I am new to Functional Analysis .Please review the following proof:
Let $X$ be a Banach space. Let $T:X\to X$ be a invertible linear operator and $M>0$ be such that $\|T^{-k}\|<M$ for all $k\ge 1$.
Prove that $\inf _{n\ge 1} \|T^n(x)\|>0$ for all $x\neq 0$ in $X$
My try:
In order to show that $\inf _{n\ge 1} \|T^n(x)\|>0$ for all $x\neq 0$ in $X$ ;let us fix $x(\neq 0)\in X$ and then we have to show that $\|T^n(x)\|>0$ for all $n\ge 1$.
Without loss of generality let $\|x\|=1$
Again $T$ is invertible so there exists some $T_1$ such that $TT_1=T_1T=I$.
Since $\|T^{-k}\|<M\implies \sup _{\|x\|=1,x\in X}\|T^{-k}(x)\| <M\implies \sup {\|(T^k)^{-1}(x)\|}<M$ $\implies \|T^{-k}(x)\|<M$ for all $x$ with $\|x\|=1$
Now for any linear operator $P$; $P^{-1}$ is defined as $P^{-1}(y)=x$ where $P(x)=y$
Then we have $T^{-k}(x)=y$ where $T^k(y)=x$
Now $\|T^{-k}(x)\|<M$ for all $x$ with $\|x\|=1\implies \|y\|<M$ whenever $\|T^k(y)\|=1$ for all $k\ge 1$
Thus $\inf _{n\ge 1} \|T^n(x)\|>0$ for all $x\neq 0$.I think that there will be some problems as I have not used the fact that $X$ is Banach.
Please check this proof and suggest edits if required. Looking forward to you all.
By assumption there exists $M>0$ such that for all $y\in X$, $n\ge 1$:
$$ \|T^{-n} (y)\|\le M\|y\|.$$
Let $x$ be given. Set $y=T^n x$. To obtain
$$ \|x\| = \|T^{-n} (T^n x) \| \le M \|T^n x\| .$$
Therefore
$$\inf_{n} \|T^n x\| \ge \frac{\|x\|}{M}.$$