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!



convexity in linear metric spaces
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 $$d(z, w)>1\geqslant \alpha d(z,x)+(1-\alpha)d(z,y).$$ 


