Proving integral inequality for all $f\in H^1(D)$

Question

LET $D\subset\mathbb R^n$ and $s\ge 0$ be some constant. I want to show that there exists a constant $C$ such that

for all $f\in H^1(D)$ that satisfies $s\le \sharp\{t \mid f(t)=0\}$.

Can someone please give me some ideas?

Answer 1

I'll outline the ideas of an argument that I think should do the job:

• By contradiction assume that such a constant doesn't exist. You can find a sequence $f_{k}$ in $H^1(D)$ s.t. $$\int_Df_k^2> k \int_D |\nabla f_k|^2$$ and with $f_{k}=0$ on a set of measure $\geq s$.

• Set $g_k=\frac{f_{k}}{||f_{k}||^2_{L^2(D)}}$. You have $||g_{k}||^2_{L^2(D)}=1$ and $\nabla g_k=\frac{\nabla f_{k}}{||f_{k}||^2_{L^2(D)}}$.

• $$1=||g_{k}||^2_{L^2(D)}>\ k\ \ \frac{\int_D |\nabla f_k|^2}{||f_{k}||^2_{L^2(D)}}=k\ \ \int_D |\nabla g_k|^2.$$

• Using Rellich-Kondrachov you get a subsequence of $g_k$ (we will still denote it $g_k$) which converges in $L^2(D)$ to some $g$.
• You have $1=||g||^2_{L^2(D)}$ and you can show (easy) that the gradient of $g$ is $0$. Thus $g$ is constant and since it is $0$ on a set of measure $\geq s$, it must be identically $0$. But since the norm of $g$ is $1$ we get a contradiction. ($g=0$ on a set of measure $\geq s$ because passing to a further subsequence you have also conergence a.e.).

Just a note: I'm not sure the result is true if you don't require D to be connected.