Convex Combination of Contraction Maps is a Contraction Map

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.

• It's not the part you're focusing on in this question, but to show that $F_\varepsilon$ maps $K$ into $K$, I'm pretty sure you will need to assume that $K$ is also a convex subset of $H$. – Aweygan May 6 at 14:48

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 $$(1-\epsilon)\alpha_F+\epsilon\alpha_W=\alpha_F-\epsilon(\alpha_F-\alpha_W)<\alpha_F\le 1$$

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 $$q <1.$$