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.
Best Answer
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.