## Bound of norm on gradient implies bound on function

$\mathbf{Problem:}$ Suppose $p\in L^2_{loc}(B_1)$, $\nabla p\in L^2(B_1)$, prove that $p\in L^2(B_1)$ where $B_1\subset\mathbb{R}^n$ is a ball with radius 1.

We are going to use the following lemma.

Lemma 7.16 on GT’s book. p162

Let $\Omega$ be convex and $u\in W^{1,1}(\Omega)$. Then

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|Du(y)|}{|x-y|^{n-1}}dy$,   $a.e. \Omega$

where $\displaystyle u_S=\frac{1}{|S|}\int_S udx$, $d=diam\,\Omega$.

$\mathbf{Proof:}$ For any $B_r\subset B_1$ concentric with $B_1$, $p\in W^{1,1}(B_r)$, apply the lemma 7.6

$\displaystyle |p(x)-p_S|\leq \frac{r^n}{n|S|}\int_{B_r}\frac{|\nabla p(y)|}{|x-y|^{n-1}}dy\leq \frac{1}{n|S|}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy\int_{B_1}\frac{1}{|x-y|^{n-1}}dy$

Since $\displaystyle \int_{B_1}\frac{1}{|x-y|^{n-1}}dy$ is uniformly bounded for $\forall x\in\mathbb{R}^n$, then the above inequality implies that

$\displaystyle |p(x)-p_S|\leq C\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy$

$\displaystyle \int_{B_r}|p(x)-p_S|^2dx\leq C\int_{B_1}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dydx\leq C\int_{B_1}|\nabla p|^2dy$

Let $r\to 1^+$, we get $||p||_{L^2}\leq C(p_S+||\nabla p||_{L^2})$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ For any bounded domain with locally lipschitz boundary, this proposition is true. Because the locally boundary enable the domain to have locally star-shaped property, we can apply this proposition locally.