Ngô Quốc Anh

November 5, 2009

An equivalent criterion for absolutely continuous functions

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:29

In mathematics, absolute continuity is a smoothness property which is stricter than continuity and uniform continuity.

Definition. A finite function f on a finite interval [a,b] is said to be absolute continuous if and only if for given \varepsilon > 0, there exists \delta > 0 such that

\displaystyle\sum_k |f(b_k) - f(a_k)| < \varepsilon

for any collection (finite or not) \{[a_k, b_k]\} of non-overlapping subintervals of [a, b] with \sum (b_k - a_k) < \delta.

We now prove the following result.

Statement. Show that f is absolutely continuous on [a, b] if and only if given \varepsilon > 0, there exists \delta > 0 such that

\displaystyle \Big|\sum_k (f(b_k) - f(a_k)) \Big| < \varepsilon

for any finite collection \{[a_k, b_k]\} of non-overlapping subintervals of [a, b] with \sum (b_k - a_k) < \delta.

Proof. If f is absolutely continuous on [a, b], then the result is easily obtained by using the definition and the fact that |x + y| \leq |x| + |y| for every x,y \in \mathbb R.

Now we prove that

for given \varepsilon > 0, there exists \delta > 0 such that \sum |f(b_k) - f(a_k)| < \varepsilon for any finite collection \{[a_k, b_k]\} of non-overlapping subintervals of [a, b] with \sum (b_k - a_k) < \delta.

Indeed, we split the collection \{[a_k, b_k]\} into two types:

  • type A are all k such that f(b_k) - f(a_k) \geq 0 and
  • type B are all k such that f(b_k) - f(a_k) < 0.

For given \varepsilon > 0, there exists \delta > 0 such that | \sum (f(b_k) - f(a_k))| < \frac{\varepsilon}{3} for any finite collection \{[a_k, b_k]\} with \sum (b_k - a_k) < \delta. Then for k \in A we also have

\displaystyle\sum_{k \in A} \left( f(b_k) - f(a_k) \right) = \Big| \sum_{k \in A} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}.

Similarly,

\displaystyle\sum_{k \in B} \left( f(a_k) - f(b_k) \right) = \Big| \sum_{k \in B} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}.

From the following inequality a + b \leq |a-b| + b + b with a,b \geq 0 we deduce that

\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} \right|} = \underbrace {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} }_a + \underbrace {\sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} }_b

with the fact that

\displaystyle a+b \leqslant \left| {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} - \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} } \right| + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)}.

Note that the right hand side of the above inequality is bounded from above by

\displaystyle\frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon.

Thus, we have proved that for given \varepsilon >0, there exists \delta>0 such that

\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {b_k } \right) - f\left( {a_k } \right)} \right|} < \varepsilon

for any finite collection \{[a_k, b_k]\} of non-overlapping subintervals of [a, b] with \sum (b_k - a_k) < \delta. Letting n \to \infty we can claim that f is absolutely continuous.

2 Comments »

  1. I think you are excellent. Your web is very useful for people who studies analysis, study = learn and do research.

    Much achievement! Congratulations

    Comment by ULF — November 12, 2009 @ 5:10

    • Thank you.

      Comment by Ngô Quốc Anh — November 12, 2009 @ 17:05


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