What is the justification for this equivalence: $ \sum_{m=-\infty}^{\infty} |x(m - n)|= \|x\|_1 $

Question

In Berberian's Lectures in Functional Analysis and Operator Theory, I'm having trouble following the Aperitif on Wiener's Theorem and some of lapses in logic present within; one in particular.

On the proof of the convergence of $ \sum_{n=-\infty}^{\infty} x(m - n)y(n) $, the author asserts the following equivalence:

$ \sum_{m=-\infty}^{\infty} x(m - n) = \|x\|_1 $ - note the change in index.

Given that x $ \in l^1(\mathbb{Z}) $ for $ x = \sum_{n=-\infty}^{\infty} x(n) e_n $

where $e_n (m) = \delta_{mn}$

and $(xy)(m)= \sum_{n=-\infty}^{\infty} x(m - n)y(n) $ for $ m \in \mathbb{z}$

Why does $ \sum_{m=-\infty}^{\infty} |x(m - n)|$ = $\|x\|_1$ ?

Why does a "one" just magically appear as a subscript?

Note on edits: I had $y(n)$ originally multiplied to the sum, but that was not what I was intending to ask about; my apologies.

Answer 1

In case we are talking about numbers, clearly $$ \sum_{m=-\infty}^{\infty} |x(m - n)y(n)| = \|x\|_1y(n), $$ as it must be in the book (it cannot be as you wrote).

Answer 2

"Why does a "one" just magically appear as a subscript?" It's the definition of the $\ell^{1}(\mathbb{Z})$-norm. Given a sequence $a : \mathbb{Z} \to \mathbb{C}$, we define the norm $\|\cdot\|_{1}$ by \begin{equation*} \|a\|_{1} = \sum_{n = - \infty}^{\infty} |a(n)|. \end{equation*} Of course, this is only a norm on the vector space of sequences for which the sum is finite. This is precisely the definition of the Banach space $\ell^{1}(\mathbb{Z})$.

Now given a sequence $x : \mathbb{Z} \to \mathbb{C}$ and an integer $n \in \mathbb{Z}$, substitution implies that \begin{equation*} \sum_{m = - \infty}^{\infty} |x(m - n)| = \sum_{m = - \infty}^{\infty} |x(m)| = \|x\|_{1}. \end{equation*}