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I am trying to show that a function is a Lipschitz $M$ continuous if and only if it is absolutely continues and $|f'(x)| \leq M$.

I think I am on the right track:

Proof: (=>) Let f be Lipschitz M continues ie if $|f(x)-f(y)| \leq M|x-y|$ for all $x,y \in E=[a,b]$. now we want to show abs cont: $\sum^n_{i=1}|f(x'_i)-f(x_i)|< \epsilon$ if $\sum^n_{i=1}|x'_i-x_i|< \delta$ for all any finite collection of disjoint intervals $(x'_i,x_i)$.

Consider we let $\sum^n_{i=1}|x'_i-x_i|< \delta$, then we observe that $\sum^n_{i=1}|f(x'_i)-f(x_i)| \leq M \sum^n_{i=1}|x'_i-x_i|$ by the Lipschitz M continues and triangle inequality. Then define $\epsilon = M \delta$ and we are done.

To see that $|f'(x)| \leq M$ we can just let $x'_i=x_i+h$ and have the following $|f(x_i+h)-f(x_i)| \leq M|x_i+h-x_i|$ thus we have $\frac{|f(x_i+h)-f(x_i)|}{|h|}\leq M$ and if we take the limit and take it inside the absolute values we are done.

(<=) Now assume abs continuity and $|f'(x)|=M$ now by another theorem we know that f is absolute continues if and only if it is an indefinite integral $f(x)= \int_a ^x f'(t)dt +f(a)$ we can manipulate this to $f(x)-f(a) \leq \int_a ^x Mdt$ . Now we can integrate the RHS to get $\int_a ^x Mdt = M (x-a)$ now we have $f(x)-f(a) \leq M (x-a)$ we can take the absolute calues of both sides and we are done.

Does this seem correct?

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  • $\begingroup$ There is a minor technicality concerning a.e. vs everywhere. Otherwise, everything looks good. The derivative of an absolutely continuous function exists a.e., and the bound $|f'(x)| \le M$ will hold a.e.. if the function is Lipschitz with Lipschitz constant $M$. You might want to try to clean up the a.e. issues. $\endgroup$ – DisintegratingByParts Dec 6 '15 at 2:43
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Mostly correct. However, in the proof of $\Rightarrow$,

  • "define $\epsilon=M\delta$" is not logical, since $\epsilon$ is given. You define $\delta=\epsilon/M$.
  • Before proving $|f'(x)|\le M$ one has to discuss the existence of $f'(x)$. Cite the theorem saying that an absolutely continuous function is differentiable almost everywhere. The argument for $|f'(x)|\le M$ applies at the points of differentiability.

And in the proof of $\Leftarrow$, "we can take the absolute calues of both sides" is too hasty: $A\le B$ does not imply $|A|\le |B|$. Instead, follow the chain of inequalities again starting with $f'(x)\ge -M$ and arriving at $f(x)-f(a) \geq -M (x-a)$.

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