Convex Combination of Contraction Maps is a Contraction Map

Question

I'm working on the following problem:

Let $K$ be a closed bounded subset of a Hilbert Space $H$ and let $F: K \rightarrow K$ be a nonexpansive mapping $\bigg(d(F(x),F(y)) \leq \alpha d(x,y), \alpha \in [0,1]\bigg)$. Let $W$ be a contraction $\bigg(\alpha \in [0,1)\bigg)$on $H$ such that $W(K) \subset K$. Show that for $0 < \epsilon < 1$, $$F_{\epsilon} = (1-\epsilon)F + \epsilon W$$ is a contraction mapping from $K$ to $K$. Further let $x_{\epsilon}$ be the unqiue fixed point of $F_{\epsilon}$. Show that $x_{\epsilon}$ converges to a fixed point of $F$ as $\epsilon$ tends to $0$.

To show that $F_{\epsilon}$ is a contraction: $$d(F_{\epsilon}(x),F_{\epsilon}(y)) \\ = d(F(x)-\epsilon F(x) + \epsilon W(x),F(y)-\epsilon F(y) + \epsilon W(y))$$ Then I want to break this up using some kind of triangle inequality, but the I need something like: $$d(x+y,z) \leq d(x,z) + d(y,z)$$ but I don't think this is true, so I'm getting stuck. Needless to say, the second part eludes me as well.

Answer 1

Well, this is a normed space, therefore $d(a+b,c+d)\le d(a,c)+d(b,d)$ does hold.

Therefore \begin{align}&d((1-\epsilon)F(x)+\epsilon W(x),(1-\epsilon)F(y)+\epsilon W(y))\le\lvert 1-\epsilon\rvert d(F(x),F(y))+\lvert\epsilon\rvert d(W(x),W(y))\\&\le ((1-\epsilon)\alpha_F+\epsilon\alpha_W)d(x,y)\end{align}

If $\alpha_W\ge\alpha_F$, then $$(1-\epsilon)\alpha_F+\epsilon\alpha_W\le\alpha_W<1$$ If $\alpha_W<\alpha_F$, then $$(1-\epsilon)\alpha_F+\epsilon\alpha_W=\alpha_F-\epsilon(\alpha_F-\alpha_W)<\alpha_F\le 1$$

Answer 2

We have $d(F_{\epsilon}(x),F_{\epsilon}(y))=||F_{\epsilon}(x)-F_{\epsilon}(y)||$, where $|| \cdot ||$ is the norm on $H$.

With the triangle inequality for the norm you should get:

$||F_{\epsilon}(x)-F_{\epsilon}(y)|| \le q ||x-y||$ , where $q=(1- \epsilon) \alpha + \epsilon \beta$.

Here is $\alpha$ the constant for $F$ and $ \beta $ is the constant for $W$.

Show that $q <1.$