Define $\mathcal{F}$ as the following set of continuous functions:

$$ \mathcal{F} := \left\{ f: \mathbb{R} \rightarrow \mathbb{R}^n \mid f(\cdot) \ \text{contin.}, \ f(x) \in K(x) \subset \mathbb{R}^n \ \forall x \in \mathbb{R}, \ K(x) \neq \varnothing \ \text{compact, } K(\cdot) \ \text{contin.} \right\} $$

I am looking for examples of contractions operator $\mathcal{C}: \mathcal{F} \rightarrow \mathcal{F}$.

  • $\begingroup$ I don't understand what $K$ is. From the way you have set it up it is a map $\mathbb{R} \rightarrow \{A \subset \mathbb{R}: A$ is compact $\}$. What does it mean for such a function to be continuous? $\endgroup$ – Frank Apr 1 '14 at 19:36
  • $\begingroup$ I mean that $K: \mathbb{R} \rightrightarrows \mathbb{R}^n$ is a continuous, compact-valued set-valued mapping. $\endgroup$ – user693 Apr 1 '14 at 19:39
  • $\begingroup$ What topology do you endow the set of compact sets with? $\endgroup$ – Frank Apr 1 '14 at 19:43

Try $C:F\to F$ such that $C(f)=\frac{1}{2}f$, then $\|C(f)-C(g)\|=\|\frac{1}{2}(f-g)\|\lt\|f-g\|$, contraction?

  • $\begingroup$ I'd suggest maybe an integral operator or something, but there are no assumptions about integration or differentiation in the set $F$. $\endgroup$ – Ellya Apr 1 '14 at 19:31
  • $\begingroup$ I am not sure about $C[f(\cdot)] := \frac{1}{2} f(\cdot)$. In fact, it can be that - for some $x \in \mathbb{R}$ - $f(x) \in K(x)$ while $\frac{1}{2} f(x) \notin K(x)$. $\endgroup$ – user693 Apr 1 '14 at 19:42
  • $\begingroup$ is that because (as a hypothetical example) let $K\subset \mathbb{R}$ s.t. $K=[2,3]$ then let $f(x)=5/2\in K$, but $\frac{1}{2}f=5/4$ which is not in $K$? $\endgroup$ – Ellya Apr 1 '14 at 19:46
  • $\begingroup$ Yes, exactly. Probably I can think of a certain function in $\mathcal{F}$ an try to "reduce the distance" from it. $\endgroup$ – user693 Apr 1 '14 at 19:50
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    $\begingroup$ Then you would have: $\|C(f)-C(h)\|=\|g-g\|=\|0\|=0\lt\|f-h\|$, so this seems like a good candidate, but not a very interesting one. $\endgroup$ – Ellya Apr 2 '14 at 6:32

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