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Difference between revisions of "User:Boris Tsirelson/sandbox1"

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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
 
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}
 
\begin{equation}\label{e:metric}
d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\,
+
d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a< b\in I\,
 
\end{equation}
 
\end{equation}
 
(cp. with ). This theorem motivates the following
 
(cp. with ). This theorem motivates the following

Revision as of 13:28, 10 August 2012


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:
Boris Tsirelson/sandbox1. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Boris_Tsirelson/sandbox1&oldid=27476