Suppose $u$ is harmonic on an open set $D \subseteq R^n$, then for any $B_R(x_0) \subseteq D$ and $\alpha = ({\alpha _1},...,{\alpha _n}) \in {Z^n}$, $k = |\alpha |$, why the following inequality holds
Thank you!
This is a step in a proof. The steps before this inequality is shown in the following notes, which might be helpful. I just don't understand why the integral is no larger than the norm. Thank you!
This is just monotonicity of the integral: Notice that, since $x\in B_{R/2}(x_0)$, then $B_{R/2}(x) \subset B_{R}(x_0)$. Furthermore $u(y)\leq |u(y)$|. Combining this we obtain $$ \int_{B_{R/2}(x)} u(y)dy \leq \int_{B_{R/2}(x)}|u(y)|dy=\| u\|_{L^1(B_{R/2}(x))} \leq \| u\|_{L^1(B_R(x_0))}. $$