On Absolutely Continuous Functions

Question

I would like to know if we can extend the concept of absolute continuity to functions $f:[a,b]\to X$, where $X$ is a topological vector space. I browsed some books on Topological Vector Spaces but can't find the definition of absolutely continuous functions defined on $[a,b]$ and take values on $X$. I would be greatful if someone can provide me a definition (if possible) of absolute continuous function $f:[a,b]\to X$. Thanks in advance...

Answer 1

For general Topological Vector Spaces: no.

There are two definitions of absolute continuity: one is for functions, the other for measures.

Definition (AC for functions): Let $I$ be an interval in $\mathbb{R}$. A function $f:\mathbb{R}\to\mathbb{R}$ is said to be absolutely continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that if $(a_k,b_k) = I_k \subset I$, $k \in \{1\ldots n\}$ are pairwise disjoint subintervals of $I$ satisfying $$ \sum_1^n |I_k| = \sum_1^n |b_k - a_k| < \delta $$ then $$ \sum_1^n |f(b_k) - f(a_k)| < \epsilon.$$

This definition only requires that we can evaluate $$ \sum_1^n |f(b_k) - f(a_k)| $$ and naturally we can extend this to the case of the codomain being any metric space $(X,d)$ by replacing $$ \sum_1^n |f(b_k)- f(a_k)| \implies \sum_1^n d(f(b_k),f(a_k)) $$

A general TVS may not even be metrizable, and I don't see how you can have a reasonable generalisation of the notion of absolute continuity if you don't have a reasonable replacement for $\sum_1^n|f(b_k) - f(a_k)|$.

See also the EOM entry.

Definition (AC for measures): Let $(\Omega,\mathcal{S},\mu)$ be a measure space (in the interval case we usually take $\Omega = [a,b]$, $\mathcal{S},\mu$ corresponding to the Lebesgue measure). A measure $\lambda$ is said to be absolutely continuous with respect to $\mu$ if every $\mu$-null set is $\lambda$-null.

The usual generalisation of this concept is to the case where $\lambda$ is a vector measure (a measure "taking values" in a Banach space). See, e.g., Chapter 29 of Schecter's HAF.