absolutely continuous function

The basic idea behind a continuous function is that the output of the function can be made to change by only a small amount so long as the input is allowed to change by only a small amount. There are, of course, different ways to make this precise, including uniformly continuous functions and Lipschitz continuous functions. With an absolutely continuous function, you allow multiple changes to multiple inputs to be combined into a single total change (and you consider the absolute values of the changes, so that they won't cancel).

The result is a notion of function that gets along well with the fundamental theorem of calculus in the context of the Lebesgue integral on the real line. Absolute continuity is weaker than Lipschitz continuity but stronger than mere (pointwise) continuity.

The first definition below is the most elementary; that the others are equivalent are important theorems.

Let $a$ and $b$ be real numbers, and let $f$ be a real-valued function on the interval $[a,b]$. Then $f$ is **absolutely continuous** on $[a,b]$ iff:

Given any positive number $\epsilon$, for some positive number $\delta$, given any natural number $n$ and any $2n$-tuple of elements of $[a,b]$, interpreted as an increasing $n$-tuple of nonoverlapping subintervals of $[a,b]$, if the total length of the intervals is less than $\delta$, then the total variation of $f$ on the intervals is less than $\epsilon$. That is (after $\epsilon$ and $\delta$), given $a \leq a_1 \leq b_1 \leq a_2 \leq b_2 \leq \cdots \leq a_n \leq b_n \leq b$, if

$\sum_{i = 1}^n (b_i - a_i) \lt \delta ,$

then

$\sum_{i = 1}^n {|{f(b_i) - f(a_i)}|} \lt \epsilon .$

Various trivial variations of this may be met with: the comparison with $\delta$ and/or $\epsilon$ may be weak instead of strict; the number of subintervals may be infinite (so long as they are still nonoverlapping), since an infinite sum (of nonnegative numbers, as we have here) is simply a supremum of finite sums; and of course we may start by specifying that the $2n$ numbers come in order as the endpoints of the $n$ subintervals, rather than starting with any $2n$ numbers and then putting them in order and forming the subintervals from those. (Note that putting them in order is fine even in constructive analysis, since choosing the $i$th element in order from a list of rational numbers is continuous, so may be extended constructively to real numbers, although we can't assume that the final list is a permutation of the original list.)

For the next definition, fix a model of nonstandard analysis.

Given any hypernatural number $n$ in the model and any $2n$-tuple of elements of the nonstandard extension of $[a,b]$, interpreted as an increasing $n$-tuple of nonoverlapping subintervals of $[a,b]$, if the total length of the intervals is infinitesimal, then the total variation of $f$ on the intervals is infinitesimal. That is, given hyperreal numbers $a \leq a_1 \leq b_1 \leq a_2 \leq b_2 \leq \cdots \leq a_n \leq b_n \leq b$, if

$\sum_{i = 1}^n (b_i - a_i) \approx 0 ,$

then

$\sum_{i = 1}^n {|{f^*(b_i) - f^*(a_i)}|} \approx 0 .$

See Tuckey 1993, pages 34–36.

That the next definition is equivalent is the fundamental theorem of calculus for the Lebesgue integral on the real line.

There exists a Lebesgue-integrable function $g$ on $[a,b]$ such that $f(x) = f(a) + \int_{x=a} g(x) \,\mathrm{d}x$ for $x \in [a,b]$. (This is a semidefinite integral.)

In this case, $g$ must equal the derivative $f'$ almost everywhere on $[a,b]$. (So in particular, $f$ is differentiable almost everywhere with a Lebesgue-integrable derivative, although this is not enough without requiring that $f$ be an indefinite integral of its derivative.)

The function $f$ is uniformly continuous on $[a,b]$, $f$ is of bounded variation? on $[a,b]$, and the direct image under $f$ of any null subset of $[a,b]$ is null.

The Stieltjes measure? $\mathrm{d}f$ is absolutely continuous with respect to Lebesgue measure on $[a,b]$.

The last of these is the source of the term ‘absolutely continuous’ as applied to measures.

One may easily generalize the codomain of the elementary definition of absolutely continuous functions to any metric space.

The cube-root function is absolutely continuous (on any bounded interval) but not Lipschitz continuous (on any interval containing $0$).

The Cantor function? is *not* absolutely continuous, even though it is continuous, and differentiable almost everywhere, with a Lebesgue-integrable derivative.

- Curtis Tuckey. 1993. Nonstandard Methods in the Calculus of Variations. CRC Press. Google books snippet.

Last revised on April 14, 2018 at 02:21:06. See the history of this page for a list of all contributions to it.