# Absolutely continuous function with bounded derivative on an open interval is Lipschitz

I’ve come across a question which states that one can prove $$f:mathbb{R} rightarrow mathbb{R}$$ is Lipschitz iff $$f$$ is absolutely continuous and there exists $$M in mathbb{R}$$ such that $$|f'(x)|≤ M$$ almost everywhere.

I’ve only ever seen this fact proven for functions on closed and bounded intervals. Is it possible to show this on all of $$mathbb{R}$$? I’m skeptical as the main fact I’d like to use which represents $$f$$ as an indefinite integral ( since $$f$$ is absolutely continuous) requires the domain to be a closed and bounded interval. However I can’t find a counterexample.

Mathematics Asked on November 18, 2021

If $$f$$ is absolutely continuous then by the fundamental theorem of calculus Lebesgue version) $$f'$$ exists (although here we are assuming this already), $$f'$$ is integrable (in any compact interval $$[x,y]$$ and $$f(y)-f(x)=int^y_xf'(t),dt,quad xleq y$$ If $$|f'|leq M$$ almost surely, then $$|f(y)-f(x)|leq M|y-x|$$ which means $$f$$ is Lipchitz

The converse is much easier. Suppose $$|f(x)-f(y)| for all $$xleq y$$, then for any finite number of finite disjoint intervals $$[a_j,b_j]$$ we have $$sum_j|f(b_j)-f(a_j)|leq For $$varepsilon>0$$, let $$delta=varepsilon/M$$ so that if $$sum_j|b_j-a_j|, then $$sum_j|f(b_j)-f(a_j)|.

Answered by Jean L. on November 18, 2021

$$Rightarrow:$$ Suppose that $$f$$ is Lipschitz continuous, then there exists $$M>0$$ such that $$|f(x)-f(y)|leq M|x-y|$$ for all $$x,yinmathbb{R}$$. Let $$varepsilon>0$$ be given. Define $$delta=frac{varepsilon}{2M}>0$$. Let $${I_{n}=(a_{n},b_{n})mid ninmathbb{N}}$$ be a countable family of pairwisely disjoint open intervals such that $$sum_{n=1}^{infty}mu(I_{n}). (Here, $$mu(A)$$ denotes the Lebesgue measure of a set $$A$$.) $$begin{eqnarray*} & & sum_{n=1}^{infty}|f(b_{n})-f(a_{n})|\ & leq & sum_{n=1}^{infty}Mmu(I_{n})\ & < & varepsilon. end{eqnarray*}$$ This shows that $$f$$ is absolutely continuous. From the standard theory of any real analysis textbook, for each $$Ninmathbb{N}$$, $$f'(x)$$ exists a.e. for $$xin[-N,N]$$. Let $$A_{n}={xin[-N,N]mid f'(x)mbox{ does not exist}.}$$, then $${xinmathbb{R}mid f'(x)mbox{ does not exist}}=cup_{N}A_{N}$$ which has measure zero. That is, $$f'(x)$$ exists a.e. Let $$xinmathbb{R}$$ for which $$f'(x)$$ exists. For any $$y>x$$, we have $$|frac{f(y)-f(x)}{y-x}|leq M$$. Letting $$yrightarrow x+$$, then we have $$|f'(x)|leq M$$.

$$Leftarrow:$$ Suppose that $$f$$ is absolutely continuous and there exists $$M>0$$ such that $$|f'(x)|leq M$$ a.e.. Let $$x_{1},x_{2}inmathbb{R}$$ with $$x_{1}. From the standard theory with $$f$$ restricted on $$[x_{1},x_{2}]$$, we have that $$f(x_{2})-f(x_{1})=int_{x_{1}}^{x_{2}}f'(x)dx$$. Hence $$begin{eqnarray*} & & |f(x_{2})-f(x_{1})|\ & leq & int_{x_{1}}^{x_{2}}|f'(x)|dx\ & leq & M(x_{2}-x_{1}). end{eqnarray*}$$ Therefore, $$f$$ is Lipschitz continuous.

Answered by Danny Pak-Keung Chan on November 18, 2021

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