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In what follows, $\mu^n$ denotes $n$-dimensional Lebesgue measure, and $B_r(x)$ is the open ball with radius $r$ centered at $x$.


In elementary calculus, if we have a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ which is Fréchet differentiable at a point $p \in \mathbb{R}^n$, then $f$ is automatically Gateaux differentiable at $p$ also.

However in geometric measure theory, we have the notion of approximate limits and differentiability. The definitions are valid as long as $f$ is a measurable function, and are given as follows:

  1. $f$ is approximately Fréchet differentiable at $p$ if there exists a linear map $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$, such that for all $\varepsilon \in \mathbb{R}^{>0}$: $$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{x \in B_r(p) : \frac{|f(x) - f(p) - L(x-p)|}{|x-p|} > \varepsilon \})}{\mu^n(B_r(p))} = 0 $$
  1. $f$ is approximately Gateaux differentiable at $p$ if for all unit vectors $u \in \mathbb{R}^n$ there exists $L_u \in \mathbb{R}^m$, such that for all $\varepsilon \in \mathbb{R}^{>0}$: $$\lim_{r \rightarrow 0^+} \frac{\mu^1(\{t \in (-r,r) : |\frac{f(p+tu) - f(p)}{t} - L_u| > \varepsilon \})}{2r} = 0 $$

Notice that here approximately Gateaux differentiable means that all the approximate directional derivatives exist, not just almost all of them. In fact it is this distinction which prevents Fréchet necessarily implying Gateaux pointwise.


Now my question is, what if we have everywhere approximately differentiable and not just pointwise? Specifically, if $f$ is everywhere approximately Fréchet differentiable, is $f$ necessarily approximately Gateaux differentiable?

For what it's worth, given any fixed unit vector $u$, it's known that the approximate $u$-directional derivative exists almost everywhere (see theorem 2 here https://encyclopediaofmath.org/wiki/Approximate_differentiability). My question is equivalent to asking if this can be extended to existing everywhere, once we have everywhere approximately Fréchet differentiable?

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  • $\begingroup$ "In fact it is this distinction which prevents Fréchet necessarily implying Gateaux pointwise [...] Now my question is, what if we have everywhere approximately differentiable and not just pointwise?" I'm a bit confused...aren't "pointwise" and "everywhere" synonymous? For me solving pointwise an ODE, for instance, means solving it classically at every point (and not just weakly). $\endgroup$ Commented Sep 16, 2022 at 21:03
  • $\begingroup$ Sorry if I wasn't clear. The pointwise statement would be "$f$ is A.F.D. at $x$ implies $f$ is A.G.D. at $x$" vs the everywhere statement of "$f$ is A.F.D. everywhere on its domain implies $f$ is A.G.D. everywhere on its domain". If I'm still not being clear, take the difference between the statements "$f$ is $0$ at $x$ implies $f'$ is $0$ at $x$" vs "$f$ is everywhere $0$ implies $f'$ is everywhere $0$". $\endgroup$ Commented Sep 16, 2022 at 22:02

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I have an example which does not strictly answer the question, but a slightly weaker one.

Theorem 1. There is a function $f:\mathbb{R}^n\to\mathbb{R}$ that is approximately Fréchet differentiable almost everywhere, but $f$ is approximately Gateaux differentiable on a set of measure zero.

The construction is based on the existence of Nikodym sets:

Theorem 2. (Nikodym) There is a Borel set $N\subset\mathbb{R}^2$ such that $\mu^2(N)=0$ and for every $x\in\mathbb{R}^2$ there is a line $L$ though $x$ for which $L\setminus\{x\}\subset N$.

This is Theorem 11.7 in Mattila's "Fourier Analysis and Hausdorff Dimension".

Proof of Theorem 1. Let $A=\mathbb{R}^2\setminus N$ and let $f=\chi_A$ be the characteristic function of $A$. Since the complement of $A$ has measure zero, $f$ is approximately Fréchet differentiable at every point of $A$ with $Df=0$. On the other hand if $x\in A$ and $L$ is the linie form Theorem 1, then the function $f$ restricted to that line equals $1$ at $x$ and zero everywhere else so it is not approximately differentiable in the direction of $L$ and hence $f$ is not approximately Gateaux differentiable at any point $x\in A$. $\Box$

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  • $\begingroup$ Thank you for this reply. Unfortunately it is not quite what I am looking for as you are aware, but I still appreciate this input. $\endgroup$ Commented Oct 1, 2022 at 18:44

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