Show $\sum_{n \in \mathbb{Z}} \sum_{k \in \mathbb{Z}} a_k b_{n-k} z^n =(\sum_{n \in \mathbb{Z}}a_n z^n)(\sum_{n \in \mathbb{Z}}b_n z^n)$

Question

First, let me tell the definition of series I use. Let $S$ be any set. Let $f: S \to \mathbb{C}$ be a function. We say $\sum_{n \in S}f(n)$ converges to $F\in \mathbb{C}$ if the following condition is satisfied:

For all $\epsilon > 0$, there is a finite subset $T_0$ of $S$ such that if $T\supseteq T_0$ and $T$ is a finite subset of $S$, then

$$\left|\sum_{n \in T} f(n)-F\right| < \epsilon$$

I know the basics about this kind of summation. I want to show the following:

$$\sum_{n \in \mathbb{Z}} \sum_{k \in \mathbb{Z}} a_k b_{n-k} z^n =\left(\sum_{n \in \mathbb{Z}}a_n z^n\right)\left(\sum_{n \in \mathbb{Z}}b_n z^n\right)$$

where $z\in \mathbb{C}$ and $|z|=1$, $(a_n)_n, (b_n)_n \in l^1(\mathbb{Z}) = \{(x_n)_{n\in \mathbb{Z}}: \sum_n|x_n| <\infty\}$.

I have shown that the two series on the right side converge (they converge absolutely). So, let $\epsilon > 0$ be given.Denote their sums with $A$ and $B$, resp. Choose finite subsets $T_0, T_1$ such that ($\subseteq_f$ means finite subset)

$$T_0 \subseteq T \subseteq_f \mathbb{Z} \implies \left|\sum_{n \in T} a_n z^n - A\right| < \epsilon$$ $$T_1 \subseteq T \subseteq_f \mathbb{Z} \implies \left |\sum_{n \in T} a_n z^n - A\right| < \epsilon$$

Then, I'm unsure how to proceed. The double sum is confusing me, as it means I will have to deal with two infinite sums at once.

Any help is appreciated!

Answer 1

Let $\alpha=\sum_{k\in{\mathbb Z}}|a_k|$ and $\beta=\sum_{k\in{\mathbb Z}}|b_k|$. Let also $s_n=\sum_{k\in \mathbb Z} a_kb_{n-k}$ (I presume that you already know how to show that $s_n$ is absolutely convergent, with $|s_n| \leq \alpha\beta$, so I'll skip that part. I'll add it in if you ask).

Let $\varepsilon > 0$. Let $\eta$ be another positive number, chosen according to $\varepsilon$, in a way to be determined later. We know that there is a $N$ such that

$$ \sum_{|k|\gt N} |a_k| \leq \eta, \sum_{|l|\gt N} |b_l| \leq \eta. \tag{1} $$

From the first inequality in (1), we deduce that $\sum_{|k|\gt N} |a_kb_l| \leq \eta |b_l|$, and summing on $l$ we deduce $\sum_{|k|\gt N, l\in{\mathbb Z}} |a_kb_l| \leq \eta \beta$. Similarly, we have $\sum_{|l|\gt N, k\in{\mathbb Z}} |a_kb_l| \leq \eta \alpha$. Adding up the two, we deduce

$$ \sum_{|k|>N \textrm{or} |l|>N} |a_kb_l| \leq (\alpha+\beta) \eta \tag{2} $$

A consequence of (2) is that for any $S\subseteq {\mathbb Z}^2$, we have

$$ \Bigg|\sum_{(k,l)\in S} a_kb_lz^{k+l} - \sum_{(k,l)\in S\cap [-N,N]^2} a_kb_lz^{k+l} \Bigg| \leq (\alpha+\beta) \eta \tag{3} $$

As a special case, we deduce that for any finite subset $T$ of $\mathbb Z$,

$$ \Bigg| \sum_{k+l \in T} a_kb_lz^{k+l} - \sum_{k+l \in T, |k| \leq N, |l| \leq N} a_kb_{l}z^{k+l} \Bigg| \leq (\alpha+\beta) \eta \tag{4} $$

Notice that when $T$ contains $T_0=[-2N,2N]$, in the second big sum, the condition $k+l\in T$ follows automatically from the two others, and may therefore be omitted.

So when $T \supseteq T_0$, (4) may be rewritten as

$$ \Bigg| \sum_{n\in T} s_nz^{n} - \sum_{|k| \leq N, |l| \leq N} a_kb_{l}z^{k+l} \Bigg| \leq (\alpha+\beta) \eta \tag{5} $$ Now, particularizing (3) a second time,

$$ \Bigg| AB - \sum_{|k| \leq N, |l| \leq N} a_kb_{l}z^{k+l} \Bigg| \leq (\alpha+\beta) \eta \tag{6} $$

Adding (5) and (6), we deduce $|\sum_{n\in T} s_nz^{n}-AB| \leq 2(\alpha+\beta) \eta$. Taking $\eta=\frac{\varepsilon}{2(\alpha+\beta)}$ finishes the proof.