Is $\log(x)$ uniformly continuous on $(\frac{1}{2},\infty)$?

Question

I think the answer will be true.

Here's an attempt:

Let us divide the interval in $(\frac{1}{2},1)$ and $[1,\infty)$.

In the domain $[1,\infty)$ the derivatives of $log(x)$ is bounded so it is uniformly continuous.

In the domain $(\frac{1}{2},1)$ the function is continuous and also $lim_{x \to 1^{-}} log(x)$ exists finitely and $lim_{x \to \frac{1}{2}^{+}} log(x)$ also exists finitely.

Hence the function is uc on the entire domain.

Answer 1

Yes, and you don't even need to split. On your interval you have $|\log'x|≤2$. Hence, by the Mean Value Theorem, $$ |\log x-\log y|=|\log'\xi|\,|x-y|≤2|x-y|,\qquad\qquad x, y≥\frac12. $$