Is a uniform limit of absolutely continuous functions absolutely continuous?

[AxiomOfChoice]

I was reading a Ph.D. thesis this morning and came across the claim that "a uniform limit of absolutely continuous functions is absolutely continuous." Is this true? What about the sequence of functions that converges to the Cantor function on [0,1]? Each of those functions is absolutely continuous, right? And they converge uniformly to the Cantor function, right? But the Cantor function is the canonical example of a continuous, increasing function that's not absolutely continuous!

Just so you know what I mean, I'm talking about the sequence \{ g_n(x) \}, where g_n is constant on the middle-third that is removed in stage n of constructing the Cantor set, and linear everywhere else. For example:

<br /> g_1(x) = \begin{cases} \frac{3x}{2}, &amp; x\in [0,1/3],\\ \frac 12, &amp; x\in (1/3,2/3),\\ \frac{3x}{2} - \frac 12, &amp; x\in [2/3,1]. \end{cases}<br />

This guy is clearly the integral of his derivative, so I think it's reasonable to conclude that each of the other ones is, too.

 

[AxiomOfChoice]

(As an aside...isn't it true that any function that is some piecewise combination of constant and linear functions is absolutely continuous on a closed, bounded interval?)

 

[mathman]

I haven't checked it out, but is the Cantor function a UNIFORM limit?

 

[TylerH]

This Cantor function isn't absolutely continuous. http://en.wikipedia.org/wiki/Cantor_function

 

[Citan Uzuki]

(As an aside...isn't it true that any function that is some piecewise combination of constant and linear functions is absolutely continuous on a closed, bounded interval?)

Yes, that is correct (assuming that the segments meet at the endpoints, of course). And your reasoning in the OP is correct as well. Uniform convergence of absolutely continuous functions does not, by itself, imply convergence. So unless your Ph. D. included some other condition on the functions (such as the derivatives also converging uniformly, or at least being uniformly bounded above by some integrable function), then (s)he is quite in error. Hopefully their thesis did not depend on that assertion!

 

[mathman]

This Cantor function isn't absolutely continuous. http://en.wikipedia.org/wiki/Cantor_function

You misunderstood my comment. I was wondering iif the Cantor function was a UNIFORM limit of absolutely continuous functions. I am well aware of the fact that it is not absolutely continuous.

 

[mathman]

I checked further. The Cantor function is the uniform limit of absolutely continuous functions.