MHB How do we apply the proposition?

[evinda]

Hello! (Wave)

I am reading the proof that if the function $v$ is subharmonic in $\Omega$ then $ H_{B_0}[v]$ is also subharmonic in $\Omega$.

($B_0 $ is a ball in $\Omega$)

(We say that the function $v$ is subharmonic in $\Omega$ if for every ball $B \subset \Omega$ it holds that $v \leq H_B[v]$.)

$H_B[v]$ is defined as follows:

$$\\v(x) \in C^0(\Omega), B \subset \Omega \text{ arbitrary ball} \\ \\
H_B[v]=\left\{\begin{matrix}
\text{harmonic for } & x \in B\\
v \text{ for } & x \in \Omega \setminus{B}
\end{matrix}\right.$$

We have to show that for each ball $B\subset \Omega $ it holds that $w \leq H_B[w]$ where $w=H_{B_0}[v]$.

So we compare the functions
$ w(x)=\left\{\begin{matrix}
\text{ harmonic} & , x \in B_0\\
v & , x \in \Omega \setminus{B_0}
\end{matrix}\right.$

and$H_B[w](x)=\left\{\begin{matrix}
\text{ harmonic } & , x \in B\\
w & , x \in \Omega \setminus{B}
\end{matrix}\right.$

where $v$ is a subharmonic function.

We distinguish cases.At the case $ B_0 \subset B $:

in $\Omega \setminus{B}$ we have that $ H_B[w]=H_{B_0}[v]$.

In $B$ we have that $H_B[w]$ is harmonic and $w$ is subharmonic (since $w$ is harmonic in $B_0$ and $w=v$ in $B\setminus{B_0}$, so it is subharmonic).

So we have that $ w-H_B[w]$ is subharmonic, and so

$ w-H_B[w]|_{\partial{B}}=0 \Rightarrow w-H_B|_B=0 $

We get this from the proposition $(\star)$.

$(\star)$: The subharmonic in $\Omega$ function does not achieve its maximum in the inner points of $\Omega$ if it is not constant.We apply the proposition for $\Omega=B$. But do we have an inner point of $B$ in which $w-H_B[w]$ achieves its maximum, in order to come to the conclusion that $w-H_B|_B=0 $? (Thinking)

 

[I like Serena]

We apply the proposition for $\Omega=B$. But do we have an inner point of $B$ in which $w-H_B[w]$ achieves its maximum, in order to come to the conclusion that $w-H_B|_B=0 $? (Thinking)

Hey evinda! (Smile)

If there would be a maximum in an inner point of $B$, it follows that $w-H_B[w]$ is constant, and since it's zero at the boundary, that maximum would be $0$.
Still, I don't get it yet, because it seems to me there could still be a minimum that is less than $0$.

 

[evinda]

In order to apply the proposition don't we need an additional inequality for $w-H_B[w]$ if we are not given that the maximum or minimum is achieved at an internal point? (Sweating)

 

[evinda]

In some other notes, there is the following proof:

• $B \cap B_0=\varnothing$
  
  $$H_B[w]|_{\partial{B}}=w|_{\partial{B}} \Rightarrow H_B[w]=w \text{ in } B$$
  
  (2 harmonic functions are equal at the boundary and so they are equal at the whole space)My question about this: How can we use the proposition above although the functions $w$ and $H_B[w]$ are harmonic in different spaces?
  
• $B \subset B_0$: in $\Omega \subset B$ we have $H_B[w]=H_{B_0}[v]$
  in $B$ we have that $H_B[w], H_{B_0}[v]$ are harmonic
  in $\partial{B}$ $H_B[w] \geq H_{B_0}[v]$In this case, I think that we have $H_{B}[w]=w$ in $\partial{\Omega}$. Am I wrong?
  
• $B \not\subset B_0$ and $B \cap B_0 \neq \varnothing$
  
  if $x \in B \setminus{B_0}$ we have that $w=H_{B_0}[v]=v \leq H_B[v] \leq H_B[w]$
  
  if $x \in B \cap B_0$ we have that $w$ and $H_B[w]$ are harmonic
  in $\partial{B \cap B_0}$ it holds that $w \leq H_B[w] \Rightarrow w \leq H_B[w]$ in $B \cap B_0$Why do we have that in $\partial{B \cap B_0}$ it holds that $w \leq H_B[w] $ ?
• $x \in \Omega \setminus{ B \cup B_0}$: $w=H_B[w]$
  
  Do we get this from the definition of $H_B[w]$?
  

 

[evinda]

Hey evinda! (Smile)

If there would be a maximum in an inner point of $B$, it follows that $w-H_B[w]$ is constant, and since it's zero at the boundary, that maximum would be $0$.
Still, I don't get it yet, because it seems to me there could still be a minimum that is less than $0$.

From the maximum prinviple, we get that either the maximum is achieved at the boundary and so it is equal to $0$ or the function is constant. Right?

So don't we get that $w-H_B[w] \leq 0$? (Thinking)