Showing that $F(x) = x + f(x)$ defines a homeomorphism when $f : E \to E$, and where $E$ is a Banach space.

Question

Let $E$ be a Banach space and $f : E \to E$ a contraction. Show that the equation $F(x)=x+f(x)$ defines a homeomorphism $F:E \to E$ that is Bilipschitz.

Since $f$ is a contraction the following to properties hold $$\|f(x)-f(y)\|\le q\|x-y\|$$ and by Banach fixed point theorem there exists $x \in E$ such that $f(x)=x$. So starting from the Bilipschitz part I have that $$\|F(x)-F(y)\| = \|x+f(x)-y-f(y)\| \le \|x-y\|+\|f(x)-f(y)\| \le (q-1)\|x-y\|$$

Similarly $$\|F(x)-F(y)\| = \|x+f(x)-y-f(y)\| \ge | \|x-y\| + \|f(x)-f(y)\|| \ge (q+1)\|x-y\|$$

which would conclude that $F$ is Bilipschitz. Now I'm bit stuck with the homeomoprhism part. How can I approach this?

Answer 1

First, two corrections: \begin{align*} \|F(x)-F(y)\| & = \|x+f(x)-y-f(y)\| \\ & \le \|x-y\|+\|f(x)-f(y)\| \le (1+q)\|x-y\|, \\ \|F(x)-F(y)\| &= \|x+f(x)-y-f(y)\| \\ & \ge \|x-y\| - \|f(x)-f(y)\| \ge (1-q)\|x-y\|. \end{align*} What you wrote wasn't correct.

There are two steps left.

• Show that $F$ is onto. Hint. Pick $z \in E$ and apply the Banach fixed point theorem to the map $x \mapsto z-f(x)$.
• If you put it all together, you should easily check that $F$ is a bilipschitz homeomorphism.