# Pseudo differentiable functions

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:

1. Note that $$f$$ is differentiable at $$x$$ if and only if $$\mathcal Rf(x)$$ is $$0$$.

2. 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 $$W^{1,1}_\text{loc}$$.

3. In the definition of pseudo differentiable, the condition $$Rf(x) < \infty$$ holds for all $$x$$! (Instead of merely almost all $$x$$)

• A slightly easier way to state this condition is to say that $f$ is pseudo-differentiable at $x$ if $\limsup |f(y)-f(x)|/|y-x|<\infty$ (or maybe it would be more descriptive to call this locally Lipschitz continuous at $x$). Jun 18, 2022 at 15:26

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 $$x=0$$.
• Or $f=x^2\sin (1/x^3)$, which is even differentiable everywhere (but not $C^1$, obviously, or it would be in $W^{1,1}$). Jun 18, 2022 at 15:51
• Ah so a function can be differentiable everywhere yet it’s derivative fails to be in $L^1$. Thanks for the example! Jun 18, 2022 at 15:53