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So I've been reading An Introduction to Complex Analysis in several variables and got stuck on one of the proofs in the book. Let me state the lemmas, and then the proof I'm having issues with.

Theorem 1.6.13 :Let $v_k$ be a sequence of subharmonic functions in $\Omega$ which are uniformly bounded form above on every compact subset of $\Omega$, and assume that $\overline{\lim}_{k\rightarrow \infty} v_k(z)\leq C$ for every $z\in \Omega$. For every $\epsilon>0$ and every compact subset $K\subset \Omega$, one can then find $k_0$ so that $$v_k(z)\leq C+\epsilon,\;\;\;\;\;z\in K,k>k_0$$

Lemma 2.2.11 Let $u$ be a complex valued function i the polydisc $D={z;|z_j-z_j^0|<R,j=1,\cdots,n}$, assume that $u$ is analytic in $z'=(z_1,...,z_{n-1})$ if $z_n$ is fixed and that $u$ is analytic and bounded in $D'={z;|z_j-z_j^0|<r,j=1,...,n-1,|z_n-z_n^0|<R}$ for some $r>0$ Then $u$ is analytic in $D$.

Proof of lemma 2.2.11 :We may assume that $z^0=0$. Choose $R_1$ and $R_2$ with $0<R_1<R_2<R$. By theorem 2.2.6 we have $$(2.2.6)\;u(z)=\sum_\alpha a_\alpha(z_n)z'^\alpha,\;\;\;z\in D$$,

where $\alpha$ only runs through multi-orders with $n-1$ places, $$ a_\alpha(z_n)=\partial^\alpha u(0,z_n)/\alpha!$$

is analytic in $z_n$ and

$$(2.2.7)\;|a_\alpha(z_n)|R_2^{|\alpha|}\rightarrow 0\;\;\text{when}\;|\alpha|\rightarrow \infty\;\text{for fixed}\;z_n\text{ with}\;|z_n|<R$$

Further, Cauchy'sinequality gives $$(2.2.8)\;|a_\alpha(z_n)|r^{|\alpha|}\leq M$$

if $M$ is a bound for $|u|$ in $D'$. Now apply Theorem 1.6.13 to the subharmonic functions

$$z_n\rightarrow \frac{1}{|\alpha|}\log|a_\alpha(z_n)|$$

In view of (2.2.8), these functions are uniformly bounded from above when $|z_n|<R$ and by (2.2.7) the upper limit when $|\alpha|\rightarrow \infty$ is $\leq\log(1/R_2)$ for fixed $z_n$. Hence Theorem 1.6.13 shows that for large $|\alpha|$

$$\frac{1}{|\alpha|}\log|a_\alpha(z_n)|\leq \log\frac{1}{R_1}\;\;\text{if}\;|z_n|<R_1$$ that is, $$|a_\alpha(z_n)|R_1^{|\alpha|}\leq 1\;\;\text{for large}\; |\alpha|\;\text{if}\;|z_n|<R_1$$ This proves that the series (2.2.6) converges normally in D, and, since the terms are analytic, we conclude that $u$ is analytic.

Comments I have bolded the parts that I don't quite see why it follows. If anyone could explain to me those parts I would really appreciate it. Thanks.

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