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.

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