Let $f: \mathbb R^n \to \mathbb R$ be a measurable function. Let $\mathcal L$ be the set of linear functions $\mathbb R \to \mathbb R$.
Define the roughness $\mathcal Rf(x)$ of $f$ at $x \in \mathbb R^n$ by
$$\inf_{L \in \mathcal L} \limsup_{y \to x} \left | \frac{f(y) – f(x) – L(y-x)}{|y – x|} \right |.$$
We say that $f$ is pseudo differentiable if $\mathcal Rf(x) < \infty$ for all $x \in \mathbb R^n$.
In other words, $f$ is pseudo differentiable if the difference quotients approximate the function to within $O(|y – x|)$ everywhere.
Question: Is it true that if $f$ is pseudo differentiable, then $f$ is in the Sobolev space $W^{1, 1}_\text{loc}?$
Remarks:
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Note that $f$ is differentiable at $x$ if and only if $\mathcal Rf(x)$ is $0$.
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In one dimension, the condition that $f$ is pseudo differentiable is equivalent to the upper and lower Dini derivatives being finite everywhere. In this case I believe pseudo differentiable implies (locally) absolutely continuous, and hence $W^{1,1}_\text{loc}$.
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In the definition of pseudo differentiable, the condition $Rf(x) < \infty$ holds for all $x$! (Instead of merely almost all $x$)
Best Answer
The function $$ f(x) = \begin{cases} x\sin 1/x^2 & x\not= 0 \\ 0 & x=0 \end{cases} $$ gives a counterexample. We have $f\in C^{\infty}(U)$ when we restrict to $U=\mathbb R\setminus \{ 0\}$, so if $f$ had a distributional derivative in $L^1_{\textrm{loc}}$, it would have to be its classical derivative $f'=-(2/x^2)\cos (1/x^2)+\sin (1/x^2)$, but this fails to be integrable near $x=0$.