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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!

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1 Answer 1

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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 $d_{0} = \frac{d}{|\alpha|}$.

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    $\begingroup$ This was great! Thank you very much! $\endgroup$ Commented Jan 31, 2016 at 16:53

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