Convergence of $\sum_{\alpha\in \Bbb N^n} \frac{z^\alpha}{\xi(\alpha)}$ on $\operatorname{int} U$ where $U \subset\Bbb C^n$ is bounded

Question

Problem. Let $U\subset \Bbb C^n$ be bounded. For $\alpha\in \Bbb N^n$ and $z\in \Bbb C^n$, let $z^\alpha:= z_1^{\alpha_1} z_2^{\alpha_2} \ldots z_n^{\alpha_n} \in \Bbb C$. Prove that $$\sum_{\alpha\in \Bbb N^n} \frac{z^\alpha}{\xi(\alpha)}$$ converges for every $z\in \operatorname{int} U$, where $\xi(\alpha) := \sup_{z\in U} |z^\alpha|$.

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The convergence of power series in several complex variables is defined in the following way:

Let $\{c_\alpha\}_{\alpha\in\Bbb N^n} \subset\Bbb C$. The power series $\sum_{\alpha\in \Bbb N^n}c_\alpha (z-a)^\alpha = \sum_{\alpha\in \Bbb N^n} c_\alpha (z_1-a_1)^{\alpha_1} \ldots (z_n-a_n)^{\alpha_n}$ in $n$ variables $z:= (z_1,z_2,\ldots,z_n)$ centered at $a\in \Bbb C^n$ converges at $w\in \Bbb C^n$ if there exists $M> 0$ such that $$\sum_{\alpha\in F} |c_\alpha| |(w-a)^\alpha| \le M$$ for all finite subsets $F\subset\Bbb N^n$.

My work. Since $z\in \operatorname{int} U$ (an open set), there exists $w \in \operatorname{int} U$ such that $|w_j| > |z_j|$ for all $1\le j\le n$. Does this help? Also, if we are able to uniformly bound (in $z$) the summands by $q^{|\alpha|}$ for some $0 < q < 1$, we should be able to repeat the strategy in this answer.

Please give me hints on how to proceed. Thanks!

Answer 1

Let $||z||=\max_{k=1}^n|z_k|$ and choose a polydisc $B=\Pi(|w_k-z_k| <\delta) \subset U$ (hence the poly disc is contained in the interior of $U$ by definition) and note that $z_k(1+\frac{\delta}{||z||+1}) \in B$ hence $\xi(\alpha) \ge |z_1^{\alpha_1}...z_n^{\alpha_n}|(1+\frac{\delta}{||z||+1})^{|\alpha|}$ so $|\frac{z^\alpha}{\xi(\alpha)}| \le (1+\frac{\delta}{||z||+1})^{-|\alpha|}$ and from here the result follows by an easy counting argument of how many $\alpha$ have $|\alpha|=k$ (which is $c_k=O(k^n)$ so the series $\sum c_k (1+\frac{\delta}{||z||+1})^{-k}$ converges and we are done!

Answer 2

Just wanted to add my solution.

For any $z$ in the interior of $U$, fixing $z_k$, $k\neq i$ while varying $z_i$, $\partial U$ defines a boundary for $z_i$ and it is clear that we can always find $z_i'\in \mathbb{C}$ at the boundary of larger modulus which will increase $|z^\alpha|$. Hence we have $$\xi(\alpha):=\sup_{z\in U} |z^\alpha| = \sup_{z\in \partial U} |z^\alpha| \,,$$ i.e. the supremum is attained at the boundary.

Using CS $$\left|\sum_{\alpha\in \Bbb N^n} \frac{z^\alpha}{\xi(\alpha)}\right| \leq \sum_{\alpha\in \Bbb N^n} \frac{|z^\alpha|}{\xi(\alpha)} = \sum_{k=1}^\infty \sum_{\substack{\alpha\in \Bbb N^n \\ \alpha_1+...+\alpha_n=k}} \left(\frac{|z^{\alpha/k}|}{[\xi(\alpha)]^{1/k}}\right)^k \tag{1}$$ and the fact that $[\xi(\alpha)]^{1/k}$ is also the supremum for $|z^{\alpha/k}|$, we can easily check that for fixed $z$ in the interior $$\frac{|z^{\alpha/k}|}{[\xi(\alpha)]^{1/k}} = \frac{|z_1|^{\frac{\alpha_1}{\alpha_1+...+\alpha_n}}\cdots |z_n|^{\frac{\alpha_n}{\alpha_1+...+\alpha_n}}}{[\xi(\alpha)]^{\frac{1}{\alpha_1+...+\alpha_n}}}=c_\alpha(z)<1$$ which holds true for $k\rightarrow \infty$. Of all of these $c_\alpha$ there is a maximal $c_\alpha=c<1$ and we can continue with (1) $$\leq \sum_{k=1}^\infty c^k \sum_{\substack{\alpha\in \Bbb N^n \\ \alpha_1+...+\alpha_n=k}} 1 \leq \sum_{k=1}^\infty c^k \sum_{\substack{\alpha\in \Bbb Z_{\geq 0}^n \\ \alpha_1+...+\alpha_n=k}} 1 = \sum_{k=1}^\infty c^k \binom{k+n-1}{k}$$ which implies convergence.