Is $\limsup_{\epsilon \to 0^+}\frac{1}{\epsilon}\sup_{y \in B(x; \epsilon)}|f(y)-f(x)|$ related to any standard notion of gradient?

Question

Let $f:\mathbb R^d \to \mathbb R$ be a function and fix scalar $\epsilon > 0$. Given a pint $x \in \mathbb R^d$, define $$ \Delta_f(x;\epsilon) := \sup_{y \in B(x; \epsilon)}|f(y)-f(x)|, $$ where $B(x;\epsilon):=\{y \in \mathbb R^d \mid \|y-x\| \le \epsilon\}$.

Finally, define $\partial^{-} f(x) := \liminf_{\epsilon \to 0^+}\frac{\Delta_f(x;\epsilon)}{\epsilon}$ and $\partial^+ f(x) :=\limsup_{\epsilon \to 0^+}\frac{\Delta_f(x;\epsilon)}{\epsilon}$

Question 1. Is there a way to link the quantities $\partial^{\pm} f(x)$ with any known notion of derivative ?

The strong slope of $f$ at $x$, denoted $|\partial^+ | f(x)$ is defined by $$ |\partial^+| f(x) := \limsup_{y \to x}\frac{(f(y)-f(x))_+}{\|y-x\|}, $$ where $(t)_+ := \max(t,0)$. This notion of gradient is common in gradient-flow literature, and can be defined for functions on arbitrary metric spaces (e.g spaces of probability measures).

Question 2. Is $|\partial^+| f(x)$ related in any way to $\partial^{\pm} f(x)$ ?

Answer 1

Q1: Assume $f$ is differentiable at $x$, then $ \partial^+f(x)=\partial^-f(x)=\|f'(x)\|$, where $f'(x)=\nabla f(x) = (\frac{\partial f}{\partial x_1}(x), \cdots, \frac{\partial f}{\partial x_d}(x))$ is the gradient (a row vector).

Proof. As $f$ is differentiable, $f(y) = f(x) + f'(x) (y-x) + o(\|y-x\|)$ around $x$. Therefore for suffciently small $\epsilon$ (all the sups are taken over $B(x;\epsilon)$), $$\begin{align}\Delta_f(x;\epsilon) &=\sup\mid f'(x)(y-x)+o(\|y-x\|)\mid \\ &\le \sup|f'(x)(y-x)| + \sup o(\|y-x\|) \\ &=\|f'(x)\|\epsilon + \sup o(\|y-x\|) \\ \frac{\Delta_f(x;\epsilon)}{\epsilon}&\le\|f'(x)\|+ \sup \frac{ o(\|y-x\|)}{\epsilon}\end{align}$$

Here $\sup|f'(x)(y-x)|=\|f'(x)\|\epsilon$ follows from Cauchy-Schwarz. Note that $\sup \frac{ o(\|y-x\|)}{\epsilon}\le\sup \frac{ o(\|y-x\|)}{\|y-x\|}\rightarrow 0$, We have shown that $\partial^+f(x)\le \|f'(x)\|$. Similarly $$\begin{align} \Delta_f(x;\epsilon) &=\sup\mid f'(x)(y-x)+o(\|y-x\|)\mid \\ &\ge\|f'(x)\|\epsilon - o(\epsilon) \end{align}$$

Hence $\partial^-f(x)\ge \|f'(x)\|$.

Q2: From Q1, it's easy to show that $|\partial^+|f(x)=\|f'(x)\|$ as well.