# Difference between revisions of "Talk:Absolute continuity"

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: Why not? --[[User:Boris Tsirelson|Boris Tsirelson]] 13:21, 30 July 2012 (CEST) | : Why not? --[[User:Boris Tsirelson|Boris Tsirelson]] 13:21, 30 July 2012 (CEST) | ||

: Fine by me [[User:Camillo.delellis|Camillo]] 14:08, 30 July 2012 (CEST) | : Fine by me [[User:Camillo.delellis|Camillo]] 14:08, 30 July 2012 (CEST) | ||

+ | |||

+ | Between metric setting and References I would like to type the following lines. But for some reason which is misterious to me, any time I try the page comes out a mess... [[User:Camillo.delellis|Camillo]] 10:45, 10 August 2012 (CEST) | ||

+ | |||

+ | |||

+ | if for every $\varepsilon$ there is a $\delta > 0$ such that, | ||

+ | for any $a_1<b_1<a_2<b_2<\ldots < a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have | ||

+ | \[ | ||

+ | \sum_i d (f (b_i), f(a_i)) <\varepsilon\, . | ||

+ | \] | ||

+ | The absolute continuity guarantees the uniform continuity. As for real valued functions, there is a characterization through an appropriate notion of derivative. | ||

+ | |||

+ | '''Theorem 1''' | ||

+ | A continuous function $f$ is absolutely continuous if and only if there is a function $g\in L^1_{loc} (I, \mathbb R)$ such that | ||

+ | \begin{equation}\label{e:metric} | ||

+ | d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\, | ||

+ | \end{equation} | ||

+ | (cp. with ). This theorem motivates the following | ||

+ | |||

+ | '''Definition 2''' | ||

+ | If $f:I\to X$ is a absolutely continuous and $I$ is compact, the metric derivative of $f$ is the function $g\in L^1$ with the smalles $L^1$ norm such that \ref{e:metric} holds (cp. with ) |

## Revision as of 10:45, 10 August 2012

I moved some portions of the old article in Signed measure. I have not had the time to add all Mathscinet and Zentralblatt references. Camillo 22:54, 29 July 2012 (CEST)

Could I suggest using $\lambda$ rather than $\mathcal L$ for Lebesgue measure since

- it is very commonly used, almost standard
- it would be consistent with the notation for a general measure, $\mu$
- calligraphic is being used already for $\sigma$-algebras

--Jjg 12:57, 30 July 2012 (CEST)

- Why not? --Boris Tsirelson 13:21, 30 July 2012 (CEST)
- Fine by me Camillo 14:08, 30 July 2012 (CEST)

Between metric setting and References I would like to type the following lines. But for some reason which is misterious to me, any time I try the page comes out a mess... Camillo 10:45, 10 August 2012 (CEST)

if for every $\varepsilon$ there is a $\delta > 0$ such that,
for any $a_1<b_1<a_2<b_2<\ldots < a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have
\[
\sum_i d (f (b_i), f(a_i)) <\varepsilon\, .
\]
The absolute continuity guarantees the uniform continuity. As for real valued functions, there is a characterization through an appropriate notion of derivative.

**Theorem 1**
A continuous function $f$ is absolutely continuous if and only if there is a function $g\in L^1_{loc} (I, \mathbb R)$ such that
\begin{equation}\label{e:metric}
d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\,
\end{equation}
(cp. with ). This theorem motivates the following

**Definition 2**
If $f:I\to X$ is a absolutely continuous and $I$ is compact, the metric derivative of $f$ is the function $g\in L^1$ with the smalles $L^1$ norm such that \ref{e:metric} holds (cp. with )

**How to Cite This Entry:**

Absolute continuity.

*Encyclopedia of Mathematics.*URL: http://www.encyclopediaofmath.org/index.php?title=Absolute_continuity&oldid=27468