5
$\begingroup$

Question:

Consider the set of continuous, differentiable a.e. functions from $\mathbb R \to \mathbb R$. Can we characterise the subset of these that satisfy $f’(x) = f(x)$ for almost every $x \in \mathbb R$?

Remarks:

1) The problem can be thought of as a weakening of the defining ODE for the exponential function - $f’(x) = f(x)$ everywhere, which is solvable only by piecewise combinations of $Ce^x$ and the zero function.

2) If $f$ is a solution, then so is $f + g$ for any $g$ continuous and differentiable a.e. with $g’ = 0$ a.e.

3) The full measure subset on which $f’ = f$ can in general be smaller than the full measure set on which $f’$ is defined.

Improve this question
$\endgroup$
16
  • 10
    $\begingroup$ This holds if and only if $g(x) = e^{-x} f(x)$ satisfies $g' = 0$ almost everywhere. Or, in other word, $f(x) = e^x g(x)$ with $g' = 0$ almost everywhere. Is this an answer you were looking for? $\endgroup$
    – Mateusz Kwaśnicki
    Oct 9, 2021 at 7:20
  • 2
    $\begingroup$ For slight clarification: when you say "continuous, differentiable a.e.", does that mean that it is differentiable a.e. and continuous, or that a.e. applies to both continuity and differentiability? From the "piecewise combinations" thing, it sounds like you mean a.e. to apply to both. $\endgroup$
    – user44191
    Oct 9, 2021 at 7:54
  • 1
    $\begingroup$ So set of solutions is dense in $C^0$ (I erroneously believed a bootstrap argument started from the fact that f' was equal a.e. to a continuous function...) $\endgroup$
    – Pietro Majer
    Oct 9, 2021 at 7:56
  • 1
    $\begingroup$ @user44191 Ah I mean the former, sorry for the confusion. $\endgroup$
    – Nate River
    Oct 9, 2021 at 8:05
  • 1
    $\begingroup$ @NateRiver: I may be wrong, but, for a given $x$, $f'(x)$ exists iff $g'(x)$ exists, and $f'(x)=f(x)$ if and only if $g'(x)=0$, so the answer really seems to be that simple. Does this make sense? (It's Saturday morning here, so I may need another cup of coffee to get things right.) $\endgroup$
    – Mateusz Kwaśnicki
    Oct 9, 2021 at 8:42

1 Answer 1

Reset to default
11
$\begingroup$

Let $g(x) = e^{-x} f(x)$, so that $f(x) = e^x g(x)$. For a given $x$, $f'(x)$ exists if and only if $g'(x)$ exists, and $g'(x) = e^{-x} (f'(x) - f(x))$. In particular, $f'(x) = f(x)$ if and only if $g'(x) = 0$.

It follows that $f$ is necessarily of the form $f(x) = e^x g(x)$, where $g$ satisfies $g'(x) = 0$ almost everywhere.

Improve this answer
$\endgroup$
2
  • $\begingroup$ $g'(x)=0$ almost everywhere (a.e.) if and only if $g(x)=C$ a.e. for some constant $C$. Therefore, the solutions of $y'=y$ a.e. are given by $y=Ce^x$ a.e. for some constant $C$. $\endgroup$
    – Black Mild
    Oct 15, 2021 at 16:28
  • $\begingroup$ @BlackMild: That is true if, for example, $g$ is absolutely continuous (by Lebesgue integration), or if $g$ is differentiable everywhere (by Henstock–Kurzweil integration). However, if we only require that $g'$ exists a.e. and it is zero a.e., $g$ may fail to be constant. An example is provided by the Cantor's (or Devil's) staircase function. $\endgroup$
    – Mateusz Kwaśnicki
    Oct 15, 2021 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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