Prove that $\log(\max(\left | z_1 \right |,\left | z_2 \right |,\ldots,\left | z_n \right |)) \in MPSH(\Omega)$

Question

This's an example:

For $u(z_1,z_1,\ldots,z_n)=\log(\max(\left | z_1 \right |,\left | z_2 \right |,\ldots,\left | z_n \right |))$, where $z=(z_1,z_1,\ldots,z_n) \in \Omega=\mathbb{C}^n \setminus\{0\} $. Prove that $u \in MPSH(\Omega)$.

And here's the solution in my textbook:

$\bullet$ Suppose that $v \in PSH(\Omega)$, $G \subset \subset \Omega$ and $v \le u$ on $\partial G$. For all $z \in G$, we consider an open set $A=\{t \in \mathbb{C}: tz \in G\}$ in $\mathbb{C}$.

We have $u(tz)$ and $v(tz)=\log\left | t \right |+\log(\max(\left | z_1 \right |,\left | z_2 \right |,\ldots,\left | z_n \right |)) \in PSH(\Omega), \forall z=(z_1,z_1,\ldots,z_n) $.

Moreover, $\partial A \subset \partial G$. Hence, $v(tz)|_{\partial A} \le u(tz)|_{\partial A}$.

Because $u$ be a harmonic then $v(tz)|_{A} \le u(tz)|_{A}$.

Therefore, $v(z) \le u(z)$ in $G$.////

But I don't understand the solution above. Ex Why we have $\partial A \subset \partial G$? Why $v(tz)|_{\partial A} \le u(tz)|_{\partial A}$?

Can anyone help me (explain clearly)! Thanks.

Answer 1

The statement $\partial A \subset \partial G$ makes no sense. ($A \subset \mathbb{C}$ but $G \subset \mathbb{C}^n$). But if $z \in G$ is fixed and $t\in\partial A$, then $tz \in \partial G$, which is probably what is meant.

By assumption, $v \le u$ on $\partial G$, so $v(tz) \le u(tz)$ for $t \in \partial A$. Finally $t \mapsto v(tz)$ is subharmonic and $t \mapsto u(tz) = \log|t| + \log\{\max|z_1|,\ldots,|z_n|\}$ is harmonic on $\mathbb{C}\setminus\{0\}$. By definition of subharmonicity, $v(tz) \le u(tz)$ for $t \in A$. In particular, $1 \in A$, so $v(z) \le u(z)$. But since $z \in G$ was arbitrary, $v \le u$ on $G$, showing that $u$ is indeed maximal.

Answer 2

Solution:

$\bigstar$ We'll prove that $u \in MPSH(\Omega)$, where $\Omega=\Bbb{C}^n \setminus \{0\}$.

• We have$$\begin{align*} \\ \log \left |z_j \right |\in PSH(\Omega), \forall j=\overline{1,n} &\implies \max\left \{ \log \left |z_j \right | \right \}\in PSH(\Omega),\forall j=\overline{1,n}\\ &\implies u=\log\left ( \max\left \{ \left | z_1 \right |,\ldots,\left | z_n \right | \right \} \right )\in PSH(\Omega) \end{align*}$$ $\bigstar$ With $v \in PSH(\Omega)$, $G \subset \subset \Omega, v$ is upper - semicontinuous in $\overline{G}$ and $v \le u$ on $\partial G$. We'll prove that $v \le u $ in $G$.

• Take $\omega \in G$ was arbitrary. We consider $A=\left \{ t\in \Bbb C: t\omega \in G \right \}$.

We have $\forall t_0 \in A\implies t_0\omega\in G\implies \exists B_{\Bbb C^n}(t_0\omega,r)\subset G$.

On the other hand, $$\forall t \in B_{\Bbb C}\left ( t_0,\dfrac{r}{\left |\omega \right |} \right )\implies \left | t-t_0 \right |\left | \omega \right |<r \implies \left | t\omega - t_0\omega \right |<r\implies t\omega\in G \implies t \in A$$

So, $t_0\in B_{\Bbb C}\left ( t_0,\dfrac{r}{\left |\omega \right |} \right )\subset A$.

Therefore, $A$ is an open set.

• We have $v(t\omega) \in SH(A)$, because $v \in PSH(\Omega) $

• We have $u(t\omega)\in SH(A)$. Indeed:

We have $$\begin{align*} \\ u(t\omega)&=\log\left | t \right |+\log\left ( \max\left \{ \left | \omega_1 \right |,\ldots, \left | \omega_n \right |\right \} \right )\\&=Re \log t + \log\left ( \max\left \{ \left | \omega_1 \right |,\ldots, \left | \omega_n \right |\right \} \right )\in SH(A)\end{align*}$$

• We'll show that $v(t\omega) \le u(t\omega), \forall t \in \partial A$.

Indeed, we take $t \in \partial A\implies \exists \left \{ t_n \right \}\subset A, t_n \to t \in \Bbb C\implies t_n \omega \to t\omega \in \Bbb C^n$.

Moreover, because $\left \{ t_n \omega \right \}\subset G \implies t \omega \in \overline{G}\overset{t \omega \notin {G}}{\rightarrow}t \omega \in \partial G$.

Therefore, $v(t\omega) \le u(t\omega), \forall t \in \partial A$.

Now, we let $$h(t):=v(t\omega)-u(t\omega)\in SH(A)$$ We have $h$ is upper - semicontinuous in $\overline{A}$ - compact. So, $\exists t_0 \in \overline{A}$ such that $$h(t_0)=\max_{t \in \overline{A}} h(t)$$

• If $t_0 \in A \implies h\mid_\Omega=0$

• If $t_0 \in \partial A \implies \lim_{t \to \xi} \sup h(t) \le h(t_0), \forall \xi \in \partial A \implies h \le 0$ in $A \implies v(t \omega ) \le u (t \omega )$ in $A$.

In particular, $1 \in A$, we take $t=1$ then $v(\omega ) \le u (\omega )$.

However, since $\omega \in G$ was arbitrary, $v \le u$ on $G$, showing that $u$ is indeed maximal $\blacksquare $.

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@mrf: My solution is correct?

Can you help me show that $u \notin MPSH(\Bbb C^n)$? Thanks!