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Takahashi introduced the concept of convex structure in a metric space $(X,d)$ as a mapping $\mathcal{W}:X^2\times[0,1]\longrightarrow X$ satisfying $$d\left(z,\mathcal{W}(x,y,\alpha)\right)\leq\alpha d(z,x)+(1-\alpha)d(z,y)$$ for all $x,y,z\in X$ and $\alpha\in[0,1]$. If the metric $d$ is induced by the norm $\|.\|_X$, then

$$\mathcal{W}(x,y,\alpha)=\alpha x+(1-\alpha)y\qquad \forall x,y\in X$$

defines a convex structure on $X$. The problem arises when $d$ is not induced by a norm. By assuming $X$ to be a linear metric space (linear space with $d$ defined), can we show that $\mathcal{W}(x,y,\alpha)$ defined above is a convex structure on $X$.

I am at a loss. Any kind of help/suggestion needed. Thanks!

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Suppose $(X,d)$ is a linear metric space such that $B=\{x\in X: d(0,x)\leq 1\}$ is not convex (such as $L_p$ or $\ell_p^2$ when $0<p<1$). Since $B$ is not convex, there exist $x,y\in B$ and $0<\alpha<1$ such that $w=\alpha x+(1-\alpha)x\notin B$. Then with $z=0$, $$d(z,w)=d(0,w)>1,$$ since $w\notin B$. But $$d(z,x), d(z,y)\leqslant 1,$$ since $x,y\in B$, so $$\alpha d(z,x)+(1-\alpha)d(z,y) \leqslant \alpha+1-\alpha=1.$$ Therefore $$d(z, w)>1\geqslant \alpha d(z,x)+(1-\alpha)d(z,y).$$

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  • $\begingroup$ So, can we infer that in a linear space, the notion of convex structure and convexity coincide? $\endgroup$
    – mark haokip
    Commented Jun 29, 2018 at 4:54
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    $\begingroup$ If $(X,d)$ is a linear metric space, convexity of the sets $B(z,r):=\{x\in X: d(z,x)\leq r\}$ for every $z\in X$ and $r>0$ is a necessary but insufficient condition for $\mathcal{W}(x,y,\alpha)=\alpha x+(1-\alpha)y$ to be a convex structure. To see that it's necessary, just modify the previous example. But the condition that $B(z,r)$ is convex for each $z,r$ only yields that $$d(z,\alpha x+(1-\alpha)y)\leqslant \alpha d(z,x)+(1-\alpha)d(z,y)$$ in the particular case that $d(z,x)=d(z,y)$, in which case you just take $r=d(z,x)$. Then $\alpha x+(1-\alpha)y\in B(z,r)$ by convexity. $\endgroup$
    – user114263
    Commented Jun 29, 2018 at 13:33
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    $\begingroup$ For a specific counterexample, let $X=\mathbb{R}^2$ and let $\|\cdot\|$ be the $\ell_1^2$ norm $\|(a,b)\|=|a|+|b|$. Now define $$d(x,y)=\sqrt{\|x-y\|}.$$ This is a metric, because the inequality $\sqrt{a+b}\leqslant \sqrt{a}+\sqrt{b}$ for all $a,b\geqslant 0$ yields the triangle inequality, and the other conditions are easy to verify. For any $z\in X$ and $r>0$, $$B(z,r)=\{x\in X: d(z,x)\leqslant r\}=\{x\in X: \|z-x\|\leqslant r^2\}$$ is convex, since $\|\cdot\|$ is a norm. $\endgroup$
    – user114263
    Commented Jun 29, 2018 at 13:38
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    $\begingroup$ But now take $x=(16, 0)$, $y=(0,4)$, $w=\frac{x}{2}+\frac{y}{2}=(8,2)$, and $z=0$. Then $$d(0,w)= \sqrt{10},$$ $$\frac{d(0,x)}{2}=\frac{4}{2}=2,$$ $$\frac{d(0,y)}{2}= \frac{\sqrt{4}}{2}=1,$$ and $$d(0,w)=\sqrt{10}>3 = \frac{1}{2}d(0,x)+\frac{1}{2}d(0,y).$$ $\endgroup$
    – user114263
    Commented Jun 29, 2018 at 13:41

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