Maximum principle for subharmonic functions

[evinda]

Hello! (Wave)

I have a question about the proof of the maximum principle for subharmonic functions.

The maximum principle is the following: The subharmonic in $\Omega$ function $v$ does not achieve its maximum at the inner points of $\Omega$ if it is not constant.

Proof: We suppose that at the point $x_0 \in \overline{\Omega} \setminus{\partial{\Omega}}$ the subharmonic function $v$ achieves its maximum. We consider the ball $B_{x_0}(\rho) \subset \Omega$ ($\rho< \delta=dist(x_0, \partial{\Omega})$) and the function $w=H_{B_{x_0}(\rho)}[v]$.

We have $\max w|_{\partial{B_{x_0}(\rho)}}=\max v|_{\partial{B_{x_0}(\rho)}} \leq v(x_0)$

($w$ and $v$ coincide in $\Omega \setminus{B_{x_0}(\rho)}$, in $x_0$ $v$ has its maximum ).

Taking into consideration that $v(x_0) \leq H_{B_{x_0}(\rho)}[v](x_0)=w(x_0)$, we have for harmonic $w(x)$

$$\max w|_{\partial{B_{x_0}(\rho)}} \leq w(x_0)$$

so (from the maximum principle for harmonic functions) $w(x)$ is constant in $B_{x_0}(\rho)$ and consequently $v$ is constant at its boundary $\partial{B_{x_0}(\rho)}$.

We notice that $\rho$ is arbitrary between $0$ and $dist(x_0, \partial{\Omega})$, by changing $\rho$ at the interval $(0,\delta)$ we deduce that $v$ is constant at the ball $B_{x_0}(\delta)$.

The fact that $v$ is constant everywhere in $\Omega$ can be proven as in the case of harmonic functions.

First of all, $w$ is not harmonic in $\partial{B_{x_0}(\rho)}$. Why can we say the following?

we have for harmonic $w(x)$

$$\max w|_{\partial{B_{x_0}(\rho)}} \leq w(x_0)$$Secondly, how do we deduce that $v$ is constant at the ball $B_{x_0}(\delta)$ ?

Also how can we show that $v$ is constant everywhere in $\Omega$ ?

 

[I like Serena]

Hey evinda! (Smile)

and the function $w=H_{B_{x_0}(\rho)}[v]$.

What is $H$? (Wondering)

First of all, $w$ is not harmonic in $\partial{B_{x_0}(\rho)}$.

How so?
Since $w$ coincides with $v$ in the mini ball, and since $v$ is harmonic (domain is an open set and $\Delta v=0$), doesn't that imply that $w$ is harmonic as well?

Why can we say the following?

we have for harmonic $w(x)$

$$\max w|_{\partial{B_{x_0}(\rho)}} \leq w(x_0)$$

Since $w$ coincides with $v$ in the ball, and since $v(x_0)$ is a global maximum, doesn't it follow that it holds true for $w$ as well on a subset of the domain? (Wondering)

Secondly, how do we deduce that $v$ is constant at the ball $B_{x_0}(\delta)$ ?

Also how can we show that $v$ is constant everywhere in $\Omega$ ?

If we pick some other point that we can connect with a curve to $x_0$, we can create a string of mini balls, and repeat the same argument for every mini ball.
If there's a point $y \in \Omega$ that we can't connect to $x_0$, that means that $\Omega$ is not connected.
Still, we can repeat the argument with a new maximum $y_0$ within the subset of $\Omega$ that is connected to $y$, so in that open subset we'll have a constant function as well. (Nerd)

 

[evinda]

What is $H$? (Wondering)

In general, we have the following:

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

How so?
Since $w$ coincides with $v$ in the mini ball, and since $v$ is harmonic (domain is an open set and $\Delta v=0$), doesn't that imply that $w$ is harmonic as well?

We have that $w$ coincides with $v$ in $\Omega \setminus{B_{x_0}(\rho)}$.
How do we know that $v$ is harmonic?

We have that a function $v(x)$ is subharmonic in $\Omega$ if $v \leq H_B[v], \forall B \subset \Omega$.

Since $w$ coincides with $v$ in the ball, and since $v(x_0)$ is a global maximum, doesn't it follow that it holds true for $w$ as well on a subset of the domain? (Wondering)

I understood how we got the inequality, I just didn't understand why we can say for harmonic w...

If we pick some other point that we can connect with a curve to $x_0$, we can create a string of mini balls, and repeat the same argument for every mini ball.

And we use the fact that $v$ is constant in $\partial{B_{x_0}(\rho)}$ ?

If there's a point $y \in \Omega$ that we can't connect to $x_0$, that means that $\Omega$ is not connected.
Still, we can repeat the argument with a new maximum $y_0$ within the subset of $\Omega$ that is connected to $y$, so in that open subset we'll have a constant function as well. (Nerd)

So you mean that if $\Omega$ is not connected, then we will have local maxima?
If so, would it hold that $x_0>y_0$ ?
Why will we have a constant function as well?

 

[I like Serena]

In general, we have the following:

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

We have that $w$ coincides with $v$ in $\Omega \setminus{B_{x_0}(\rho)}$.
How do we know that $v$ is harmonic?

We have that a function $v(x)$ is subharmonic in $\Omega$ if $v \leq H_B[v], \forall B \subset \Omega$.

Ah... ok. (Thinking)

I understood how we got the inequality, I just didn't understand why we can say for harmonic w...

I think we can leave out the word harmonic of that sentence.
It seems to me it only means that $w$ is harmonic in the mini ball, which is implicit from $w=H_{B_{x_0}(\rho)}[v]$.

And we use the fact that $v$ is constant in $\partial{B_{x_0}(\rho)}$ ?

Well... we use that we found that $v$ is constant in ${B_{x_0}(\rho)}$... (Thinking)

So you mean that if $\Omega$ is not connected, then we will have local maxima?
If so, would it hold that $x_0>y_0$ ?
Why will we have a constant function as well?

Indeed, we will have a local maximum.
We would have $x_0 \ge y_0$.
If we only look at the connected subset of $\Omega$ that contains $x_0$, and if we can prove that $v$ has to be constant, the same reasoning applies to this other subset with local maximum $y_0$, meaning $v$ has to be constant there as well, although possibly at a different constant value. (Thinking)

 

[evinda]

Ah... ok. (Thinking)
I think we can leave out the word harmonic of that sentence.
It seems to me it only means that $w$ is harmonic in the mini ball, which is implicit from $w=H_{B_{x_0}(\rho)}[v]$.
Well... we use that we found that $v$ is constant in ${B_{x_0}(\rho)}$... (Thinking)

Ok... (Nod)

Indeed, we will have a local maximum.
We would have $x_0 \ge y_0$.
If we only look at the connected subset of $\Omega$ that contains $x_0$, and if we can prove that $v$ has to be constant, the same reasoning applies to this other subset with local maximum $y_0$, meaning $v$ has to be constant there as well, although possibly at a different constant value. (Thinking)

If we pick some other point that we can connect with a curve to $x_0$, we can create a string of mini balls, and repeat the same argument for every mini ball.

But how do we show that at the other mini balls $v$ is also constant? (Thinking)

 

[I like Serena]

Ok... (Nod)

But how do we show that at the other mini balls $v$ is also constant? (Thinking)

We have proven that $v$ is constant at $x_0$ in the first mini ball and on its edge.
We pick the second mini ball such that it overlaps with the first mini ball.
So we have a point at $x_0$ in the second mini ball, which will again be the maximum of $v$.
According to the same argument, $v$ must be constant in all of the second mini ball.
And so on... (Thinking)

The fact that $v$ is constant everywhere in $\Omega$ can be proven as in the case of harmonic functions.

