# 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.

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 ;-) Commented Sep 18, 2019 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. Commented Sep 18, 2019 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? Commented Sep 18, 2019 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? Commented Sep 18, 2019 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 $$\varphi = -\tfrac{1}{2}$$: