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I have a question about the proof on the german wikepedia page:

The proof is stated as follow:

Let $A= \sum_{k=0}^{\infty}a_k$ and $B=\sum_{k=0}^{\infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.

Definition of the Cauchy-Product: $C=\sum_{k=0}^{\infty}c_k,c_k=\sum_{j=0}^{k}a_jb_{k-j}$

Without loss of generality let A be the absolutely convergent series and $S_n=\sum_{k=0}^{n}c_k$

1: $AB=(A-A_n)B+\sum_{k=0}^{n}a_kB$

2: $S_n=\sum_{k=0}^{n}a_kB_ {n-k}$

1-2=$AB-S_n=(A-A_n)B+\sum_{k=0}^{n}a_k(B-B_{n-k})$

$(A-A_n)B \rightarrow 0$ and with $N:=[\frac{n}{2}]$ the other series can be splitted into two parts with:

$\sum_{k=0}^{N}a_k(B-B_{n-k})+\sum_{k=N+1}^{n}a_k(B-B_{n-k})$

Then

$|\sum_{k=0}^{N}a_k(B-B_{n-k})|\leq \sum_{k=0}^{N}|a_k(B-B_{n-k})|=\sum_{k=0}^{N}|a_k||(B-B_{n-k})|\leq\max\limits_{N \leq k \leq n}|B-B_k|\sum_{k=0}^{N}|a_k|\rightarrow 0$

Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<C\forall k \in \mathbb{N}$

Hence

$|\sum_{k=N+1}{n}a_k(B-B_{n-k})|\leq \sum_{k=N+1}{n}|a_k||(B-B_{n-k})|\leq C\sum_{k=N+1}{n}|a_k|\rightarrow 0 \square$

I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.

Thank you for your time, I would appreciate your help very much.

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1 Answer 1

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The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'

Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.

As for the max. The task is to estimate a sum of the form $\sum_{k} |f_k| \ |g_k|$ where one knows that $\sum_{k} |f_k|$ tends to $0$. To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| \le G $ for each $k$, and estimates

$$\sum_{k} |f_k| \ |g_k| \le \sum_{k} |f_k| G = G \sum_{k} |f_k| $$

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