I would like to show the following:
Let $f\in H^1_0(B_R)$, where $B_R \subset \mathbb{R}^n$ denotes the $R$-ball centered at the origin. Then if $\|f\|_{2}=1$, there exists a universal constant such that $\|\nabla f\|_2^2 \leq C\cdot n^2/R^2$.
My idea was to use Green's identity, the boundary condition and then Cauchy-Schwarz inequality to show that $\|\nabla f\|_2^2 \leq \|f\|_2^2\|\Delta f\|_2^2=\|\Delta f\|_2^2$, and then use the fact that the function is radial to compute $\|\Delta f\|_2^2$ explicitely, using that for a radial function, $\Delta f(x)= \frac{1}{r^{n-1}} \frac{d}{dr} r^{n-1}f'(r)$, but the result is inconclusive. Some help would be appreciated.
Edit: it should be $\|f\|_{H^1}=1$. Therefore, the above thinking is nonesense.
This cannot be true. It would imply that the existence of a constant $c$ such that $\|\nabla f\| \le c \|f\|_{L^2}$ for such radial functions.
Just take $R=1$, $f_k(x) = \sin( k \pi \|x\|)$. Then $(f_k)$ is bounded in $L^2$ but $(\nabla f_k)$ is not.