My question is how to prove that the function $$ f(x)=-\log^{-1}(x) $$ for $x>0$ and $f(0)=0$ on $[0,1/2]$ is absolutely continuous/uniform continuous.
I feel like both proofs would be similar, but I am having a hard time bounding $$ |f(a)-f(b)| = \left|\frac{\log(a/b)}{\log(a)\log(b)}\right|. $$ I think I am missing some obvious inequality and cannot seem to crack it. Could someone nudge me in the right direction?
Thanks.
For $x\in(0,\frac{1}{2})$, we have that $f'(x)=\frac{1}{x(\ln x)^{2}}.$ We go to verify that $f(x)-f(0)=\int_{0}^{x}f'(t)dt$ for each $x\in[0,\frac{1}{2}].$ Once it is proved, $f$ is absolutely continuous because it is a function defined by an integral.
Fix $x\in(0,\frac{1}{2}]$. Note that $f'\geq0$. By Monotone Convergence Theorem, we have that $\int_{0}^{x}f'(t)dt=\lim_{n\rightarrow\infty}\int_{0}^{x}1_{[\frac{1}{n},x]}(t)f'(t)dt$. However, since $f'$ is continuous on $[\frac{1}{n},x]$, by the classical Fundamental Theorem of Calculus, \begin{eqnarray*} & & \lim_{n\rightarrow\infty}\int_{0}^{x}1_{[\frac{1}{n},x]}(t)f'(t)dt\\ & = & \lim_{n\rightarrow\infty}\int_{\frac{1}{n}}^{x}f'(t)dt\\ & = & \lim_{n\rightarrow\infty}\left[f(x)-f(\frac{1}{n})\right]\\ & = & f(x)-f(0). \end{eqnarray*}