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?



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, $R(I+A)=H$.


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.


Banach's fixed point theorem states that if an operator $T \colon X \to X$ on the non-empty complete metric space $(X, d)$ is a contraction, then $T$ has a unique fixed point $x^* \in X$ such that $Tx^* = x^*$. This means that if you can show that $T$ is a contraction, i.e. $d(Tx, Ty) \leq Kd(x, y)$ for some constant $0 \leq K < 1$, then $T$ has a fixed point. If you can show that (2) is indeed a contraction, then there will exist a unique solution.


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