# Is $f(x)=\frac{1}{\ln x},\quad x\in(0,\,\frac{1}{2}]$ Hölder continuous?

Is $$f(x)=\frac{1}{\ln x},\quad x\in(0,\,\frac{1}{2}]$$ Hölder continuous?

That is, are there constant $$C>0$$ and $$\alpha>0$$ such that $$|f(x_1)-f(x_2)|\leq C|x_1-x_2|^\alpha \text{ ?}$$

Since $$f$$ is bounded and continuous on $$(0,\,\frac{1}{2}]$$, I can only prove it's uniformly continuous. But I don't know how to handle $$\bigg|\frac{\ln x_1-\ln x_2}{\ln x_1\,\ln x_2}\bigg|\leq C|x_1-x_2|^\alpha.$$

• It is not Holder continuous. See fourth Example in en.wikipedia.org/wiki/H%C3%B6lder_condition Commented Oct 9, 2019 at 8:09
• Thanks. Can you give a proof?
– Knt
Commented Oct 9, 2019 at 9:22

Suppose there exist $$C,\alpha>0$$, such that for all $$x_1,x_2\in(0,\dfrac{1}{2}]$$, we have $$\lvert\dfrac{1}{\log x_1}-\dfrac{1}{\log x_2}\rvert\leq C\lvert x_1-x_2\rvert^{\alpha}$$. Let $$x_1\to0$$, we have $$\lvert\dfrac{1}{\log x_2}\rvert\leq C\lvert x_2\rvert^\alpha$$, which is $$\lvert \dfrac{1}{x_2^\alpha\log x_2}\rvert\leq C$$. Taking $$x_2\to0$$, by L'hopital rule, we see that $$\lvert \dfrac{1}{x_2^\alpha\log x_2}\rvert\to +\infty$$, a contradiction.