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 $a=\lim\limits_{\varepsilon \rightarrow 0}\inf\limits_{C_{\varepsilon}}\| x\|$.

I want please to prove that the intersection of all sets $C_\varepsilon$ is nonempty; see line 11, page 382 of this paper.

Thank you.

  • 4
    $\begingroup$ Where is $a$ used? $\endgroup$ – Meisam Soleimani Malekan Jul 9 '18 at 11:07
  • $\begingroup$ Please can you see the paper. $\endgroup$ – Youssef Sabar Jul 10 '18 at 23:44
  1. Each $C_\epsilon$ is closed and non-empty ( since $\inf\{\|x-fx\|: x\in K\}=0$).
  2. If $\bigcap_\epsilon C_\epsilon=\emptyset$, then $a>0$; by definition there exists a $y_\epsilon\in C_\epsilon$ such that $\|y_\epsilon\|\leq a(C_\epsilon)+\epsilon$, and $\|y_\epsilon-fy_\epsilon\|\leq\epsilon$, so if $a=0$, then $\lim_\epsilon y_\epsilon=0$, whence $0\in\bigcap_\epsilon C_\epsilon$.
  3. For $u_1,u_2\in C_\epsilon$, $\frac12(u_1+u_2)\in C_{\phi(\epsilon)}$; because of the computation in the paper.
  4. By 3. for $u_1,u_2\in D_\epsilon$, we have $\|u_1\|,\|u_2\|\leq a+\epsilon$ and $\frac12\|u_1+u_2\|\geq a(C_{\phi(\epsilon)})$, so by the lemma in the paper ($u=u_1$, $v=u_2$, and $w=0$), and the computation there, we get $\lim_{\epsilon\rightarrow0}\text{diam}(D_\epsilon)=0$ (this is where $a>0$ is used), so by Cantor's intersection theorem $\bigcap_\epsilon D_\epsilon\neq\emptyset$, and this is the contradication, since $\bigcap_\epsilon D_\epsilon\subset\bigcap_\epsilon C_\epsilon$ and we assume at first that $\bigcap_\epsilon C_\epsilon=\emptyset$.

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