I'm learning PDE from the book of Trudinger and Gilbarg and I'm attempting to prove the following theorem:
Let $\hspace{0.1ex}u\hspace{0.1ex}$ be harmonic in $\hspace{0.1ex}\varOmega \subset \mathbb{R}^n\hspace{0.1ex}$ (open, connected, bounded) and let $\hspace{0.1ex}\varOmega^{\hspace{0.1ex}'}\hspace{0.1ex}$ be any compact subset of $\hspace{0.1ex}\varOmega$. Then for any multi-index $\hspace{0.1ex}\alpha\hspace{0.1ex}$ we have
\begin{align}
\sup_{\varOmega{\hspace{0.1ex}'}} \big\lvert \,D^{\,\alpha}\hspace{0.075ex}u\,\big\rvert
&\leq
\left( \dfrac{n \left\lvert\alpha\right\rvert}{d} \right)^{\left\lvert\alpha\right\rvert} \!\sup_{\varOmega} \left\lvert\hspace{0.1ex} u\hspace{0.1ex}\right\rvert, &d&=d\left(\varOmega^{\hspace{0.1ex}'}, \partial \hspace{0.1ex}\varOmega\right).
\end{align}
They say the theorem is proved by applying the estimate
\begin{align}
\big\lvert\,D\hspace{0.3ex}u\left(\hspace{0.1ex}y\hspace{0.1ex}\right)\big\rvert
\leq
\frac{n}{d_y} \sup_{\varOmega} \left\lvert\hspace{0.1ex} u\hspace{0.1ex}\right\rvert, \quad d_y
=
d\left(\hspace{0.1ex}y, \hspace{0.1ex}\partial \hspace{0.1ex}\varOmega\hspace{0.1ex}\right)
\end{align}
successively to equally spaced nested balls.
The natural thing to do is induction on $\hspace{0.1ex}k = \left\lvert\alpha\right\rvert$.
The case $\hspace{0.1ex}k=1\hspace{0.1ex}$ follows from the above estimate.
So assume the theorem holds for every compact subset $\hspace{0.1ex}\varLambda\hspace{0.05ex}$ of $\hspace{0.1ex}\varOmega\hspace{0.1ex}$ and multi-index of order $\hspace{0.1ex}k-1\hspace{0.1ex}$ and let $\hspace{0.1ex}\alpha\hspace{0.1ex}$ be a multi-index with $\hspace{0.1ex}\left\lvert\alpha\right\rvert = k$.
Given $\hspace{0.1ex}x \in \varOmega^{\hspace{0.1ex}'}$ we have $D^{\hspace{0.1ex}\alpha}\hspace{0.1ex}u\left(x\right) = \dfrac{\partial \hspace{0.1ex}D^{\hspace{0.1ex}\beta}\hspace{0.1ex}u}{\partial \hspace{0.1ex}x_i}\,\left(\hspace{0.1ex}x\hspace{0.1ex}\right)\,$ for some $\hspace{0.1ex}i\hspace{0.1ex}$ and $\hspace{0.1ex}\beta\hspace{0.1ex}$ with $\hspace{0.1ex}\left\lvert\hspace{0.1ex}\beta\hspace{0.1ex}\right\rvert = k-1$.
Applying the mean value theorem for $\hspace{0.1ex}D^{\hspace{0.1ex}\alpha}\hspace{0.1ex}u\hspace{0.1ex}$ in a ball $\hspace{0.1ex}B\hspace{0.05ex}\left(\hspace{0.05ex}x,\hspace{0.15ex} R\hspace{0.1ex}\right)$ with $\hspace{0.1ex}R < d\left(\hspace{0.1ex}x,\hspace{0.1ex} \partial \hspace{0.1ex}\varOmega\hspace{0.05ex}\right)$ we obtain
$$\big\lvert \,D^{\hspace{0.1ex}\alpha}\hspace{0.1ex}u\left(\hspace{0.1ex}x\hspace{0.1ex}\right)\big\rvert \leq \frac{n}{R} \sup_{\partial \hspace{0.05ex}B\hspace{0.1ex}\left(\hspace{0.075ex}x,\hspace{0.2ex} R\hspace{0.1ex}\right)} \big\lvert \,D^{\hspace{0.1ex}\beta}\hspace{0.1ex}u\,\big\rvert . $$
Using the induction hypothesis with $\hspace{0.1ex}\varLambda = \partial \hspace{0.1ex}B\left(\hspace{0.1ex}x,\hspace{0.1ex} R\hspace{0.1ex}\right)$ we thus have
$$
\big\lvert\,D^{\hspace{0.1ex}\alpha}\hspace{0.1ex}u\left(\hspace{0.1ex}x\hspace{0.1ex}\right)\big\rvert
\leq \frac{n}{R} \,
\left( \frac{n \left\lvert\hspace{0.1ex}\beta\hspace{0.1ex}\right\rvert}{d\hspace{0.1ex}\big(\hspace{0.1ex}\partial\hspace{0.1ex} B\left(\hspace{0.1ex}x,\hspace{0.1ex} R\hspace{0.1ex}\right), \hspace{0.2ex}\partial\hspace{0.1ex} \varOmega\hspace{0.1ex}\big)} \right)^{\left\lvert\hspace{0.1ex}\beta\hspace{0.1ex}\right\rvert}\sup_{\varOmega} \left\lvert\hspace{0.05ex} u\hspace{0.1ex}\right\rvert,$$
leading to nothing. Any suggestions? Where do the nested balls come in?
Thanks!
Best Answer
Your solution is practically there. You're right that the evenly spaced nested balls is what's missing.
Fix $y \in \Omega$ and fix some multi-index $\alpha$ with $| \alpha | = k$. Let $d = dist(y, \partial \Omega)$. Let $d_{0} = \frac{d}{|\alpha|}$.
Then consider the ball $B = B(y, d_{0})$. Let $\alpha_{1}$ be a multi-index such that $|\alpha_{1}| = k-1$ and $\alpha_{1} < \alpha$. Based off the estimate you already know, we have that $$ |D^{\alpha} u(y)| \le \frac{n}{d_{0}} \sup_{B(y,d_{0})} |D^{\alpha_{1}} u|. $$
However, for every $y_{1} \in B(y, d_{0})$ we can apply the same estimate on the ball $B(y_{1}, d_{0})$ to gain
$$ |D^{\alpha} u(y)| \le \frac{n}{d_{0}} \sup_{y_{1} \in B(y,d_{0})} \left[ \frac{n}{d_{0}} \sup_{B(y_{1}, d_{0})} |D^{\alpha_{2}}u| \right] \le \frac{n}{d_{0}} \left[ \frac{n}{d_{0}} \sup_{B(y,2 d_{0})} |D^{\alpha_{2}} u| \right], $$
where $|\alpha_{2}| = k-2$, and $\alpha_{2} < \alpha_{1}$.
Repeating inductively, we attain the desired result
$$ | D^{\alpha} u(y) | \le \left( \frac{n}{d_{0}} \right)^{|\alpha|} \sup_{B(y, |\alpha| d_{0})} |u| \le \left( \frac{n |\alpha|}{d} \right)^{|\alpha|} \sup_{\Omega} |u|, $$
where the equality follows since we chose $d_{0} = \frac{d}{|\alpha|}$.