TITLE: Using the Banach fixed point Theorem
QUESTION [5 upvotes]: In Chapter 7, The Hille-Yosida Theorem,

Functional Analysis,
  Sobolev Spaces and Partial
  Differential Equations - Brezis, 2011.

Brezis showed the following claim (in Proposition 7.1).
Claim: Suppose that $A$ is a maximal monotone operator, $(I+\lambda A): D(A) \to R(I+\lambda A)$  is a injective operator, and  $|(I+\lambda A)^{-1}u| \leq |u| $ for all $u \in R(I+\lambda A)$  for all $\lambda >0$. Then $R(I+\lambda A) = H$.
The proof of Brezis: 

We will prove that if $R(I+\lambda_{0} A) = H$ for some $\lambda_{0}>0$ then 
  $R(I+\lambda A) = H$ for every $\lambda > \frac{1}{2}\lambda_{0}$.
For some $f \in H$, we try to solve the equation 
$u+\lambda Au = f$ with  $\lambda >0$.  (1)
Equation (1) may be written as 
$u+ \lambda_{0}Au = \frac{\lambda_{0}}{\lambda}f+ \big( 1- \frac{\lambda_{0}}{\lambda}u \big)$
or alternatively
$u= (I+\lambda A)^{-1}\big[\frac{\lambda_{0}}{\lambda}f+ \big( 1- \frac{\lambda_{0}}{\lambda}u \big)\big]$ $(2)$
If $|1-\frac{\lambda_{0}}{\lambda}|< 1$, i.e.,  $\lambda > \frac{1}{2}\lambda_{0}$, we may apply the   contraction mapping principle (the Banach Fixed point Theorem) and deduce that (2) has a solution.

My question:  how to prove that (2) has a solution using the the Banach Fixed point Theorem? 
Thanks

Definitions:
An unbounded linear operator $A: D(A)\subseteq H \to H$ is said to be monotone if it satisfies 
$\langle A u, u \rangle \geq 0$ for all $u\in D(A)$.
It is called maximal monotone if, in addition, $R(I+A)=H$.

REPLY [2 votes]: There are some typos in your derivation. Specifically the correct form of $(2)$ is:
$$u=(I+\lambda_0 A)^{-1} \left[\frac{\lambda_0}\lambda f\right]+(I+\lambda_0 A)^{-1}\left[(1-\frac{\lambda_0}\lambda)u\right].$$
Your equation is of the form
$$u=v + Bu$$
for a constant $v$ and a linear map $B$. In the case that $B$ is a contraction then $v+Bu$ is a contraction and you are done. So why is $B=(1-\frac{\lambda_0}\lambda)(I+\lambda_0A)^{-1}$ a contraction? Here note that $\|(I+\lambda_0 A)^{-1}\|≤1$, use the positivity of $A$ if you wish. This gives you
$$\|B\|≤|1-\frac{\lambda_0}\lambda|<1,$$
now you are finished.