# Using the Banach fixed point Theorem

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

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.