# Intersection of Subharmonic and Harmonic Functions (also: Extension of harmonic functions)

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.

## 3 Answers

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 $$\partial C$$.

• 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 ;-) – MartinB Sep 18 '19 at 16:24
• 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. – Alexandre Eremenko Sep 18 '19 at 16:31

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 $$x_0$$).

• 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? – MartinB Sep 18 '19 at 11:44
• @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? – Mateusz Kwaśnicki Sep 18 '19 at 12:41

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