Let $B(0,r)\subset \mathbb R^3$ be a ball centered at $0$ with radius $r$. Define $$ \mathcal A_r:=\{u\in H_0^1(B(0,r)),\,\,\|u\|_{L^{q+1}}=1\}$$ where $1<q<5$. Hence we know that each $\mathcal A_r$ is weakly closed with respect to $H_0^1$ norm.
Now I am interesting in to see whether I can find a constant $C>0$ such that $$ \max_{r>0}\min_{u\in\mathcal A_r}\int_{B(0,r)}|\nabla u|^2\leq C $$
Could this be possible? If not, please provided me a counterexample as well. Thank you!
Update:
It looks to me that if $1+\epsilon<q<5$ will work, but not just $1<q<5$.
As suggested on overflow by Pietro Majer (if I did not make a computational mistake): writing $u=v(x/r)$, $u\in A_r\Leftrightarrow v\in B_r$, with $$ B_r=\{u\in H^1_0(B(0,1)) : \|u\|_{L^{q+1}} = r^{\frac{-d}{q+1}}\}, $$ and you want $$ \max_{r>0} \min_{w\in B_r} r^{d}\int_{B(0,1)}|\nabla u|^{2}\leq C. $$ Now, $B_r = r^{\frac{-d}{q+1}}B_1$, therefore you want $$ \max_{r>0} \inf_{w\in B_1} r^{d\frac{q-1}{q+1}} \int_{B(0,1)}|\nabla u|^{2}\leq C. $$ The two problems are now decoupled, so either $$ \inf_{w\in B_1} \int_{B(0,1)}|\nabla u|^{2}=0, $$ in which case the constant $C$ is zero, or not and the upper bound is infinite if $q>1$, or $q=1$, and the bound is just the first eigenvalue of the Laplacian. Now, for $1<q<5$ thanks to Sobolev's embeddings, the inf is a min and is not zero, therefore there is no such $C$.
So for $q>1$ no, and for $q=1$, yes.