Hoffman - Definition of Blaschke product

Question

I am self-studying Banach Spaces of Analytic Functions by Hoffman. I have a question regarding the following theorem:

According to the author, the set $K$ formed as in the theorem is compact. I do not see how. If we consider the $\alpha_n = 1/n$ for each $n \in \mathbb N$ then the Blaschke product makes sense because the $\prod_{n=1}^{\infty} 1/n = 0$. However, the set $K$ formed out is $K=\mathbb N$ which is not compact.

Is the author mistaken or am I doing something wrong?

Answer 1

Different authors/text books use different definitions for the infinite product.

The source that you quoted, Complex Variables by Robert B. Ash, defines

P1 The infinite product $\prod_{k=1}^\infty z_k$ is convergent if the sequence $(P_n)$ of partial products $P_n = \prod_{k=n}^\infty z_k$ is convergent. In that case the value of the infinite product is defined as $\prod_{k=1}^\infty z_k = \lim_{n \to \infty} P_n$.

The definition in Wikipedia: Infinite product is a bit more strict:

P2 The infinite product $\prod_{k=1}^\infty z_k$ is convergent if the sequence $(P_n)$ of partial products is convergent with a non-zero limit. Again, in that case the value of the infinite product is defined as $\prod_{k=1}^\infty z_k = \lim_{n \to \infty} P_n$.

It is also mentioned in that Wikipedia article that some authors allow finitely many zero factors:

P3 The infinite product $\prod_{k=1}^\infty z_k$ is convergent if there is an index $N$ such that $\lim_{n \to \infty} \prod_{k=N}^n z_k$ exists and is not zero. In that case the value of the infinite product is defined as $ \prod_{k=1}^\infty z_k = \prod_{k=1}^{N+1} z_k\cdot \lim_{n \to \infty} \prod_{k=N}^n z_k$.

• With definition P1, $\prod_{k=1}^\infty z_k$ can be zero without any factor being zero, e.g. $\prod_{k=1}^\infty k^{-1} = 0$. Also $z_k \to 1$ is not a necessary condition for the convergence of the product.
• With definition P2, $\prod_{k=1}^\infty z_k$ is never zero.
• And with definition P3, $\prod_{k=1}^\infty z_k$ is zero if and only if one of the factors $z_k$ is zero (as it is the case with finite products).

Now let us see what happens with Blaschke products if we choose $a_k = 1/k$, $k \ge 2$, as the sequence of zeros. For fixed $z$ with $|z| < 1$ is $$ \lim_{k \to \infty}\frac{z-1/k}{1-z/k} = z $$ so that $$ \lim_{n \to \infty} \prod_{k=2}^n \frac{z-1/k}{1-z/k} = 0 \, . $$ That shows that for all $z$ in the unit disk, the infinite product $\prod_{k=2}^\infty \frac{z-1/k}{1-z/k}$ is convergent in the sense of P1, and the value of the infinite product is zero. This is no surprise as there can be no non-constant holomorphic function in the unit disk with zeros at $1/k$ for all integers $k \ge 2$.

$\prod_{k=2}^\infty \frac{z-1/k}{1-z/k}$ is not convergent in the sense of P2 or P3.

So to exclude products which are identically zeros, Blaschke products are defined for sequences $(a_k)$ of non-zero numbers in the unit disk which satisfy the ”Blaschke condition”: $$ \sum_{k=1}^\infty (1-|a_k|) < \infty \, . $$ This condition guarantees that for every $z$ in the unit disk, the infinite product $$ f(z) = \prod_{k=1}^\infty \frac{|a_k|}{a_k} \frac{a_k-z}{1-\bar a_k z} $$ has only finite many zero factors, and that the product converges in the sense of P3. One can show that the converges is uniform in each disk of radius $r < 1$, so that $f$ is a holomorphic function in the unit disk, with zeros exactly at the $a_k$. (An additional factor $z^p$ can be added for a zero at the origin.)

Also (finally coming back to your question), $|a_k| \to 1$, so that the set $Z = \{ 1/\bar a_k \}$ is bounded, and ”$Z$ plus its accumulation points” is compact.