Let $X$ be uniformly convex Banach space. $f:K\rightarrow K$, such that $\parallel fx-fy\parallel \leq\parallel x-y\parallel\,\,\forall x,y\in K $, with $K$ a nonempty, closed, convex, bounded subset of $X$.

Set $C_{\varepsilon}=\{x:\parallel x-fx\parallel\leq\varepsilon\}$, where $a=\lim\limits_{\varepsilon \rightarrow 0}\inf\limits_{C_{\varepsilon}}\parallel x\parallel$.

I want please to prove that the intersection of all sets $C_\varepsilon$ is nonempty. (We have $\inf\limits_{x\in K}\parallel x-fx\parallel=0$ and why if each $C_\varepsilon$ is closed it would follow that $a>0$).

Thank you.

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    $\begingroup$ Did you want to assume $a > 0$? $0 \in \mathbb K$? $\endgroup$ – Robert Israel Jul 9 '18 at 3:19
  • $\begingroup$ It's my idea to prove this, it's not a assumption and for $0$, we can assume that $0\in K$ $\endgroup$ – Youssef Sabar Jul 9 '18 at 9:27
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    $\begingroup$ You can't prove $a > 0$. What if $0$ is a fixed point? $\endgroup$ – Robert Israel Jul 9 '18 at 18:01
  • $\begingroup$ [researchgate.net/publication/… $\endgroup$ – Youssef Sabar Jul 9 '18 at 18:54

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