How do I show that $s_n = id_E + (id_E-f)+ \dots + (id_E -f)^n$ is continuous and Cauchy sequence?

Question

Let $E$ be a a Banach space. Let $f \in (\mathcal{L}(E), |||\cdot |||)$ such that $|||\text{id}_E-f|||<1$. We put $$s_n = id_E + (id_E-f)+ (id_E-f)^2 \dots + (id_E -f)^n$$ Show that $f$ is continuous, that it's a Cauchy sequence, and that $\underset{n \rightarrow \infty}{\text{lim}}=(s_n \circ f)= id_E$.

Showing that it is continuous wasn't too hard. $s_n$ is a sum of composition of functions that are continuous, thus is continuous.

Now showing that it's Cauchy was a bit harder. I first decided to see what for $n,m \in \mathbb{N}$, $|||s_n - s_m |||$ is inferior to. So suppose $n< m$, then $s_m-s_n=(id_E-f)^{n+1} \dots + (id_E -f)^m$, thus $|||s_n - s_m ||| \leq ||| id_E-f|||^{n+1} + \dots + |||id_E - f |||^m \leq (m-n)||| id_E-f|||^{n+1}$ and here I am rather unable to decide what $N_{\epsilon} $ to pick such that if $m,n > N_{\epsilon}$ $(m-n)||| id_E-f|||^{n+1} \leq \epsilon$

And for the first part, I have no idea on how to proceed.

Answer 1

Let $I = \operatorname{id}_E$ and $\|\cdot\|$ the operator norm on $\mathcal{L}(E)$.

To show that $f$ is continuous, it suffices to show that it is bounded:

$$\|f\| = \|(f - I) + I\| \le \|f - I\| + \|I\| < 1 + 1 = 2$$

using the triangle inequality of the operator norm.

Now, the idea behind showing that $(s_n)_{n=1}^\infty$ is Cauchy is to bound it with the geometric series. Assume $m \ge n$:

\begin{align} \left\|s_m- s_n\right\| &= \left\|\sum_{k=0}^m (I - f)^k - \sum_{k=0}^n (I - f)^k\right\|\\ &= \left\|\sum_{k=n+1}^m (I - f)^k\right\|\\ &\le \sum_{k=n+1}^m \|I - f\|^k\\ &\xrightarrow{m, n \to\infty} 0 \end{align}

because $\sum_{k=1}^\infty \|I - f\|^k$ is a convergent series, since $\|I - f\| < 1$.

Explicitly: $$\sum_{k=n+1}^m \|I - f\|^k \le \sum_{k=n+1}^\infty \|I - f\|^k = \|I - f\|^{n+1}\sum_{k=0}^\infty \|I - f\|^k = \frac{\|I - f\|^{n+1}}{1 - \|I - f\|}$$

So if you pick $N \in \mathbb{N}$ such that $\|I - f\|^{N+1} < \varepsilon(1 - \|I - f\|)$ then for every $m, n \ge N$ you will have $\|s_m - s_n\| < \varepsilon.$

Finally, let's show that $f \cdot \displaystyle\lim_{n\to\infty} s_n = I$. We can write $\displaystyle\lim_{n\to\infty} s_n = \sum_{n=0}^\infty (I-f)^n$.

$$f\cdot\sum_{n=0}^\infty (I - f)^n = \sum_{n=0}^\infty (I - f)^n - (I - f)\sum_{n=0}^\infty (I - f)^n = \sum_{n=0}^\infty (I - f)^n - \sum_{n=1}^\infty (I - f)^n = (I-f)^0 = I$$

Answer 2

Your inequality is insufficient. Let $r = ||| id_E-f|||$; we have (by the formula for the partial sum of a geometric series) $$ |||s_n - s_m ||| \leq r^{n+1} + \dots + r^m = r^{n+1} \cdot \frac{1 - r^{m - n}}{1 - r} $$ To show that $f$ is continuous, it suffices to show that $|||f|||$ is finite.

In going through the proof of the latter bit, it may help you to observe that the operator $s = \lim_{n \to \infty}s_n$ is $$ s = \sum_{k=0}^\infty (\operatorname{id}_E - f)^k $$ which is suggestive of the geometric series expansion of $\frac 1{f} = \sum_{k=0}^\infty (1 - f)^k$.