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 $0<m=\underset{x\in [0,1]}{\min }f(x)$.

Since $f\in AC[0,1]$ then for any $ϵ>0$ 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 $\sum _{k=1}^n|f(y_k)-f(x_k)|<ϵm^2$.

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