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 $$ \bigg|\frac{\ln x_1-\ln x_2}{\ln x_1\,\ln x_2}\bigg|\leq C|x_1-x_2|^\alpha. $$


1 Answer 1


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.


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