Alternatively, it appears you already have a similar proof elsewhere (in the case of harmonic functions).
What is it? (Wondering)

 

[evinda]

We have proven that $v$ is constant at $x_0$ in the first mini ball and on its edge.
We pick the second mini ball such that it overlaps with the first mini ball.
So we have a point at $x_0$ in the second mini ball, which will again be the maximum of $v$.
According to the same argument, $v$ must be constant in all of the second mini ball.
And so on... (Thinking)

So you mean that we pick a ball $B_x(\epsilon) \subset \Omega$ such that $B_{x_0}(\delta) \subset B_x(\epsilon)$ ?

Which argument do you mean? (Thinking)

Alternatively, it appears you already have a similar proof elsewhere (in the case of harmonic functions).
What is it? (Wondering)

I think the proof of the maximum principle of harmonic functions is meant.

Maximum principle: The harmonic in $\Omega$ function $v$ cannot achieve neither its maximum nor its minimal value at the inner points of $\Omega$ if it is not constant.

Proof: We suppose that $u$ achieves its maximum in $x_0 \in \Omega$, $u(x_0)=M$.
Without loss of generality we can suppose that $u>0$.
We have that $$u(x_0)=\frac{1}{w_n \rho^{n-1}} \int_{|\xi-x_0|=\rho} u ds \ \forall \rho>0 \text{ if } |\xi-x_0|\leq\rho \subset \Omega$$

We suppose that $\exists \xi_0 \in \{|\xi-x_0|=\rho\}$ such that $u(\xi_0)<u(x_0)$, then $u(\xi)<u(x_0)$ in some region $\partial{\Omega_{\epsilon}} \subset \{ |\xi-x_0|=\rho\}$ ($\xi_0 \in \partial{\Omega_{\epsilon}}$).

So $M=u(x_0)=\frac{1}{w_n \rho^{n-1}} \int_{\partial{\Omega_{\epsilon}}} u ds+\frac{1}{w_n \rho^{n-1}} \int_{\{|\xi-x_0=\rho \}\setminus{\partial{\Omega_{\epsilon}}}}u ds< \frac{1}{w_n \rho^{n-1}}\int_{|\xi-x_0|=\rho}M ds=M$, i.e. $M<M$ , contradiction.

Consequently $u(\xi)=u(x_0) \ \forall \xi \in \{ |\xi-x_0|=\rho \}$. Since $\rho$ is arbitrary from the interval $(0,d), d=dist(x_0, \partial{\Omega})$, we have that $u(x)=u(x_0)M \ \forall x \in \{ |x-x_0|<d\}$.

We pick now an arbitrary $y \in \Omega$. We will prove that $u(y)=u(x_0)$. We pick a curve $l$ that is in $\Omega$ and connects the points $x_0$ and $y$. Let $\overline{d}=dist(l,\partial{\Omega})$. We cover the curve $l$ with a finite number of spaces $B_i=\{ |x-x^i|< \frac{\overline{d}}{2}\}, i=0,\dots,m$ ( balls with radius $\overline{d}$ and center at $x^i$ ) where $x^i \in l\ \forall i$ and $y \in B_m$. We have $u \equiv M$ in the spaces $B_1, B_2, \dots, B_m$ so $u(y)=M$.

First of all, why do we pick the balls $B_i=\{ |x-x^i|< \frac{\overline{d}}{2}\}, i=0,\dots,m$ ?
And how do we deduce that $u \equiv M$ in the spaces $B_1, B_2, \dots, B_m$ ?

Btw... I haven't understood how we get the inequality at the following part, and so the contradiction.

$M=u(x_0)=\frac{1}{w_n \rho^{n-1}} \int_{\partial{\Omega_{\epsilon}}} u ds+\frac{1}{w_n \rho^{n-1}} \int_{\{|\xi-x_0=\rho \}\setminus{\partial{\Omega_{\epsilon}}}}u ds< \frac{1}{w_n \rho^{n-1}}\int_{|\xi-x_0|=\rho}M ds=M$

 

[I like Serena]

So you mean that we pick a ball $B_x(\epsilon) \subset \Omega$ such that $B_{x_0}(\delta) \subset B_x(\epsilon)$ ?

