Injective operator on a Banach space Let $X$ be a Banach space, $T \in B(X)$ and $||I−T||=γ<1$. Show that $T$ is injective. My attempt:
Suppose that $Tx=Ty$, for some $x,y\in X$. Then, by the linearity of $T$, $$T(x-y)=0\implies ||T(x-y)||=0.$$ By definition of the operator norm, $$||T(x-y)|| \le ||T||\,||x-y||.$$ How can I show that this implies $||x-y||=0$?
 A: It is enough to show that the kernel of $T$ is trivial. Suppose $T(x)=0$ for some vector $x\ne 0$. Then we have:
$1>||I-T||\geq \frac{||(I-T)(x)||}{||x||}=\frac{||I(x)-T(x)||}{||x||}=\frac{||x||}{||x||}=1$
Which is of course a contradiction. 
Edit: Note that you don't really need completeness for this. On the other hand, user474986 proved a stronger statement in his answer, that $I-T$ is actually invertible in $B(X)$. For that you really need $X$ to be a Banach space. 
A: You cannot show that $\| x - y\| = 0$ from the inequality $\|T(x-y)\| \le \|T\| \| x - y\|$.
However, you can show that the operator $I - T$ is a contraction, which means that there exists a unique fixed point of the operator $I-T$, which is $0$. Now suppose that $Tx = Ty$, then 
\begin{align*}
(I-T)(x-y) = x - Tx - y + Ty = x - y + (Ty - Tx) = x -y.
\end{align*}
Thus also $x-y$ is a fixed point of $I-T$, which means that $x - y = 0$, or $x = y$ and hence $T$ is injective.
A: The space of bounded linear operators on $X$, denoted $\mathcal{B}(X)$, is a Banach space, since $X$ is complete. The norm associated with $\mathcal{B}(X)$ is the operator norm, i.e.
\begin{align*}
\|T\| = \sup_{\|x\| = 1} \|Tx\|.
\end{align*}
Recall that a normed vector space is complete if and only if every absolutely convergent series converges. 
So consider the series
\begin{align*}
\sum_{n=0}^{\infty} \|I-T\|^n
\end{align*}
Since $\|I-T\| < 1$, this series converges, hence
\begin{align*}
\sum_{n=0}^{\infty} (I-T)^n
\end{align*}
converges in $\mathcal{B}(X)$ to some element, call it $S$. To show that $S = T^{-1}$, it suffices to show that $ST= I$. Observe,
\begin{align*}
ST &= \sum_{n=0}^{\infty}  (I-T)^n T \\
&= \sum_{n=0}^{\infty}  (I-T)^n - \sum_{n=0}^{\infty} (I-T)^{n+1} \\
&= \sum_{n=0}^{\infty}  (I-T)^n - \sum_{n=1}^{\infty} (I-T)^{n} \\
&= (I-T)^0 \\
&= I
\end{align*}
Thus, $S = T^{-1}$. In particular, $T$ is injective.
A: Assume that $T$ is not injective, then there is some $x \in X$ with $\Vert x \Vert = 1$ s.t. $Tx = 0$ and thus 
$$ \Vert I - T \Vert \geq \Vert (I - T)x \Vert = \Vert x - 0 \Vert = 1 $$
which is a contradiction.
