Let $f:B^n_1 \to \Bbb{R}$ defined on the n dimensional ball of radiu $1$. If $f\in C^1$. and $f(0) = 0$.
Can we prove that $$\int_{B_1}|f|^2 \le C\int_{B_1}|\nabla f|^2 +C'$$
If assume further that $m(\{x:|f(x)|\le 1\})\ge \delta |B_1|$ for some constant $\delta >0$ which independent on the choice of $f$.
Can we prove $$\int_{B_1}|f|^2 \le C\int_{B_1}|\nabla f|^2 + C'$$ then?
I should made the statement more formal, for the second version,we have :
Theorem. For any $\varepsilon>0$ there exists a $ C=C(\varepsilon, n)$ such that for $u \in H^{1}\left(B_{1}\right)$ with $$ \left|\left\{x \in B_{1} ; u=0\right\}\right| \geq \varepsilon\left|B_{1}\right| \quad \text { there holds } \int_{B_{1}} u^{2} \leq C \int_{B_{1}}|D u|^{2} $$
The proof is similar to the proof for the poincare wirtinger inequality on Evan's PDE book. This proof can also be found on Q. Han and F. Lin, Elliptic partial differential equations. 4.8.
With slight modification, we can prove the following result :
Theorem For any $\varepsilon>0$ there exists a $ C=C(\varepsilon, n)$ such that for $u \in H^{1}\left(B_{1}\right)$ with $$ \left|\left\{x \in B_{1} ; |u|<\frac{1}{2}\right\}\right| \geq \varepsilon\left|B_{1}\right| \quad \text { there holds } \int_{B_{1}} u^{2} \leq C \int_{B_{1}}|D u|^{2} $$
Proof. Suppose not. Then there exists a sequence $\left\{u_{m}\right\} \subset H^{1}\left(B_{1}\right)$ such that $\left|\left\{x \in B_{1} ; |u_{m}|<\frac{1}{2}\right\}\right| \geq \varepsilon\left|B_{1}\right|, \quad \int_{B_{1}} u_{m}^{2}=1, \quad \int_{B_{1}}\left|D u_{m}\right|^{2} \rightarrow 0$ as $m \rightarrow \infty$ Hence we may assume $u_{m} \rightarrow u_{0} \in H^{1}\left(B_{1}\right)$ strongly in $L^{2}\left(B_{1}\right)$ and weakly in $H^{1}\left(B_{1}\right)$(Using the Sobolev embedding). Clearly $u_{0}$ is a nonzero constant,and $|u_0| > \frac{1}{2}$
which is constant since $Du =0$ by the fact that $Du_m \to 0$ in $L^2$ , it's non-zero since $\|u_m\| = 1 $ so the limit in $L^2$ can not have zero $L^2$ norm. since $\|u_0\|_{L^2(B_1)} =1$, therefore $|u_0|> \frac{1}{2}$ .
So:
$$ \begin{aligned} 0=\lim _{m \rightarrow \infty} \int_{B_{1}}\left|u_{m}-u_{0}\right|^{2} & \geq \lim _{m \rightarrow \infty} \int_{\left\{|u_{m}|<\frac{1}{2}\right\}}\left|u_{m}-u_{0}\right|^{2} \\ & \geq (|u_0| - \frac{1}{2})^2 \inf_m|\{x:|u_m|<\frac{1}{2}\}| >0 \end{aligned} $$ Contradiction.