Contractive family of functions

Question

Problem

Let $\{K_{\alpha}:C\to C\}_{\alpha \in \mathbb{R}^{n}}$ be a family of functions where $C\subset X$ is a closed subset of a Banach space X. Suppose that for some $\theta \in (0,1)$ we have that $\| K_{\alpha}(x) - K_{\alpha}(y)\| \leq \theta \| x-y\|$ and $K_{\alpha}(x)$ is continuous for some fixed x. Lets call $\overline{x} \in C $ the unique fixed point of $K_{\alpha}$ (the contraction).

Show that the function $\alpha \to \overline{x}$ is continous.

Im totally lost in this one, so any help would be really appreciated.

Thanks so much in advance guys <3

Answer 1

Let us denote the unique fixed point of $K_{\alpha}$ by $f(\alpha).$

Counterexample: $C=X= \mathbb R, n=1$ and

$K_{\alpha}(x)= \frac{\alpha}{2}x$, if $| \alpha |<1$ and $K_{\alpha}(x)= \frac{1}{2 \alpha}(1-x)$, if $| \alpha | \ge 1$. In this case we have $ \theta=1/2$ and

$f(\alpha)=0$, if $| \alpha |<1$ and $f(\alpha)=\frac{1}{2 \alpha +1}$, if $| \alpha | \ge 1$.

$f$ is not continuous !