsilon^\alpha}u(x/\epsilon) $$ then $ J(u_\epsilon) \rightarrow \infty $ as $ \epsilon \rightarrow 0 $. I am having trouble in deducing this , what I have so far is that $ p < 2^*-1 \Rightarrow 2(1+\alpha) > n  $ and using transformation $ x/\epsilon \rightarrow x$ I have 
$$  J(u_\epsilon) = \frac{1}{2\epsilon^{2(1+\alpha)-n}}\int_{\epsilon\Omega}|\nabla u|^2 - \frac{1}{p+1}\int_{\epsilon\Omega}|u|^{p+1} $$ But now I don't see anyway to get rid of $\epsilon $ in $ \epsilon \Omega$ and I am stuck. I would be extremely thankful if anyone can help me out. Thank you.
 A: I think the deduction on $J(u_{\epsilon})$ should be like this:
\begin{align}
J(u_{\epsilon})=&\frac{1}{2}\int_\Omega|\nabla u_\epsilon|^2-\frac{1}{p+1}\int_\Omega|u_\epsilon|^{p+1}
\\
=&\frac{1}{2\epsilon^{2\alpha+2}}\int_\Omega|\nabla u(x/\epsilon)|^2-\frac{1}{(p+1)\epsilon^{\alpha(p+1)}}\int_\Omega|u(x/\epsilon)|^{p+1}
\end{align}
Note that for small $\epsilon$, $x/\epsilon$ may go outside $\Omega$, but we only have $u\in C_0^{\infty}(\Omega)$, so we may assume $u(x)=0$ outside $\Omega$, then $u(x/\epsilon)=0$ outside $\epsilon\Omega$, here $\epsilon\Omega$ denotes the domain rescaled from $\Omega$ by $\epsilon$.
From the above we have
\begin{align}
J(u_{\epsilon})=&\frac{1}{2\epsilon^{2\alpha+2}}\int_{\epsilon\Omega}|\nabla u(x/\epsilon)|^2-\frac{1}{(p+1)\epsilon^{\alpha(p+1)}}\int_{\epsilon\Omega}|u(x/\epsilon)|^{p+1}
\\
=&\frac{1}{2\epsilon^{2\alpha+2-n}}\int_{\Omega}|\nabla u(x)|^2-\frac{1}{p+1}\int_{\Omega}|u(x)|^{p+1}
\end{align}
by changing variables.
Now I think everything goes through. Hope this may help :)
