Contraction Mapping Principle Let $X$ be a Banach space and $T\in\mathscr{L}(X,X)$ with $\|T\|_*<1$. Use the Contraction Mapping Principle to show (where $I$ is the identity map on $X$) that $I-T\in\mathscr{L}(X,X)$ is injective and surjective.
Attempt: Since $\mathscr{L}(X,X)$ is a normed linear space and $I,T\in\mathscr{L}(X,X)$ we must have $I-T\in\mathscr{L}(X,X)$ as well.
To show that $I-T$ is injective, let $x_1,x_2\in X$ such that 
$$
\begin{align*}
(I-T)(x_1)&=(I-T)(x_2)\\
I(x_1)-T(x_1)&=I(x_2)-T(x_2)\\
x_1-T(x_1)&=x_2-T(x_2)\\
x_1-x_2&=T(x_1-x_2)
\end{align*}
$$
So $x_1-x_2$ is a fixed point of $T$. Since $T$ is linear we must have that $T(0)=0$ and since $T$ is a contraction mapping, $0$ is the unique fixed point of $T$. Thus $x_1-x_2=0$ and we conclude that $x_1=x_2$.
So now what about the surjective part? I know that I need to find an $x$ for each $y\in X$ such that $(I-T)(x)=y$, but I am not having much luck manipulating this.
 A: You want to find $x$ such that $x=Tx+y$. The map $Cx = Tx+y$ is a contractive map because $Cx-Cx'=Tx-Tx'$.
A: I don't think the contraction mapping principle is the easiest way to to approach this question, but here goes. As you have discovered, injectivity isn't too bad. If $x\in\ker(I-T)$ then $(I-T)X=0$, that is, $x=Tx$. Since $T$ is a contraction and $T0=0$ by linearity, we must have $x=0$. This implies $I-T$ is injective.
Surjectivity is less straightforward. First note that
$$\|(I-T)x\|=\|x-Tx\|\ge\|x\|-\|Tx\|\ge\|x\|-\|T\|\|x\|=(1-\|T\|)\|x\|$$
and $1-\|T\|>0$. This implies $I-T$ has closed range. Indeed, if $y_n=(I-T)x_n$ and $y_n\to y\in X$, then $\|x_n-x_m\|\le\frac1{1-\|T\|}\|y_n-y_m\|$, so $(x_n)$ is Cauchy and hence converges, say to $x\in X$. By continuity we get $y_n=(I-T)x_n\to(I-T)x$, so $y=(I-T)x\in\operatorname{im}(I-T)$.
Now if $T'$ is the adjoint of $T$, then $\|T'\|=\|T\|<1$ so $T'$ is also a contraction. Hence by the same working as above we have that $(I-T)'=I-T'$ is injective. Assume now for a contradiction that $I-T$ is not surjective and let $x_0\in X\setminus\operatorname{im}(I-T)$. Since $\operatorname{im}(I-T)$ is closed, by the Hahn-Banach theorem there exists $f\in X'$ such that $f(x_0)=1$ and $f(y)=0$ for all $y\in\operatorname{im}(I-T)$. Then if $x\in X$, $((I-T)'f)x=f((I-T)x)=0$, so $(I-T)'f=0$. Hence $f\in\ker(I-T)'$, so $f=0$, a contradiction. Hence $I-T$ is surjective.
For the record, this question could be approached far more easily this way: since $\|T\|<1$, $R:=\sum_{n=0}^\infty T^n$ converges absolutely and hence converges in $\mathscr{L}(X)$ since $X$ is Banach. A direct calculation shows $R(I-T)=(I-T)R=I$, so $I-T$ is invertible with inverse $R$. In particular $I-T$ is both injective and surjective.
A: So I think I have an answer for the surjectivity part of the proof.
Let $y\in X$ and define $f_y(x)=T(x+y)$. First I show that $f_y$ is a contraction map. Let $\sigma$ be the metric induced by the norm on $X$ and observe that for any $x_1,x_2\in X$ we have
$$
\begin{align*}
\sigma(f_y(x_1),f_y(x_2))&=\|f_y(x_1)-f_y(x_2)\|\\
&=\|T(x_1+y)-T(x_2+y)\|\\
&=\|T(x_1+y-x_2-y)\|\\
&=\|T(x_1-x_2)\|\\
&\le\|T\|\cdot\|x_1-x_2\|\\
&<\|x_1-x_2\|
\end{align*}
$$
Thus $f_y$ is a contraction mapping. Let $x^*=f_y(x^*)$ be the unique fixed point for $f_y$ then
$$
(I-T)(x^*+y)=x^*+y-T(x^*+y)=x^*+y-f_y(x^*)=x^*+y-x^*=y
$$
Thus $I-T$ is surjective.
