Uniform convergence of an infinite product

Question

This is taken from Rudin's Real and Complex Analysis pg.314 I'm trying to understand the proof of Muntz Theorem in the text.

Let $\lambda_1, \lambda_2,..$ be positive real numbers such that $\sum\frac{1}{\lambda_k}<\infty$. Define the function $$f(z):=\frac{z}{(2+z)^3}\Pi_{k=1}^{\infty}\frac{\lambda_k-z}{2+\lambda_k+z}$$

Note that since

$$1-\frac{\lambda_k-z}{2+\lambda_k+z}=\frac{2+2z}{2+\lambda_k+z} (*)$$

The author claims that the infinite product that appears in the definition of f converges uniformly on compact subsets of $\mathbb{C}\backslash(\{-2\}\cup\{-2-\lambda_k\}_{k=0}^{\infty}$ and that each factor in the infinite product of f has absolute value less than 1 for $Re(z)>-1$

My question

1) What is the justification for the uniform convergence of the product

2) How can we prove that each factor has absolute value less than -1

For (1) there is a theorem on pg.300 that says

Suppose $f_n \in H(\Omega)$ for $n=1,2,3,...$, no $f_n$ is identically zero in any component of $\Omega$ and $$\sum|1-f_n(z)|$$ converges uniformly on compact subsets of $\Omega$. Then the product $$f(z)=\Pi_{n=1}^{\infty}f_n(z)$$ converges uniformly on compact subsets of $\Omega$.

So for (1) my real question is how (*) implies $\sum|1-f_n(z)|$ converges uniformly. This might have to do with the Weierstrass M-Test but I don't see how.

Any help would be appreciated

Answer 1

$1)$ For $f_k(z)=\frac{\lambda_k-z}{2+\lambda_k+z}$ we prove that $$\sum|1-f_k(z)| =\sum\left|\frac{2+2z}{2+\lambda_k+z}\right|$$ converges uniformly on compact subsets of $\mathbb{C}\setminus\bigcup_{k=1}^{\infty}\{-2-\lambda_k\}$. Suppose a compact $K \subset \mathbb{C}\setminus\bigcup_{k=1}^{\infty}\{-2-\lambda_k\}$ is given. Take a large $R$ so that $\{|z|<R\}$ contains $K$. Since $\lambda_k \to \infty$ as $k\to \infty$, there is an integer $N$ such that $\lambda _k\ge 2R$ for all $k\ge N+1.$ Then for $z\in K$ we have $$ \left|\frac{2z+2}{2+\lambda _k+z}\right|< \frac{2(R+1)}{2+\lambda_k -R}< \frac{2(R+1)}{\lambda _k-R}< \frac{4(R+1)}{\lambda _k} \quad(k\ge N+1), $$ since $|z|<R$. Therefore we have for $z\in K$ \begin{align} \sum_{k=1}^\infty\left|\frac{2+2z}{2+\lambda_k+z}\right|&=\sum_{k=1}^{N}\left|\frac{2+2z}{2+\lambda_k+z}\right|+\sum_{k=N+1}^\infty\left|\frac{2+2z}{2+\lambda_k+z}\right|\\ &\le \sum_{k=1}^{N}\left|\frac{2+2z}{2+\lambda_k+z}\right|+\sum_{k=N+1}^\infty\frac{4(R+1)}{\lambda _k} \end{align} and by the Weierstrass M-Test we conclude the uniform convergence of $\sum_{k=1}^\infty\left|\frac{2+2z}{2+\lambda_k+z}\right|$, since $\sum_{k=N+1}^\infty 1/\lambda _k<\infty.$

$2)$ It is easy to see that $$ \operatorname{Re} z>-1\implies |z|<|z+2|,\; |z+2|>1\implies \left|\frac{z}{(z+2)^3}\right|=\left|\frac{z}{z+2}\right| \cdot\left|\frac{1}{(z+2)^2}\right|<1. $$ Also we see \begin{align} \left| \frac{\lambda _k-z}{2+\lambda _k+z}\right|^2&=\frac{{\lambda _k}^2-2\lambda _k\operatorname{Re}z+|z|^2}{(2+\lambda _k)^2+2(2+\lambda _k)\operatorname{Re}z+|z|^2}\\ &<\frac{{\lambda _k}^2+2\lambda _k+|z|^2}{(2+\lambda _k)^2-2(2+\lambda _k)+|z|^2}\\ &=\frac{{\lambda _k}^2+2\lambda _k+|z|^2}{{\lambda _k}^2+2\lambda _k+|z|^2}=1. \end{align}