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Let $\Omega \subset \mathbb{R}^n$ be open, convex and bounded with smooth boundary. Define

$$\mathcal{A}(\Omega) = \left\{ C \subset \Omega \ \big\vert \begin{array}{l} \text{for any subharmonic function } \varphi : \Omega \rightarrow \mathbb{R} \\ \text{there exists a harmonic function } h_{\varphi} : \Omega \rightarrow \mathbb{R} \\ \text{such that } h_{\varphi} \leq \varphi \text{ on } C \text{ and } h_{\varphi} \geq \varphi \text{ on } \Omega \setminus C \end{array} \right\}.$$

Maybe it is helpful to restrict to closed $C \subset \Omega$.

If $n = 1$, then $\mathcal{A}(\Omega)$ is the set of all (closed) intervals in $\Omega$. For $n \geq 2$ and sufficiently nice $C$, the restriction of $h$ to $\overline{C}$ has to be the solution of the Dirichlet problem with boundary conditions given by $\varphi$. But then the (non-trivial?) task of extending the solution from $\overline{C}$ to $\Omega$ remains.

Are there any characterizations of (the elements in) $\mathcal{A}(\Omega)$ for $n \geq 2$? Do you know references where this kind of question is studied?


Edit: As pointed out below, it holds

$$ \mathcal{A}(\Omega) = \begin{cases} \text{the set of (closed) intervals in } \Omega & n = 1 \\ \{ \emptyset, \Omega \} & n \geq 2 \end{cases} $$

because for any $C$ there exist subharmonic functions on $\Omega$ that can't coincide with the values of a harmonic funtion on $\partial C$. Thank you for your help so far!

Let $B(C) = \{ \varphi : \Omega \rightarrow \mathbb{R} \ \vert \ \exists \text{ harmonic } h : \Omega \rightarrow \mathbb{R} \text{ s.t. } h = \varphi \text{ on } \partial C \}$. Then we could consider

$$\mathcal{A}'(\Omega) = \left\{ C \subset \Omega \ \big\vert \begin{array}{l} \text{for any subharmonic function } \varphi \in B(C) \\ \text{there exists a harmonic function } h_{\varphi} : \Omega \rightarrow \mathbb{R} \\ \text{such that } h_{\varphi} \leq \varphi \text{ on } C \text{ and } h_{\varphi} \geq \varphi \text{ on } \Omega \setminus C \end{array} \right\}.$$

Is it still true that $\mathcal{A}'(\Omega)$ is trivial for $n \geq 2$? Perhaps this question goes too far ... I'm actually a little bit surprised that there is no (easy) analogue to the one-dimensional case.

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There are no such $C$ at all, when $n>2$. Indeed your conditions imply $h_\phi=\phi$ on $\partial C$. But $h$ is bounded and continuous on $\partial C$, while when $n\geq 2$ there is always a subharmonic function which is equal to $-\infty$ on a dense subset of $\partial C$.

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  • $\begingroup$ I see your point, but isn't this similar to the regularity issue raised by @Mateusz Kwasnicki? For the sake of argument we could restrict to a subclass of subharmonic $\varphi$ which are well-behaved enough to have a chance to be harmonic on $\partial C$. I probably should edit the question ;-) $\endgroup$ – MartinB Sep 18 '19 at 16:24
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    $\begingroup$ It is similar. I just wanted to strengthen his answer to: NO such C exist. It is difficult to define in this context what does it mean "well behaved enough", because harmonic functions (unlike subharmonic ones) are not only continuous but in fact analytic. $\endgroup$ – Alexandre Eremenko Sep 18 '19 at 16:31
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There is no such "nice" $C$ (other than $\Omega$ and $\varnothing$) in dimensions other than $1$.

Suppose otherwise, that a non-empty, smooth, open set $C$ in $\mathcal{A}(\Omega)$ exists. If $C \ne \Omega$, then there is $x_0 \in \Omega \cap \partial C$. If $h_\varphi$ exists, then $\varphi$ is equal to a real-analytic function $h_\varphi$ on $\partial C \cap B(x_0, \varepsilon)$. This is clearly not the case for the subharmonic function $\varphi(x) = |x - x_0|$ (given that the boundary of $C$ is sufficiently smooth near $x_0$).

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    $\begingroup$ I really like your argument, thank you! However, maybe the problem is just not well-posed. If we restrict to a subclass of test functions, e.g. real-analytic subharmonic functions, there wouldn't be a problem with regularity, right? Or is there a more fundamental argument that separates the one-dimensional case from the higher-dimensional one? $\endgroup$ – MartinB Sep 18 '19 at 11:44
  • $\begingroup$ @MartinB: Not sure if this changes the answer. In this case regularity indeed is no longer an issue. Also, subharmonicity stops playing an essential role: for any real-analytic $\varphi$, the function $\varphi(x) - M |x|^2$ is subharmonic if $M$ is large enough. So the question essentially becomes: when a real-analytic function on a curve extends to a harmonic function in the neighbourhood of the curve? $\endgroup$ – Mateusz Kwaśnicki Sep 18 '19 at 12:41
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The answer to the new question is still no.

Let us identify $\mathbb{R}^2$ with $\mathbb{C}$. Consider $\varphi(z) = \log |\sin z|$ for $z \in \mathbb{C}$. This is clearly a subharmonic function. Let $\Omega = B(0, \pi)$ be a disk of radius $\pi$, and let $C = \{z \in B(0, \tfrac{\pi}{2}) : \varphi(z) \leqslant -\tfrac{1}{2}\}$. Clearly, $h_\varphi(z) = -\tfrac{1}{2}$ for every $z \in \Omega$. However, it is not true that $\varphi(z) \geqslant -\tfrac{1}{2}$ for all $z \in \Omega \setminus C$: $\varphi(z)$ goes to $-\infty$ when $z \to -\tfrac{\pi}{2}$.

Here is a 3-D plot of $\varphi$:

3-D plot of phi

And a contour plot of the level line $\varphi = -\tfrac{1}{2}$:

Contour plot of a level line of phi

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