2

I have a question regarding well - definedness.

Suppose $X$ is a banach space $\mathcal{l}^{1}(\mathbb{Z})$ given by the norm $||(x_{n})_{n}||_{1} := \sum_{n \in \mathbb{Z}} |x_{n}|$

If we define the product $xy$ as $(xy)_{n} = \sum_{m} x_{m}y_{n-m}$.

Then is it correctly understood that showing

$\sum_{n} |(xy)_{n}| \leq ||x|||y||$ and that $X$ has a unit implies that
$(xy)_{n} = \sum_{m} x_{m}y_{n-m}$ is a WELL DEFINED PRODUCT which makes $X$ a Banach algebra?

Chengdu
  • 451
  • 1
    Yes, this is a well defined product and $X$ becomes a Banach algebra when endowed with it. Is that enough of an answer? If not, please explain what is confusing you and what have you done so far :) – Just dropped in Nov 15 '21 at 19:57
  • @JustDroppedIn Thanks a lot what I was confused about was how do we know that this product is actually well defined? and what do we have to prove in order to show that the product is well defined? Is it enough to simply show that $\sum_{n} |(xy)_{n}| \leq ||x|||y||$? – Chengdu Nov 15 '21 at 20:12
  • 1
    Well, in order to prove that this is well-defined, one needs to verify that this is indeed in $\ell^1$. In other words, you should verify that $\sum_{n}|(xy)_n|<\infty$. After that, one trivially checks that this well-defined multiplication is an associative operation, i.e. $(x\cdot y)\cdot z=x\cdot(y\cdot z)$ and that it is distributive, i.e. $x(y+z)=xy+xz$. Finally, in order to verify that $\ell^1$ becomes a Banach algebra, one should verify that $|xy|\leq|x||y|$. – Just dropped in Nov 15 '21 at 20:42
  • @JustDroppedIn I added my own answer can you check please. – Chengdu Nov 15 '21 at 22:04
  • Where is your answer? – Just dropped in Nov 15 '21 at 22:08
  • @JustDroppedIn it is there now. – Chengdu Nov 15 '21 at 22:09

1 Answers1

2

So first of all I show that the product is indead in $\mathcal{l}^{1}(\mathbb{Z})$

\begin{align*} \sum_{n} |(xy)_{n}| &\leq \sum_{n} \sum_{m} |x_{m}| |y_{n-m}|\\ &= \sum_{m} |x_{m}| \sum_{n} |y_{n-m}|\\ &= \sum_{m} |x_{m}| ||y||\\ &= ||x|| \ ||y|| \end{align*} which implies that $xy \in \mathcal{l}^{1}(\mathbb{Z})$ which shows that $\sum_{n} |(xy)_{n}| < \infty$. Furthermore,

\begin{align*} x(y+z)_{n} &= x_{m} (\sum_{m} y_{n-m} + \sum_{m} z_{n-m})\\ &= \sum_{m} x_{m} y_{n-m} + \sum_{m} x_{m} z_{n-m}\\ &= (xy)_{n} + (xz)_{n} \end{align*}

and

\begin{align*} ((x \cdot y) \cdot z)_{n} &= \sum_{m} (x_{m} y_{n-m}) z_{m})\\ &= \sum_{m} x_{m} (y_{n-m} z_{m})\\ &= (x(y \cdot z))_{n} \end{align*}

which shows that the product is well defined

Chengdu
  • 451
  • 2
    everything is fine. Note that you didnt just prove that $|xy|<\infty$, but you also proved that $|xy|\le|x||y|$, i.e. the norm is submultiplicative (which is needed in the definition of a Banach algebra). Nice! – Just dropped in Nov 15 '21 at 22:11