# Bijectivity and Lipschitz continuity of a function on a Banach space

I don't really know how to solve the following exercise, I need a little help:

a) Let $(X,\|\cdot \|)$ be a Banach space and $F\colon X \to X$ Lipschitz-continuous (i.e. $|F(x) - F(y)| \le L|x-y|$) with Lipschitz constant $L < 1$. Show that $G\colon X \to X$ with $G(x) = x + F(x)$ is bijective and its inverse function $G^{-1}\colon X \to X$ is Lipschitz-continuous with Lipschitz constant $\frac 1{1-L}$.

b) Use a) to show that there is a unique function $f \in C[0,1]$ for which the equation $$f(t) + \int_0^1e^{\tau+t-3}f(\tau)d\tau = 1 \quad \forall t \in [0,1]$$ holds. Determine $f$. Use the fact that $(X,\|\cdot \|_{\infty})$ is a Banach space.

In a) I think using the Banach fixed point theorem is the way to go in order to prove F is bijective. However, I don't know how to start or how to prove the Lipschitz continuity of $G^{-1}$.
For b) I don't know how to start, I'm thinking of the intermediate value theorem?

• Isn't $X'$ in fact $X$? – Alex M. Jun 7 '16 at 13:35
• Of course, I edited it. – lasik43 Jun 7 '16 at 13:37
• Where is this exercise from? What book are you using? – Jack Jun 7 '16 at 14:17
• It's from an exercise sheet we got from our University. I'm not using a specific book, just my professors script. – lasik43 Jun 7 '16 at 14:22

1. The reverse triangle inequality yields $|G(x) -G(y)|\ge (1-L)|x-y|$. This gives the Lipschitz continuity of the inverse map, as soon as $G$ is shown to be surjective. For surjectivity, you are right to use the Banach fixed point theorem: the equation $x+F(x)=y$ has a solution for $x$ because the map $x\mapsto y-F(x)$ has a fixed point.
2. The supremum norm of $\int_0^1e^{\tau+t-3}f(\tau)d\tau$ is at most $$\|f\|_\infty \int_0^1e^{\tau+1-3} d\tau = \|f\|_\infty (e^{-1}-e^{-2}) = L\|f\|_\infty$$ where $L = e^{-1}-e^{-2} <1$. This implies the existence and uniqueness of $f$, by virtue of $f$. To determine $f$, rewrite the equation as $$f(t) = 1 - e^t \int_0^1e^{\tau-3}f(\tau)d\tau$$ which says that $f$ is of the form $f(t)=1-Ce^t$. Plug this in the equation to find $C$.
Proposition. Suppose that $X$ is a Banach space, and that $f \colon X \to X$ has the form $f(x)=x+r(x)$ where $\operatorname{Lip}(r)<1$. Then $f$ is a lipeomorphism and $\operatorname{Lip}(f^{-1}) \leq (1-\operatorname{Lip}(r))^{-1}$.
Proof. Pick $u \in X$ and define $\xi_u$ by $\xi_u(x) = u-r(x)$. Thus $f(x)=u$ if and only if $\xi_u(x)=x$. Of course $\operatorname{Lip}(\xi_u)=\operatorname{Lip}(r)<1$, and therefore there is a unique fixed point $x$ of $\xi_u$. In particular, $f$ is bijective. Moreover \begin{align} \|f^{-1}(u)-f^{-1}(v) \| &= \|x-y\| \\ &= \|\xi_u(x)-\xi_v(y)\| \\ &=\|(u-r(x)-(v-r(y))\| \\ &\leq \|u-v\| + \operatorname{Lip}(r) \|x-y\| \\ &\leq \|u-v\| + \operatorname{Lip}(r) \|f^{-1}(u)-f^{-1}(v)\| \end{align} where $f(x)=u$ and $f(y)=v$. This proves that $$\|f^{-1}(u)-f^{-1}(v) \| \leq \frac{1}{1-\operatorname{Lip}(r)} \|u-v\|.$$