We pick $B_x(\epsilon) \subset \Omega$ with$B_{x_0}(\delta) \cap B_x(\epsilon) \ne \varnothing$. (Thinking)

Which argument do you mean? (Thinking)

Since $v$ takes a maximum value of $v(x_0)$ somewhere in $B_x(\epsilon)$, it must be constant in $B_x(\epsilon)$.

We pick now an arbitrary $y \in \Omega$. We will prove that $u(y)=u(x_0)$. We pick a curve $l$ that is in $\Omega$ and connects the points $x_0$ and $y$. Let $\overline{d}=dist(l,\partial{\Omega})$. We cover the curve $l$ with a finite number of spaces $B_i=\{ |x-x^i|< \frac{\overline{d}}{2}\}, i=0,\dots,m$ ( balls with radius $\overline{d}$ and center at $x^i$ ) where $x^i \in l\ \forall i$ and $y \in B_m$. We have $u \equiv M$ in the spaces $B_1, B_2, \dots, B_m$ so $u(y)=M$.

First of all, why do we pick the balls $B_i=\{ |x-x^i|< \frac{\overline{d}}{2}\}, i=0,\dots,m$ ?
And how do we deduce that $u \equiv M$ in the spaces $B_1, B_2, \dots, B_m$ ?

Turns out it's the same argument as I was already giving, just more formalized.
Btw, apparently it's already assumed that $\Omega$ is a connected open set, since it says that we can pick a curve that connects $x_0$ to an arbitrary point $y$.
We make a string of overlapping mini balls on the curve, and from the fact that $v(x)=v(x_0)$ for every point in the first ball, we must have a point in the second ball that also has the value of $v(x_0)$, which must still be the maximum. Therefore it's constant as well. And so on. (Nerd)

Btw... I haven't understood how we get the inequality at the following part, and so the contradiction.

$M=u(x_0)=\frac{1}{w_n \rho^{n-1}} \int_{\partial{\Omega_{\epsilon}}} u ds+\frac{1}{w_n \rho^{n-1}} \int_{\{|\xi-x_0=\rho \}\setminus{\partial{\Omega_{\epsilon}}}}u ds< \frac{1}{w_n \rho^{n-1}}\int_{|\xi-x_0|=\rho}M ds=M$

We assumed that $u$ was not constant with maximum $u(x_0)$ and that there was a point $\xi_0$ on the boundary with a lower value $u(\xi_0)$.
If there is, there must be a mini subset around it where $u(\xi) < u(x_0)$, while at the remainder of the boundary we only know $u(\xi) \le u(x_0)$.
Integrating it, we can make an upper estimate for the first integral that is $<M=u(x_0)$, and for the second integral that is $\le M=u(x_0)$. Together they give an estimate that is $<M$. (Thinking)

 

[evinda]

We pick $B_x(\epsilon) \subset \Omega$ with$B_{x_0}(\delta) \cap B_x(\epsilon) \ne \varnothing$. (Thinking)
Since $v$ takes a maximum value of $v(x_0)$ somewhere in $B_x(\epsilon)$, it must be constant in $B_x(\epsilon)$.
Turns out it's the same argument as I was already giving, just more formalized.
Btw, apparently it's already assumed that $\Omega$ is a connected open set, since it says that we can pick a curve that connects $x_0$ to an arbitrary point $y$.
We make a string of overlapping mini balls on the curve, and from the fact that $v(x)=v(x_0)$ for every point in the first ball, we must have a point in the second ball that also has the value of $v(x_0)$, which must still be the maximum. Therefore it's constant as well. And so on. (Nerd)

I see... But could you maybe explain to me why we take these $B_i$ ?

Integrating it, we can make an upper estimate for the first integral that is $<M=u(x_0)$, and for the second integral that is $\le M=u(x_0)$. Together they give an estimate that is $<M$. (Thinking)

So we have that $\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} u ds \leq M w_n \rho^{n-1}$, right?

 

[I like Serena]

I see... But could you maybe explain to me why we take these $B_i$ ?

To prove that $u$ is not only constant and equal to $u(x_0)$ in $B_\epsilon(x_0)$ but also in any point $y\in\Omega$ that can be connected to $x_0$. (Thinking)

So we have that $\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} u ds \leq M w_n \rho^{n-1}$, right?

