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For $f:[0,1]\to\mathbb{R}$, let $f'_{app}(x)$ denote the approximate derivative (that is, the derivative calculated along some set with density $1$ at $x$, if such a thing exists). Assume that $f'_{app}$ exists and is continuous on $[0,1]$. Does this imply that $f'$ exists on $[0,1]$ (of course immediately $f'$ would also be continuous, since $f'_{app}=f'$ when $f'$ exists).

Put succinctly:

Does $C^1_{app}[0,1]=C^1[0,1]$?

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  • $\begingroup$ I am not sure I understand what you call an approximate derivative. Doesn't the indicator function of $\mathbb{Q}$ have approximate derivative $0$? $\endgroup$ – abx Oct 20 '17 at 14:22
  • $\begingroup$ @abx No, at any rational point, the approximate derivative of the indicator function of $\mathbb{Q}$ does not exist. The difference quotient from a rational point $x_1$ to an irrational point $x_2$ approaches $\pm\infty$ as $x_2\to x_1$ from the right or left. $\endgroup$ – Trevor Richards Oct 20 '17 at 14:47
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Yes. Let $g$ be an antiderivative of $f'_{app}$, then $(f-g)'_{app}=0$ , this implies that $h:=f-g=\text{const}$.

Sketch of the proof of this fact. It suffices to prove that $h(x)=h(0)$ a.e. Indeed, if $h(a)\ne 0$, than this would imply $h'_{app}(a)=\infty$. We may fix $c>0$ and $\rho\in (0,1)$ and prove that the measure of $x\in [0,1]: h(x)-h(0)<cx$ is at least $\rho$. This, again, suffices. We say that a point $t\in [0,1]$ is nice if $\mu\{x\in [0,t]:h(x)-h(0)\geqslant cx\}\leqslant (1-\rho)t$, and $t$ is very nice if $t$ is a density point for the set of nice points. Let $a$ be a supremum of very nice points in $[0,1]$. If $a<1$, we get $h(a)-h(0)\leqslant ca$, else $h'_{app}$ would be infinite. Next, there exist a segment $[a,a+\varepsilon]$ such that $h(x)-h(a)<c(x-a)$ for more than $\rho\cdot \varepsilon$ points in this segment. This allows to find a very nice point greater than $a$. A contradiction.

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    $\begingroup$ You are using a (very plausible) result that if $h'_{app}=0$ on $[0,1]$, then $h$ is constant? I see a (slightly complicated) way to prove this using the Lebesgue density theorem, but is there a more or less trivial way to see this? Thanks $\endgroup$ – Trevor Richards Oct 20 '17 at 14:51
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    $\begingroup$ The way I have in mind also uses Lebesgue density theorem. $\endgroup$ – Fedor Petrov Oct 20 '17 at 15:05

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