Absolute Continuity

**manjohn12** · December 19th 2008, 08:41 AM

How would you show that Lipschitz functions are absolutely continuous?

Suppose $f: X \to Y$ is a function. Let $k$ be a positive integer. If $d_{Y}(f(a), f(b)) \leq k d_{X}(a,b)$ for all $a,b \in X$ then $f$ is Lipschitz.

Here is the definition for absolute continuity. So choose $\delta$ so that it depends on the diameter of our interval $I$ . This exists by the least upper bound property?

**manjohn12** · December 19th 2008, 12:56 PM

Suppose $f: X \to Y$ is a function. Let $k$ be a positive integer. If $d_{Y}(f(a), f(b)) \leq k d_{X}(a,b)$ for all $a,b \in X$ then $f$ is Lipschitz.

A function $I \to X$ is absolutely continuous on $I$ if for every positive number $\varepsilon$ , there is a positive number $\delta$ such that whenever a sequence of pairwise disjoint sub-intervals $[x_{k}, y_{k}]$ of $I$ satisfies $\sum_{k} |y_{k}-x_{k}| < \delta$ then $\sum_{k} d(f(y_{k}), f(x_{k})) < \varepsilon$ .

Since $I \subseteq \mathbb{R}$ it has a finite diameter. Call this $p$ . Let $\varepsilon > 0$ . Choose $\delta = \frac{\varepsilon}{kp}$ . Then:

$\sum_{k} |y_{k}-x_{k}| < \delta \implies \sum_{k} d(f(y_{k}), f(x_{k})) \leq kd_{X}(a,b) < \frac{\varepsilon}{p}$ . Hence $f$ is absolutely continuous.

Is this correct?

**Plato** · December 19th 2008, 01:21 PM

Suppose that $f$ is Lipschitzian on $\left[ {a,b} \right]$ .
That means that there is a constant $M$ such that $\forall x,t \in \left[ {a,b} \right]\left[ {\left| {f(x) - f(t)} \right| \leqslant M\left| {x - t} \right|} \right]$ .

To show that $f$ is absolutely continuous on $\left[ {a,b} \right]$ :
if $\varepsilon > 0$ take your $\delta = \frac{\varepsilon }{M}$ .

Don’t make it any more complicated.

**manjohn12** · December 19th 2008, 01:24 PM

was I correct?

**Plato** · December 19th 2008, 02:13 PM

Originally Posted by manjohn12

was I correct?

Not really.
If $\mathbb{I} = \left\{ {\left[ {s,r} \right] \subset \left[ {a,b} \right]} \right\}$ is a collection of non-overlapping subintervals such that $\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < \delta$ then
$\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\left| {f(r) - f(s)} \right|} \leqslant \sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {M\ell \left( {\left[ {s,r} \right]} \right)} = M\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < M\delta = \varepsilon$

**manjohn12** · December 19th 2008, 04:08 PM

Originally Posted by Plato

Not really.
If $\mathbb{I} = \left\{ {\left[ {s,r} \right] \subset \left[ {a,b} \right]} \right\}$ is a collection of non-overlapping subintervals such that $\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < \delta$ then
$\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\left| {f(r) - f(s)} \right|} \leqslant \sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {M\ell \left( {\left[ {s,r} \right]} \right)} = M\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < M\delta = \varepsilon$

Is there any way you can classify continuity in terms of the spaces they define?

Like regular continuity is defined point wise, uniform continuity is defined space wise, and absolute continuity is defined interval wise. Could we use this information to order continuity to strongest to weakest as follows:

1. Absolute continuity
2. Uniform Continuity
3. Continuity

**manjohn12** · December 20th 2008, 10:54 AM

But you could have any number of intervals. For example you could have two intervals or you could have a sequence of infinite number pairwise disjoint intervals. Will $\delta = \varepsilon/k$ still work? Recall that in proving a Lipschitz funtion is uniformly continuous, $\delta = \varepsilon/(k+1)$ will work. We define $k$ as the Lipschitz constant.

We know that $\sum_{k} |y_{k}-x_{k}|$ will always be less than or equal to the length of the interval. So $\delta$ is independent of how you choose your intervals?

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