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