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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

2 Answers

2 Answers

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

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 $sum_{n=1}^{infty}mu(I_{n})<delta$. (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 $x_{1}<x_{2}$. 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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