absolutely continuous

liyi

$f$ is absolutely continuous on $\mathbb{R}$
Prove that there exist two non-negative constants $A$ and $B$ such that
$|f(x)|\leq A|x|+B$ for all $x\in \mathbb{R}$

**Posted:** Thu Jan 27, 2005 12:14 am

alekk

**Posted:** Thu Jan 27, 2005 7:29 am

_________________
Please, could you rephrase your question in the form of the answer?

liyi

maybe you mean that
$|y-x|<1/A$ ?

**Posted:** Thu Jan 27, 2005 5:28 pm

Leon Parsek · P versus NP Offline Joined: 30 Jan 2005 Posts: 21

Maybe I dind't understand your question.

Otherwise the statement is wrong, since e.g. $f(x)=e^x$ or $f(x)=x^2$ are absolutely continuous on $\mathbb R$ but not bounded by $A|x|+B$ for any $A,B\in\mathbb R$ ?

Absolute continuity - without any reference to a specific measure - means, that $f(x)$ can be written as an (parameter-)integral w.r.t Lebesgue-measure, e.g. $x^2=81+\int_9^{x} 2u\;du$ ...

It seems that your question originates from complex analysis, and 'absolutely continuous' has to be changed into 'entire + ...' or something else (I can't remember)?

Hm,
Leon

**Posted:** Sun Jan 30, 2005 1:52 pm

grobber

I'm pretty sure liyi was referring to something like this, in this case. $f(x)=x^2$ is not absolutely continuous, and neither is $e^x$ . First of all, the function should be uniformly continuous, and your examples are not.

**Posted:** Sun Jan 30, 2005 2:08 pm

Leon Parsek · P versus NP Offline Joined: 30 Jan 2005 Posts: 21

Dear grabber

The definitions given on the linked page coincide with my (read theorem 2): an absolutely continuous function is a function with (Lebesgue- or Radon-Nikodym-) derivative.

Absolutely continuity on $\mathbb R$ does not imply uniform continuity. If instead liyi means** a.c. on a closed interval $[a,b]$ , the statement is trivial, since we may chose arbitrary large constants $A$ and $B$ .

Sincerely,
Leon

*(do you [like me] hate people that throw notions like an octopus squirts ink?)
**(please excuse my english)

**Posted:** Sun Jan 30, 2005 2:36 pm

liyi

**Posted:** Sun Jan 30, 2005 5:16 pm

liyi

**Posted:** Sun Jan 30, 2005 5:19 pm

Leon Parsek · P versus NP Offline Joined: 30 Jan 2005 Posts: 21

Dear Liyi, dear grabber

I have now discovered our/my/your problem.

1. Absolute continuity (w.r.t. Lebesgue measure $\lambda$ ) is a notion of measure theory, where it is used for a measure $\mu$ if $\lambda(A)=0$ implies $\mu(A)=0$ for measurable sets $A$ . Using this definition even unbounded measures $\mu$ are allowed - I first supposed, that this definition was meant. The Radon-Nikodym theorem (theorem 2 of the link) says that $\mu(A)=\int_A f\;d\lambda$ for some function $f$ that is locally $L_1(\mu)$ (here we have our problem). Thus my examples are a.c. in this sense (see e.g. Mr.Rudin's book).

2. But you are both right, if you restrict your attention to finite (real-valued) measures $\mu$ . For these measures the Radon-Nikodym theorem says, that $\mu(A)=\int_A f\;d\lambda$ for some proper $L_1(\mu)$ -function $f$ . Moreover in this case the alternative definition, grabber gave in his link, applies: $\mu$ is a.c. iff for every $\epsilon>0$ there is a $\delta>0$ , so that $|\lambda(A)|<\delta$ implies $|\mu(A)|<\epsilon$ . These functions are clearly uniformly continuous.

And after sleeping $n$ nights, with $n>0$ , I now think that Liyi actually meant this definition, as grabber suspected. So let's forget the more general definition and focus on absolutely continuous functions, 2nd definition, and try to solve problem.

Regards,
Leon

**Posted:** Mon Jan 31, 2005 4:22 am

Leon Parsek · P versus NP Offline Joined: 30 Jan 2005 Posts: 21

I just recognized that alekk gave a solution to the problem 250 years ago (sorry for the confusion).

Here is my version. According to the uniform continuity of $f$ we can find an $A>0$ , so that $|f(x)-f(y)|<1$ whenever $|x-y|<1/A$ . Then letting $A|x|\leq n\leq A|x|+1$ we get $|f(x)-f(0)|=|\sum_{k=1}^n f(xk/n)-f(x(k-1)/n)|$
$\leq\sum_{k=1}^n |f(xk/n)-f(x(k-1)/n)|\leq n\leq A|x|+1$ ,
so that $|f(x)|\leq A|x|+B$ with $B=|f(0)|+1$ .

Leon

PS: Can we sharpen the result, if we use absolute continuity instead of the weaker condition of uniform continuity?

**Posted:** Mon Jan 31, 2005 5:53 am

liyi

hmm. I was wondering that why it needs "absolutely continuous" which is stricter than "uniformly continuous"...
So I posted it...

**Posted:** Mon Jan 31, 2005 5:26 pm