I am a Graduate student! I have a question.

Let's have our domain $\Omega$ as an open, path-connected subset in $R^n$. Let $x,y$ be two points in $\Omega$. Then surely there exists at least a path from $x$ to $y$. So here's my question. For such set of paths, is there any path $\gamma$ such that $dist(\gamma,\partial\Omega)$ is strictly positive?

Motivation:(which you do not have to understand to answer my question) I was struggling to prove Strong Maximum Principle of weak solution of generalized elliptic divergence form. Since the problem I'm considering is not continuous, the set $\Omega_{M}=\{x\in\Omega|u(x)=M\}$ where $M=sup_{\Omega}u$ could not be closed. What I'm having now is that for some ball $B\subset\subset \Omega$ if we have $sup_{B}u=sup_{\Omega}u\geq 0$, then the function $u$ must be constant in ball $B$ almost everywhere. so ignoring measure zero set, $\Omega_{M}$ is open in $\Omega$ and I need the closedness of $\Omega_{M}$ to get the desired result.

My reference of motivation is Gilbarg, Trudinger Elliptic PDE of Second Order Thm 8.19.

Thank you so much!

  • $\begingroup$ Hint: A continuous function on a compact set attains its minimum. Apply it to the image of any path in $\Omega$ connecting $x$ to $y$. $\endgroup$ – Moishe Kohan Feb 9 '17 at 7:57
  • $\begingroup$ Oh yes you are definitely right. shame on me.. :) Thanks! $\endgroup$ – Dongjun Kim Feb 10 '17 at 9:22

Yes. If $f:[0,1]\to \Omega$ is continuous, with image $P,$ then $P$ is compact because $[0,1]$ is compact. There are many ways to show that $\inf \{d(p,q):p\in P\land q\in \partial \Omega \}>0.$

Note that $\partial \Omega \subset R^n$ \ $\Omega$ because $\Omega$ is open, so it suffices to show that $$\inf \{d(p,q): p\in \Omega \land q \in R^n \backslash \Omega \}>0.$$

Method (1). The function $g(p)=\inf \{d(p,q):q\in R^n$ \ $P\}$ for $p\in P$ is continuous, and $P$ is compact, so the image of $g$ is compact, so $g$ attains its minimum value. Which must be positive.

Method (2). Assume $\Omega$ is not empty. For $p\in P$ let $r(p)>0$ such that the open ball $B_d(p, r(p))$ is a subset of $\Omega.$ Let $C=\{B_d(p, \frac {1}{2}r(p_i)): p\in P\}.$Then C is an open cover of the non-empty compact $P.$ So $C$ has a finite subcover $\{B_d(p_i),\frac {1}{2}r(p_i)):i=1,...,n\}.$

Let $r=\min (\frac {1}{2}r(p_i):i=1,...,n).$ Any $p\in P$ satisfies $d(p,p_i)<\frac {1}{2}r(p_i)$ for some $i,$ and we have $B_d(p_i,r(p_i))\subset \Omega.$ So the triangle inequality implies $\inf d(p,R^n$ \ $\Omega)\geq \frac {1}{2}r(p_i)\geq r.$

  • $\begingroup$ Thanks for your kind explanations. :) Have a nice day! $\endgroup$ – Dongjun Kim Feb 10 '17 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.