If an $H^1$ function vanishes on a set of positive measure, its $L^2$ norm is controlled by the gradient

Question

I am trying to solve question 15 from Evans' PDE book, chapter 5. You have a set of positive measure, subset of the unit ball $B$, such that $u$ is equal to zero on that set. Then, one can show that: $$ \int_Bu^2\le C\int_B(Du)^2 $$ for $u\in H^1(B)$. Now, if that set would somehow "cover" the boundary of the ball, it would be trivial. But if not, then I'm stuck. One idea was to map this set to the boundary, but it doesn't seem to work either. Can you give me some advice/hints? (This is not a homework, I'm just preparing for my exams by doing the whole Evans.)

Answer 1

In section 5.8.1 Evans proves that $$ \int_B (u - u_B)^2\le C \int_B (Du)^2 \tag1$$ where $u_B$ is the mean value of $u$ on $B$. Thus, it suffices to prove that under the assumption $u=0$ on $E$ (a set of positive measure) $$ \int_B u^2 \le C_1 \int_B (u - u_B)^2 \tag2$$ A geometric way to do this is to observe that the assumption $u=0$ on $E$ precludes $u$ from having near-zero angle with the constant function $v\equiv 1$. Namely, denoting this angle by $\theta$ we have $$ \cos\theta = \frac{\langle u,v\rangle}{\|u\|_2\|v\|_2} = \frac{\langle u,\chi_{B\setminus E}\rangle}{\|u\|_2\|v\|_2} \le \frac{\|u\|_2 \|\chi_{B\setminus E}\|_2 }{\|u\|_2\|v\|_2} =\sqrt{|B\setminus E|/|B|} $$ Therefore, the orthogonal projection of $u$ onto the orthogonal complement of $v$ has norm $$ \|u\|_2 \sin\theta \ge \|u\|_2 \sqrt{|E|/|B|} $$ This orthogonal projection is exactly $u-u_B$, which yields (2) with $C_1={|B|/|E|}$.