More accurately it's:
$$\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} u ds \leq M \int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds$$
(Thinking)

 

[evinda]

To prove that $u$ is not only constant and equal to $u(x_0)$ in $B_\epsilon(x_0)$ but also in any point $y\in\Omega$ that can be connected to $x_0$. (Thinking)

Yes, I see... But I haven't understood why in order to do so we take the balls with center $x^i \in l$ and radius $\frac{\overline{d}}{2}$, where $\overline{d}=dist(l, \partial{\Omega})$. (Worried)

More accurately it's:
$$\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} u ds \leq M \int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds$$
(Thinking)

I see... And $\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds=w_n \rho^{n-1}$, right?

 

[I like Serena]

Yes, I see... But I haven't understood why in order to do so we take the balls with center $x^i \in l$ and radius $\frac{\overline{d}}{2}$, where $\overline{d}=dist(l, \partial{\Omega})$. (Worried)

$\overline d$ is the smallest distance that any point of $l$ has to the boundary.
So balls with their center on $l$ and with radius $\frac{\overline{d}}{2}$ will be completely inside $\Omega$.
We pick $x^i \in l$ such that the balls together "cover" the curve $l$.
That is, each of the consecutive balls overlap with each other. (Thinking)

I see... And $\int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds=w_n \rho^{n-1}$, right?

Not quite.
We have:
$$\int_{\partial{\Omega_{\epsilon}}} ds + \int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds=w_n \rho^{n-1}$$
(Thinking)

 

[evinda]

$\overline d$ is the smallest distance that any point of $l$ has to the boundary.
So balls with their center on $l$ and with radius $\frac{\overline{d}}{2}$ will be completely inside $\Omega$.
We pick $x^i \in l$ such that the balls together "cover" the curve $l$.

I see. (Nod)

That is, each of the consecutive balls overlap with each other. (Thinking)

How do we see that each of the consecutive balls overlap with each other?

We have for example $B_1=\{ |x-x^1|< \frac{\overline{d}}{2}\}$ and $B_2=\{ |x-x^2|< \frac{\overline{d}}{2}\}$.
How do we deduce that $B_1 \cap B_2 \neq \varnothing$?

Not quite.
We have:
$$\int_{\partial{\Omega_{\epsilon}}} ds + \int_{\{ |\xi-x_0|=\rho\} \setminus{\partial{\Omega_{\epsilon}}}} ds=w_n \rho^{n-1}$$
(Thinking)

Ah, I see... (Smile)

 

[I like Serena]

How do we see that each of the consecutive balls overlap with each other?

We have for example $B_1=\{ |x-x^1|< \frac{\overline{d}}{2}\}$ and $B_2=\{ |x-x^2|< \frac{\overline{d}}{2}\}$.
How do we deduce that $B_1 \cap B_2 \neq \varnothing$?

By picking $x^2$ such that $|x^2-x^1|< \overline{d}$.
If 2 balls are closer together than the sum of their radius's, they overlap. (Thinking)

 

[evinda]

By picking $x^2$ such that $|x^2-x^1|< \overline{d}$.
If 2 balls are closer together than the sum of their radius's, they overlap. (Thinking)

I see... But we have to mention that we pick the balls $B_i=\{ |x-x^i|\}< \frac{\overline{d}}{2} \}$ such that $|x^{i+1}-x^i|< \overline{d}, i=0, \dots,m-1$, don't we? (Thinking)

 

[I like Serena]

I see... But we have to mention that we pick the balls $B_i=\{ |x-x^i|\}< \frac{\overline{d}}{2} \}$ such that $|x^{i+1}-x^i|< \overline{d}, i=0, \dots,m-1$, don't we? (Thinking)

In the proof it says: 'We cover the curve ...'.
That's what the word 'cover' means.
So it's already there. (Smirk)

 

[evinda]

In the proof it says: 'We cover the curve ...'.
That's what the word 'cover' means.
So it's already there. (Smirk)

I see... Thank you very much! (Smile)