There exists $L$ so that $\|x-\tilde{x}\|\leq \frac{1}{1-L}\cdot \sup_{z\in D} \|\phi(z) -\tilde{\phi}(z)\|\,$?

Question

Let $x=\phi(x)$ and $\tilde{x}=\tilde{\phi}(\tilde{x})$ be two fixed point equations, which meet the requirements of the Banach fixed point theorem in $D\subseteq \mathbb R^n$. Show, that there exists $0\leq L<1$, so that $$\|x-\tilde{x}\|\leq \frac{1}{1-L}\cdot \underset{z\in D}{\sup} \|\phi(z) -\tilde{\phi}(z)\|.$$

I started with $\|x-\tilde{x}\|=\|\phi(x)-\tilde{\phi}(x)\|$ and my own idea is to work with the triangle inequality.

Answer 1

I hope worth for you

$||x-\tilde{x}||=||\phi(x)-\tilde{\phi}(\tilde{x})||\leq ||\phi(x)-\phi(\tilde{x})+\phi(\tilde{x})-\tilde{\phi}(\tilde{x})+\tilde{\phi}(x)-\tilde{\phi}(x)|| \leq ||\phi(x)-\phi(\tilde{x})||+||\tilde{\phi}(x)-\tilde{\phi}(x)||+||\phi(\tilde{x})-\tilde{\phi}(\tilde{x})||\leq L_1||x-\tilde{x}||+L_2||x-\tilde{x}||+||\phi(\tilde{x})-\tilde{\phi}(\tilde{x})||$

then

$||x-\tilde{x}||(1-(L_1+L_2))\leq ||\phi(\tilde{x})-\tilde{\phi}(\tilde{x})||\leq \sup ||\phi(z)-\tilde{\phi}(z)||$. Or equivalently

$$||x-\tilde{x}||\leq \frac{1}{1-L}\sup ||\phi(z)-\tilde{\phi}(z)||$$ with $L=L_1+L_2$, and $L_i$ is the constant contraction, $i=1,2$

Answer 2

Let $\tilde{\phi}$ have Lipschitz constant $L<1$. Then $$ ||x-\tilde{x}|| = ||\phi(x)-\tilde{\phi}(\tilde{x})|| \le ||\phi(x)-\tilde{\phi}(x)|| + ||\tilde{\phi}(x)-\tilde{\phi}(\tilde{x})|| $$ $$ \le (\sup_{z \in D} ||\phi(z)-\tilde{\phi}(z)||) + L||x-\tilde{x}|| $$ $$ \Rightarrow ||x-\tilde{x}|| \le \frac{1}{1-L}\sup_{z \in D} ||\phi(z)-\tilde{\phi}(z)||. $$ This works even if $\phi$ is not a contraction, but has a fixed point $x$. The role of $\phi$ and $\tilde{\phi}$ can be interchanged. If $\phi$ is a contraction too, with constant $M<1$, say, then $$ ||x-\tilde{x}|| \le \frac{1}{1-\min\{L,M\}}\sup_{z \in D} ||\phi(z)-\tilde{\phi}(z)||. $$