A past question from a qualifying exam at my university reads:
Let $f$ be a continuous real-valued function on the real line that is differentiable almost everywhere with respect to the Lebesgue measure and satisfies $f(0)=0$ and
$$ f'(x)=2f(x)$$
almost everywhere. Prove that there exists infinitely many such functions, but that only one of them is absolutely continuous.
I have tried modifying the function $e^{2x}$, but I cannot satisfy all the conditions given.
Once one shows that there are infinitely many such functions, then if we pick 2 such functions $f_1$ and $f_2$ and fix $a>0$, we can apply the fundamental theorem of Calculus for Lebesgue Integrals on $[0,a]$ and see that if both are absolutely continuous, then
$$ f_1(x)=\int_0^x 2f(t)dt=f_2(x) $$
So this would imply that the are the same function on $[0,\infty)$. I'm not sure how to proceed with the whole real line.
Best Answer
It well known that there is a non-constant continuous function $g$ with $g'=0$ a.e. [ You can search for 'Cantor Function']. We may suppose $g(0)=0$ (by considering $g(x)-g(0)$). The function $g^{n}(x)e^{2x}$ satisfies the stated property for any positive integer $n$ so there are infinitely many solutions.
Now let $f$ and $g$ be two solutions which are absolutely continuous. I will let you verify that $h(x)=e^{-2x} (f(x)-g(x))$ is absolutely continuous on any finite interval and its derivative is $0$. Now $h(b)-h(a)=\int_a^{b} h'(t) dt=0$ whenever $a <b$. Since $h(0)=0$ we get $h(x)=0$ for al $x$ and hence $f(x)=g(x)$ for all $x$.
PS: Use the inequality $ |(FG)(b_i)-(FG)(a_i)| \leq |F(b_i)-F(a_i)||G(b_i)|+|F(a_i)||G(b_i)-G(a_i)|$ to show that the product of two bounded absolutely continuous functions is absolutely continuous.