Banach fixed point. Let $T \in B(\mathcal{B})$. If $\Vert{T}\Vert < 1$ then, given $\eta \in \mathcal{B}$, the equation $\xi-T\xi=\eta$ has a unique solution. 
In effect, defining the operator $S:\mathcal{B} \hookleftarrow$, $S\xi=T\xi+\eta$, since by definition we know that: Let $M$ a set. A fixed point of an application $A:M\hookleftarrow$ is an element $\xi \in M$ satisfying $A(\xi)=\xi$.
we have $\Vert{S\xi - S\omega}\Vert \leq \Vert{T}\Vert \Vert{\xi - \omega}\Vert$, for everything $\xi , \omega \in \mathcal{B}$, 
which is a contraction whose only fixed point $\zeta$ is the desired solution. 
How do I prove that $\zeta = \sum _{j=0} ^{\infty} T^j \eta$?.
 A: We have the following well-known fact:
$T \in B(\mathcal B), \; \Vert T \Vert < 1 \Longrightarrow \exists (I - T)^{-1} = \displaystyle \sum_0^\infty T^k = I + T + T^2 + \ldots; \tag 1$ 
this follows from the purely algebraic identities
$(I - T) \displaystyle \sum_0^n T^k = (I - T)(I + T + T^2 + \ldots + T^n) = I - T^{n + 1}, \tag 2$
which hold for every $n \in \Bbb N$; to see (2), simply write
$(I - T) \displaystyle \sum_0^n T^k = I \sum_0^n T^k - T \sum_0^n T^k = \sum_0^n T^k - \sum_1^{n + 1} T^k$
$= I + \displaystyle \sum_1^n T^k - \sum_1^n T^k - T^{n + 1} = I - T^{n + 1}; \tag 3$
once we have (2)-(3) we also have
$\left \Vert (I - T) \displaystyle \sum_0^n T^k - I \right \Vert = \Vert T^{n + 1} \Vert \le \Vert T \Vert^{n + 1} \to 0 \; \text{as} \; n \to \infty, \tag 4$
since $\Vert T \Vert < 1$; (4) shows that
$\displaystyle \lim_{n \to \infty} (I - T) \sum_0^n T^k = I, \tag 5$
or 
$\displaystyle (I - T) \sum_0^\infty T^k = I, \tag 6$
establishing 
$(I - T)^{-1} = \displaystyle \sum_0^\infty T^k. \tag 7$
Now 
$\xi - T\xi = \eta, \tag 8$
is equivalent to
$(I - T)\xi = \eta, \tag 9$
and then by (7),
$\xi = (I - T)^{-1}(I - T) \xi = (I - T)^{-1} \eta = \left ( \displaystyle \sum_0^\infty T^k \right ) \eta = \displaystyle \sum_0^\infty T^k \eta, \tag{10}$
as per request.
A: First you show that $\sum T^j \eta$ exists (because the norm of $T$ is $<1$).
Then by continuity of $T$, see that $T(\sum T^j \eta)=\sum T T^j \eta$.
