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$\{f_n\}$ are absolutely continuous functions on $[0,1]$, we know that if $f_n$ are uniformly convergent to a function $f$, then $f$ is continuous.

The question is: is the function $f$ absolutely continuous?

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1 Answer 1

Recall the Weierstrass Approximation Theorem: for every continuous function $f: [0,1] \rightarrow \mathbb{R}$, there is a sequence of polynomials $P_n$ such that $P_n$ converges uniformly to $f$ on $[0,1]$.

Therefore if $P$ is any property of a function $f: [0,1] \rightarrow \mathbb{R}$ possessed by all polynomial functions, then any continuous function $f: [0,1] \rightarrow \mathbb{R}$ is a uniform limit of functions satisfying property $P$.

Try this out with $P$ being the property of absolute continuity: if

(1) Every polynomial function is absolutely continuous on $[0,1]$ and
(2) Every uniform limit of absolutely continuous functions is absolutely continuous, then
(3) Every continuous function on $[0,1]$ would be absolutely continuous.

I leave it to you to follow up on this syllogism and solve the problem.

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A concise way of saying this: The question is essentially asking whether the set of absolutely continuous functions is closed under uniform convergence. Pete points out that this set is dense (with respect to uniform convergence) in the space of all continuous functions. If a set is both closed and dense, what must it be? –  Nate Eldredge Sep 20 '11 at 16:16
@Nate: I didn't say that absolutely continuous functions are uniformly dense in all continuous functions. Perhaps the OP will draw that conclusion from what I said -- perhaps correctly! -- but I didn't say it. :) –  Pete L. Clark Sep 20 '11 at 16:21

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