Harmonic lifts of subharmonic functions are subharmonic.

Question

Let $U$ be the harmonic lift of a subharmonic function $u$ on $B \Subset \Omega \subset_{\mathrm {domain}} \Bbb R^n$. Then $$ U(x) = \left\{ \begin{array}{ll} \bar {u} (x) & \quad x \in B \\ u(x) & \quad x \in \Omega \setminus B \end{array} \right. $$

where $\bar {u} (x)$ is harmonic on $B$ such that $\bar {u} = u$ on $\partial B$. So by subharmonicity of $u$ it follows that $u \le \bar {u}$ on $B$. So $u \le U$ on $\Omega$. Now to prove that $U$ is subharmonic we first take an open ball $B' \Subset \Omega$ and a harmonic function $h$ on $B'$ which is continuous on $\bar B'$ and $U \le h$ on $\partial B'$. We need to show that $U \le h$ on $B'$ in order to reach at the desired conclusion.

To do this we first note that $u \le U$ on $\Omega$ and hence $u \le U$ on $\partial B'$. But $U \le h$ on $\partial B'$ according to our assumption. So $u \le h$ on $\partial B'$. Since $h$ is harmonic on $B'$ so by subharmonicity of $u$ it follows that $u \le h$ on $B'$. So $U \le h$ on $B' \setminus B$ since $B' \setminus B \subset \Omega \setminus B$. Now if we can show that $U \le h$ on $B \cap B'$ then we are through. So if $B \cap B' = \emptyset$ we have nothing left for the proof. But how can I tackle the case when $B \cap B' \ne \emptyset$? My intuition says that somehow we have to use Maximum Principle here but I couldn't manage to find the way to apply this theorem in this case.

Would anybody kindly help me in overcoming this problem?

Thanks a lot in advance.

Answer 1

You have done almost everything. The remaining part can be tackled using set theory.

First note that $\overline {B \cap B'} \subset \overline B \cap \overline B' \subset \overline B'$. So

$\partial (B \cap B') = \overline {B \cap B'} \setminus (B\cap B') \subset {\overline {B'}} \setminus {(B \cap B')} = (B' \cup \partial B') \setminus (B \cap B') \subset (B' \setminus (B \cap B')) \cup (\partial B' \setminus (B \cap B')) \subset (B' \setminus B) \cup \partial B'.$

(Reason for the last inclusion $:$ Since for any two set $A$ and $B$ we know that $A \setminus (A \cap B) = A \setminus B$ and $A \setminus B \subset A$).

Now you have proved that $U \le h$ on $B' \setminus B$ and according to your hypothesis $U \le h$ on $\partial B'$. So $U \le h$ on $(B' \setminus B) \cup \partial B'$ and hence on $\partial (B \cap B')$. Now $U$ is harmonic on $B$ so is on $B \cap B'$. So by Maximum Principle as you have rightly guessed we have $U \le h$ on $B \cap B'$ and then as you have mentioned we are through.

Answer 2

Note that subharmonicity ( like harmonicity) is a local condition. That is: if there exists an open cover of the domain of the function such that the restriction to each piece is subharmonic, then the function is subharmonic. ( in fact, if the function is $C^2$, then the condition is $\Delta f\ge 0$)

It's better to use the following characterization of subharmonic functions: $v$ is subharmonic if for every $x$ in the domain there exists $R>0$ so that the average of $u$ over any ball centered at $x_0$ of radius $\rho < R$ is $\ge v(x_0)$. Conversely, if $u$ is subharmonic, the the average over any ball inside the domain is $\ge $ value at the center.

So let's show this property for $U$. Take a point $x_0$ in $\Omega$.

If the point is outside $B$, we can take $R= d(x_0, B)$ and notice that around $x_0$ $U$ equals $u$.

If the point is in the interior of $B$, around $x_0$ the function is $\bar u$, harmonic, so it has the mean value property ( with equality).

The only problem seems to come from points $x_0$ on the boundary. Take any ball centered at $x_0$. Since $\bar u\ge u$ on $B$ we see that the average of $U$ on the ball $\ge $ average of $u$ on that ball. But that is $\ge u(x_0)$. So we checked the average property also for points on the boundary.

We are done.