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

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

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

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

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!

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.

• Thanks for the answer. It was more complicated than I expected! Note also that there is a more direct way if we use Fubini's theorem. Then we can get the result by direct computation.
– user745578
Commented Mar 6, 2020 at 21:43
• @user745578 Sure. I thought that you didn't know about Fubini's theorem, as you said you only knew the basics and seemed to ask for a proof directly from the definition Commented Mar 7, 2020 at 4:19
• Yes, your answer is precisely what I was looking for! But interesting to look for alternatives nevertheless.
– user745578
Commented Mar 7, 2020 at 8:16