I was reading an article yesterday which was silent on the algebra of Banach. In that article was provided this example
If $a\in\mathcal {A},$ and $\Vert a\Vert <1$, then $e-a$ is regular element of Banach algebra $\mathcal{A}$ with unit $e$. Prove!!!
Please if someone can let them solve this example, to see how it looks choice, because it seems interesting example.
I hope that someone will help me, thank you preliminarily
I assume that by "regular" the OP means "invertable"; that is, that there exists $b \in \mathcal A$ with $b(e - a) = (e - a)b = e$. To see that such $b$ exists for $\Vert a \Vert < 1$, note that we have, for any positive integer $m$,
$(e - a) \sum_0^m a^j = e - a^{m + 1}, \tag{1}$
in which we understand that $a^0 = 1$. If we now let $m \to \infty$, the sum on the left converges because it is majorized, term by term, by the ordinary, real geometric sequence $\Vert a \Vert^j$:
$\Vert a^j \Vert \le \Vert a \Vert^j; \tag{2}$
thus, setting
$S_m = \sum_0^m a^j, \tag{3}$
we have
$\Vert S_{m + k} - S_m \Vert = \Vert \sum_{m + 1}^{m + k} a^j \Vert$ $\le \Vert a^m\Vert \sum_1^k \Vert a \Vert^j \le \Vert a^m\Vert \sum_1^\infty \Vert a \Vert^j = \Vert a \Vert^m \dfrac{\Vert a \Vert}{1 - \Vert a \Vert}; \tag{4}$
since $\Vert a \Vert^m \to 0$ as $m \to \infty$, (4) shows that $S_m$ is Cauchy; hence $S_m \to S \in \mathcal A$ as $m \to \infty$ as well. Turning now to (1), and again letting $m \to \infty$, we see that
$(e - a)S = e \tag{5}$
since $a^m \to 0$: $\Vert a^m \Vert \le \Vert a \Vert^m \to 0$ as $m \to \infty$. Furthermore, since we also have
$(\sum_0^m a^j) (e - a) = e - a^{m + 1} \tag{6}$
we in addition have
$(e - a)S = e; \tag{7}$
(5) and (7) show that
$S = \sum_0^\infty a^j \tag{8}$
is the inverse of $e - a$; thus $e - a$ is invertible or "regular" as was to be shown. QED.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
Take $e+a+a^2+\cdots$ (why converges?). Then, $$(e-a)(e+a+a^2+\cdots)=e+a-a+a^2-a^2+\cdots$$
