Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a lipschitz continuous function that satisfies $ \langle f(x)-f(y), x-y\rangle \geq \alpha\|x-y\|_2^2 \text { for all } x, y \in \mathbb{R}^n $ and fixed $\alpha>0$.
I am trying to find a range $\omega>0$ such that the fixpoint iteration $$ x=\Phi(x) \quad \text { mit } \quad \Phi(x):=x-\omega f(x) $$ converges.
I tried finding a lipschitz constant for $\Phi(x)$ such that $\Phi(x)$ is a contraction: $$\|\Phi(x)-\Phi(y)\|_2=\|x-\omega f(x)-y+\omega f(y)\|_2 \leq \|x-y\|_2+\omega\|f(x)-f(y)\|_2\leq \|x-y\|_2+\omega L\|x-y\|_2= \\ \|x-y\|_2(1+\omega L)$$
I am not to sure where to use that $f$ is strongly continous and how to choose $\omega$ accordingly. Can someone help me out?
For $x,y \in \mathbb{R}^n$ we have $$ \|\Phi(x)-\Phi(y)\|^2=\Big\langle x-y - \omega (f(x)-f(y)), x-y - \omega (f(x)-f(y)) \Big\rangle $$ $$ =\|x-y\|^2 + \omega^2\|f(x)-f(y)\|^2 - 2 \omega\Big\langle f(x)-f(y), x-y\Big\rangle $$ $$ \le (1+\omega^2 L^2) \|x-y\|^2 - 2\omega\alpha \|x-y\|^2 = (1+\omega^2 L^2- 2\omega\alpha ) \|x-y\|^2. $$ Thus, if $$ 1+\omega^2 L^2- 2\omega\alpha < 1 \iff \omega^2 L^2- 2\omega\alpha < 0 \iff \omega < \frac{2\alpha}{L^2} $$ then $\Phi$ is a contraction.
Note also that $$ \alpha \|x-y\|^2 \le \Big\langle f(x)-f(y), x-y\Big\rangle \le L\|x-y\|^2, $$ hence $\alpha \le L$. Therefore $1+\omega^2 L^2- 2\omega\alpha \ge 1+\omega^2 L^2- 2\omega L = (1-\omega L)^2 \ge 0$.