UMD Analysis Qualifying Exam/Jan11 Real

Problem 1

Let $f\in AC[0,1]$ be an absolutely continuous function on [0,1] with $f>0$. Prove that $1/f\in AC[0,1]$.

Solution

Since $f$ is (absolutely) continuous on [0,1] with $f>0$ then there exists some $0<m=\min _{x\in [0,1]}f(x)$.

Since $f\in AC[0,1]$ then for any $\epsilon >0$ there exists some $\delta >0$ such that for any finite collection of disjoint intervals $I_{k}=(x_{k},y_{k}),k=1,...,n$ such that if $\sum _{k=1}^{n}|y_{k}-x_{k}|<\delta$ then $\sum _{k=1}^{n}|f(y_{k})-f(x_{k})|<\epsilon m^{2}$.

Then for any such collection of intervals described above, we have $\sum _{k=1}^{n}|{\frac {1}{f(y_{k})}}-{\frac {1}{f(x_{k})}}=\sum _{k=1}^{n}|{\frac {f(x_{k})-f(y_{k})}{f(y_{k})f(x_{k})}}|\leq \sum _{k=1}^{n}{\frac {1}{m^{2}}}|f(y_{k})-f(x_{k})|<\epsilon$.