Namespaces
Variants
Actions

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

From Encyclopedia of Mathematics
Jump to: navigation, search
Line 7: Line 7:
 
</asy>
 
</asy>
  
−
Every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a sketch of a proof. Given $\varepsilon$, we take a compactly supported continuous $f:X\to[0,1]$ such that $\int f \rd\mu > 1-\varepsilon$. Now, consider another probability measure $\nu$ widely close to $\mu$, namely, satisfying $|\int f \rd(\mu-\nu)| < \varepsilon$ and therefore $\int f \rd\nu > 1-2\varepsilon$.
+
--------------------------------------
  
−
Claim: If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ for a given $g:X\to[-1,1]$ then $|\int g \rd(\nu-\mu)|<4\varepsilon$.
+
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.
  
−
Proof of the claim: $ |\int g \rd(\nu-\mu) - \int fg \rd(\nu-\mu)| = |\int (1-f)g \rd(\nu-\mu)| \le |\int (1-f)g \rd\nu| + |\int (1-f)g \rd\mu| \le \int (1-f) \rd\nu + \int (1-f) \rd\mu < 2\varepsilon + \varepsilon $.
+
'''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 12:51, 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=27474
Retrieved from "https://encyclopediaofmath.org/index.php?title=User:Boris_Tsirelson/sandbox1&oldid=27474"