Timeline for convexity in linear metric spaces
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2018 at 13:41 | comment | added | user114263 | 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 $$d(0,w)=\sqrt{10}>3 = \frac{1}{2}d(0,x)+\frac{1}{2}d(0,y).$$ | |
| Jun 29, 2018 at 13:38 | comment | added | user114263 | 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. | |
| Jun 29, 2018 at 13:37 | vote | accept | mark haokip | ||
| Jun 29, 2018 at 13:33 | comment | added | user114263 | 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. | |
| Jun 29, 2018 at 4:54 | comment | added | mark haokip | So, can we infer that in a linear space, the notion of convex structure and convexity coincide? | |
| Jun 27, 2018 at 20:47 | history | answered | user114263 | CC BY-SA 4.0 |