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I'm reading about subdifferentiable function at page 232 of Villani's Optimal Transport: Old and New.

Definition 10.5 (Subdifferentiability, superdifferentiability). Let $U$ be an open set of $\mathbb{R}^n$, and $f: U \rightarrow \mathbb{R}$ a function. Then:

  • (i) $f$ is said to be subdifferentiable at $x$, with subgradient $p$, if $$ f(z) \geq f(x)+\langle p, z-x\rangle+o(|z-x|) . $$ The convex set of all subgradients $p$ at $x$ will be denoted by $\nabla^{-} f(x)$.

  • (ii) $f$ is said to be uniformly subdifferentiable in $U$ if there is a continuous function $\omega: \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$, such that $\omega(r)=o(r)$ as $r \rightarrow 0$, and $$ \forall x \in U \quad \exists p \in \mathbb{R}^n ; \quad f(z) \geq f(x)+\langle p, z-x\rangle-\omega(|z-x|). $$

  • Corresponding notions of superdifferentiability and supergradients are obtained in an obvious way by just reversing the signs of the inequalities.

My understanding I think $o(|z-x|)$ is the Landau symbol, i.e., $g (x) = o(|z-x|)$ means that $\lim_{z \to x} \frac{g(x)}{|z-x|} = 0$.

Could you explain what $f(z) \geq f(x)+\langle p, z-x\rangle+o(|z-x|)$ means?

I'm not sure if it means that the limit $$ \lim_{z\to x} \frac{f(z) - f(x)-\langle p, z-x\rangle}{|z-x|} $$ exists and is non-negative.

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    $\begingroup$ My understanding is that $o(|z-x|)$ in a formula means “some function which is $o(|z-x|)$ to be inserted here”, existentially quantified in a way that is left for the reader to figure out. So, specifically, I read (i) in the definition 10.5 you quote as the same as (ii) but with the $∀x$ being brought outside of the $∃ω$. Which would match the terminology of (ii) being called “uniformly” (i). I don't think this can be expressed with a limit. $\endgroup$
    – Gro-Tsen
    Commented Dec 20, 2022 at 10:47
  • $\begingroup$ (To make my last point clear, consider something like $f(x) := |x|\,(1+\sin(1/x))$ (extended by $f(0)=0$). I'd say that $f(x) ≥ o(x)$ since in fact $f(x) ≥ 0$, but $f(x)/|x|$ doesn't exist.) $\endgroup$
    – Gro-Tsen
    Commented Dec 20, 2022 at 10:53
  • $\begingroup$ @Gro-Tsen Could you confirm if my following understanding is correct? $f$ is said to be subdifferentiable at $x$, with subgradient $p$, if there is a continuous function $\omega: \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ such that $\omega(r)=o(r)$ as $r \rightarrow 0$, and $$ \forall z \in U; \quad f(z) \geq f(x)+\langle p, z-x\rangle-\omega(|z-x|). $$ Of course, $\omega$ depends on both $x$ and $p$. $\endgroup$
    – Akira
    Commented Dec 20, 2022 at 10:55
  • $\begingroup$ Actually, drop the “continuous”: I shouldn't have said that. Maybe it's better to write it in an $ε$-$δ$ way: for all $ε>0$ there is $δ>0$ such that for all $z∈U$ with $|z-x|<δ$ we have $f(z) ≥ f(x) + \langle p,z-x\rangle - ε·|z-x|$. This should be the same as what you wrote without the “continuous” (I hope!). $\endgroup$
    – Gro-Tsen
    Commented Dec 20, 2022 at 11:01
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    $\begingroup$ Well, that's just my understanding. I'm not fully confident that someone won't come and say “oh, well, in optimal transport, it's well known that things should be interpreted in the following slightly different way”, so I suggest we leave the question open so far. $\endgroup$
    – Gro-Tsen
    Commented Dec 20, 2022 at 11:03

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Landau $o(\cdot)$ notations should be interpreted in inequalities as inferior/superior limits. In this case in particular, it $f(z)\ge f(x)+⟨p,z−x⟩+o(|z−x|)$ is equivalent to $$ \liminf_{z \to x} \frac{f (z) - f (x) - \langle p, x - z\rangle}{\vert z - x\vert} \ge 0; $$ the corresponding limit does not need to exist.

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  • $\begingroup$ I'm confused by "corresponding limit does not need to exist" because $\liminf_{z \to x}$ always exists in the extended real numbers. Could you elaborate on this point? $\endgroup$
    – Akira
    Commented Dec 20, 2022 at 12:09
  • $\begingroup$ I was meaning “the corresponding limit with $\lim_{z \to x}$ replacing $\liminf_{z \to x}$” and have edited accordingly. $\endgroup$
    – Jean Van Schaftingen
    Commented Dec 20, 2022 at 12:39
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    $\begingroup$ Ah by "corresponding limit", you meant the limit $\lim_{z\to x} \frac{f(z) - f(x)-\langle p, z-x\rangle}{|z-x|}$ I wrote, right? $\endgroup$
    – Akira
    Commented Dec 20, 2022 at 12:41
  • $\begingroup$ I have just found that your interpretation is exactly the same as in the paper Some properties of semiconcave functions with general modulus by Paolo Albano. $\endgroup$
    – Akira
    Commented Dec 21, 2022 at 1:18

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