12
$\begingroup$

For measurable functions $f(x)$, $g(x)$ on $[0,1]$ define the distance $\rho(f,g)$ as a Lebesgue measure of the set $\{x:f(x)\ne g(x)\}$. Then Luzin's famous theorem states that $C[0,1]$ is dense with respect to this metric in the set of all measurable functions.

The question is to describe the completion of $C^1[0,1]$ (that is, informally, for given function give a receipt how to recognize wether it belongs to completion or not.)

$\endgroup$
  • 3
    $\begingroup$ Related: mathoverflow.net/q/34518/5701 and the answer which shows that whatever the completion is, it is not all of $C[0,1]$. $\endgroup$ – Vaughn Climenhaga May 6 '14 at 14:58
  • 1
    $\begingroup$ Any such function must have an "$L^0$ derivative", in the sense that $\frac{1}{t}(f(\cdot + t) - f(\cdot))$ has a limit in measure as $t \to 0$. $\endgroup$ – Alexander Shamov May 6 '14 at 15:03
9
$\begingroup$

This holds precisely if $f$ is approximately differentiable a.e. This condition implies that $f|_A$ can be extended to a $C^1$ function for suitable sets $A$ of almost full measure by a Theorem from Federer's book, or see here.

The converse is obvious: If $f=g\in C^1$ on $A$, then $f$ is approximately differentiable at all points of density of $A$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.