T linear continuous function of normed spaces with $\|T\|<1$, then $\|T^n\|\le \|T\|^n$

Question

Let $E$ be a Banach space (complete normed vector space) and $T:E\to E$ linear and continuous with $\|T\|<1$, with the norm $\|f\|=\sup_{x\in S}|f(x)|$ where $S$ is the unit sphere $S = \{x\in E$ st. $|x|=1 \}$. Then $\|T^n\|\le \|T\|^n$ for any $n\ge 0$.

I'm trying by induction: for $n=0, |Id| = 1$. For $n+1$ I think I should use $\|T \cdot S\|\le \|T\|\|S\|$ but I'm stuck in proving this last inequality.

Thanks in advance for any hints.

Answer 1

To prove the first inequality: $$ \|TS\| = \sup_{\|x\| = 1}\|TSx\| \leq\|T\| \sup_{\|x\| = 1}\|Sx\| \leq \|T\|\|S\| $$ This follows from the fact that for any $x$: $$ \|Tx\| = \left\|T \left(\|x\|\frac{x}{\|x\|}\right)\right\| = \|x\|\left\|T \left(\frac{x}{\|x\|}\right)\right\| \leq \|x\|\sup_{y = 1} \|Ty\| = \|x\|\|T\| $$ And this result uses the fact that $\frac{x}{\|x\|}$ is a unit (norm 1) vector.

Answer 2

Your thoughts are correct. In fact you nearly have it. Your base case is correct, and you have the right idea for the inductive step.

To complete it, just observe the following $$\|T^{n+1}\|=\|T^nT\|\le \|T^n\|\|T\|\le \|T\|^n\|T\|=\|T\|^{n+1},$$ where the first inequality is the inequality you think you should use and the second is the inductive hypothesis.