distance between any two points in a metric space and its diameter

Question

Could anyone tell me how to prove the following?

$d(x,y)\le \text{diam }(S)^{1-\alpha}\cdot d(x,y)^{\alpha}$ where $S$ is any complete, separable metric space. Or compact metric space. $0<\alpha\le 1$? Thanks for helping.

$\text{ diam }S=\sup\{d(x,y):x,y\in S\}$. thanks for helping to proceed.

$d(x,y)\le \text{diam} (S)$

Now, If $d(x,y)<1\Rightarrow (d(x,y))^{\alpha-1}>1\Rightarrow(d(x,y))\times \text{ diam }(S)$

$\iff d(x, y)\leq \text{diam}(S)$ but this is from definition of diameter. Holds for any kind of metric space — Jakobian, Apr 11 '19 at 22:25

score 1 · Accepted Answer · answered Apr 11 '19 at 22:32

1

By the definition of diameter, $$ d(x, y) \leq \mbox{diam}(S). $$ Raise both sides to the power $(1 - \alpha)$: $$ d(x, y)^{1 - \alpha} \leq \mbox{diam}(S)^{1 - \alpha}. $$ Now multiply both sides by $$ d(x, y)^{\alpha}. $$

answered Apr 11 '19 at 22:32

avs

6,276

is that result true for any metric space? – Myshkin Apr 11 '19 at 22:33
Yes. I didn't use any special properties of the metric space, only the existence of a metric. – avs Apr 12 '19 at 02:44

distance between any two points in a metric space and its diameter

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