# Estimating continuous functions in terms of distance functions.

Let $\Omega\subset \mathbb{R}^N$ be a bounded domain and $u:\overline{\Omega}\to\mathbb{R}$ a continuous function which vanishes on the boundary $\partial\Omega$. Is there constants $a,b>0$ such that $$|u(x)|\le a\operatorname{dist}(x,\partial\Omega)+b,\ \forall x\in \overline{\Omega},\tag{1}$$

where $\operatorname{dist}$ denotes distance and $b$ does not depends on $u$?

Note that the constant $b$ must be, in general, positive, because if $u$ grows too fast near the boundary, for example, like the function $\sqrt{x}$ in $[0,1]$ then, there is no constant $a$ which satisfies $(1)$ with $b=0$

On the other hand, if $u$ is Lipschitz then $(1)$ is satisfied with $b=0$, indeed, just note that $|u(x)|\le L |x-y|$ for all $x\in \overline{\Omega}$ and $y\in \partial \Omega$, where $L$ is the Lipschitz constant.

The answer is yes. In fact, given $b>0$ there exists $a>0$ (depending on $u$ and $b$) such that the inequality holds. Let's prove it.
For any $x\in\partial\Omega$ there exists $\epsilon_x>0$ such that $|u(y)|\le b$ for all $y\in\Omega\cap B(x,\epsilon_x)$. Since $\partial\Omega$ is compact there exist $x_1,\dots,x_n\in\partial\Omega$ such that $\partial\Omega\subset\bigcup_{k=1}^nB(x_k,\epsilon_{x_k})$. Let $\epsilon=\min_{1\le k\le n}\epsilon_{x_k}>0$. If $x\in\Bigl(\bigcup_{k=1}^nB(x_k,\epsilon_{x_k})\Bigr)\bigcap\Omega$, then $|u(x)|\le b$. On the other hand, if $x\in\Omega\setminus\Bigl(\bigcup_{k=1}^nB(x_k,\epsilon_{x_k})\Bigr)$, then $\text{dist}(x,\partial\Omega)\ge\epsilon$ and $$|u(x)|\le\frac1\epsilon\,\|u\|_\infty \,\text{dist}(x,\partial\Omega).$$
• It seem that your proof, can be adapted for the case where $\Omega$ is unbounded and $u$ is uniformly continuous. Commented Jul 1, 2014 at 19:20