I came across the statement ``Every closed and convex subset of a uniformly convex b-metric space is Chebyshev'' in [1]. Here, the term `convex' is in the sense of Takahashi. I tried looking up for the proof of the same without success. I would be grateful if someone can direct me to the proof. Thanks.


[1] H. Fukhar-ud-din, One step iterative scheme for a pair of nonexpansive mappings in a convex metric space, Hacet. J. Math. Stat., Vol. 44 (2015), 1023 – 1031.



Every closed and convex subset of a uniformly convex metric space is Chebyshev?
It's not stated in the paper, but you need the metric space to be complete. This is implicit in all papers on fixed point theory since otherwise the fixed points only live in the completion. With this proviso the claim holds and the proof is the same as in Hilbert space:


Let $(X,d)$ be a complete metric space, and let $W\colon X\times X\times[0,1]\to X$ be a "convex structure" in that for any $p,x,y\in X$ and $t\in[0,1]$ we have
$$ d\left(p,W(x,y,t)\right) \leq (1-t)d(p,x) + td(p,y)\,, $$
which is "uniformly convex" in that for $\epsilon>0$ there is $0<\eta<1$ such that for any $r>0$ and $p,x,y\in X$ with $d(p,x),d(p,x)\leq r$ and $d(x,y)\geq \epsilon r$ we have
$$ d\left(p,W(x,y,\frac12)\right) \leq \eta r\,. $$


Now let $C \subset X$ be a non-empty closed convex subset (this is defined to mean that $W(x,y,t)\in C$ if $x,y\in C$) and let $p\in X$. We need to show that the function $d(p,\cdot)\colon C\to\mathbb{R}\_{\geq 0}$ achieves a unique minimum. For this let $m = \inf\_{x\in C} d(p,x)$ (this exists since distances are bounded below) and, for $\delta>0$, let $A\_\delta = \{x\in C \mid d(p,x)\leq m+\delta \}$. These sets are non-empty since $m$ is an infimum. Also, as $\delta\searrow 0$, the sets $A\_\delta$ decrease. We shall show their diameters decrease to zero which, by Cantor's Lemma will show their intersection is a point.


Accordingly let $D = \inf\_{0<\delta\leq 1} \mathrm{diam}(A\_\delta)$. If $D>0$ set $\epsilon = \frac{D}{2(m+1)}$ so that for $0<\delta\leq 1$ there are points $x,y \in A\_\delta$ with $d(x,y) \geq \mathrm{diam}(A\_\delta)/2 \geq D/2 \geq \epsilon(m+\delta)$.


In summary we have Then $d(p,x),d(p,y) \leq m+\delta$ ($x,y\in A\_\delta$) while $d(x,y)\geq \epsilon (m+\delta)$. By the uniform convexity assumption there is $\eta = \eta(\epsilon)\in [0,1)$ and a midpoint $z = W(x,y,1/2)\in X$ such that $d(p,z) \leq \eta (m+\delta)$. By the convexity of $C$, $z\in C$.


Now for $\delta < \left(\frac{1}{\eta}-1\right)m$ (recall that $\eta<1$) we have $\eta (m+\delta) < m$, and this makes $d(p,z)$ strictly less than $m$, the infimum of such distances -- a contradiction. It follows that $D=0$ as claimed.


