Let $O$ be an open bounded connected set in $R^n$ and K its boundary. Given a continuous real function $f$ defined on $K$, I would like to extend $f$ to a continuous real function $g$ (i.e. $g$ restricted to $K$ equals $f$) defined in the closure of $O$, in such a way that g is also monotone (e.g. the min and max of $g$ on each closed ball contained in $O$ is attained on the boundary of the ball). In case $K$ has the extra property of being Holder-continuous, there exists a harmonic function $g$ as intended; in particular $g$ is real-analytic in $O$ and is monotonic by the min and max principles. What about if no extra property of $K$ is known, can one still find a monotonic $g$ as intended?
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$\begingroup$ Do you mean $f$ is Holder continuous / no extra property of $f$ is known ? $\endgroup$– Arnaud MortierJan 6, 2018 at 12:36
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$\begingroup$ How about using functions harmonic with respect to the operator $u \mapsto \partial \cdot (a(x) \partial u(x))$ instead, for a suitable $a$? If $a(x) = (\operatorname{dist}(x,K))^{-p}$ for $p$ sufficiently large, every boundary point should become regular. Have you tried that? $\endgroup$– Mateusz KwaśnickiJan 7, 2018 at 7:30
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$\begingroup$ @ArnaudMortier No, I mean that the boundary has a Holder continuous parametrisation. $\endgroup$– Antonio KubikoJan 7, 2018 at 8:33
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$\begingroup$ @MateuszKwaśnicki Thanks for your nice answer. No, I have not tried that. Can you suggest some good book where I could read about these concepts you mention ? $\endgroup$– Antonio KubikoJan 7, 2018 at 8:39
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1$\begingroup$ I added some details in an answer. By the way, even for domains with Hölder boundary I believe you need some condition on the Hölder exponent to assert that all boundary points are regular, right? $\endgroup$– Mateusz KwaśnickiJan 8, 2018 at 9:51
2 Answers
I believe there must be an elementary answer to this question, but I could not find one. While the following is not a complete answer, here is what I would try.
Consider the operator $$L u(x) = \nabla \cdot (a(x) \nabla u(x)),$$ where $$a(x) = (\operatorname{dist}(x, \partial \Omega))^{-p}$$ for some $p > 0$. By general arguments there is a `solution' to the Dirichlet problem $$\begin{cases} Lu = 0 & \text{in $\Omega$,} \\ u = f & \text{on $\partial \Omega$,} \end{cases} $$ in an appropriate sense. On the stochastic processes side the argument might be the following: there is a diffusion process $X_t$ with values in $\Omega$ corresponding to $L$, and $u(x)$ is simply the expected value of $f(X_{\tau-})$, where $\tau$ is the life-time of $X$.
Clearly $u$ satisfies the (strong) maximum principle, so it has no local extrema in $\Omega$, unless constant. The question is whether $u$ is continuous at the boundary.
A standard approach in potential theory is to find barriers: superharmonic functions which vanish continuously at the boundary. In our case $$h(x) = \operatorname{dist}(x, \partial \Omega)$$ appears to be a barrier at every boundary point. It clearly vanishes continuously at the boundary, so let us see if it is superharmonic.
Fix $x \in \Omega$ and choose $z \in \partial \Omega$ so that $|x - z| = \operatorname{dist}(x, \partial \Omega)$. Define $v(y) = |y - z|$. Then $h(y) \le v(y)$ for all $y$ and $h(x) = v(x)$, so $L h(x) \le L v(x)$. However, $L v(x) \le 0$ if $p > n/2 - 1$ (if I am not mistaken; in any case, for $p$ large enough), as desired.
Now why this is not a complete solution:
- The coefficient $a(x)$ is singular near the boundary, so one needs to be careful when showing the existence of the harmonic measure (the `solution' of the Dirichlet problem).
- The coefficient $a(x)$ is not smooth, so extra care is needed when showing that $h$ is a barrier.
Perhaps some day I will find time to fill in these gaps, maybe someone else does that, or perhaps someone will come up with a simpler solution.
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$\begingroup$ Thanks again for another still nicer answer. Here is my motivation to pose this question: I was trying to prove it using elementary means; but then I got convinced that maybe this is well-known to specialists, so I decided to ask. But now I see that this is not a well-known result, and felt encouraged by your suggestion to try an elementary approach. So I tried again, and indeed already found a way to prove existence of a continuous monotone extension, into the interior, of any continuous function given on the boundary; and will ask one of my students to fill-in the details. $\endgroup$ Jan 9, 2018 at 8:27
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1$\begingroup$ @AntonioKubiko: Just wondering: do you plan to post the elementary solution? $\endgroup$ Feb 8, 2018 at 11:43
To precise better: under no extra assumption on K, I would need just a continuous monotonic g, no need for extra regularity.
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1$\begingroup$ I think to let this as a comment (since you posted the question, it's better to wait at a least one answer) $\endgroup$ Jan 6, 2018 at 23:21