Limiting subdifferential

Remarks

Taking $x_k = \bar x$ and $v_k = v$ shows that $\hat\partial f(\bar x) \subseteq \partial f(\bar x)$. If $f$ is differentiable near $\bar x$ with continuous gradient, then $\partial f(\bar x) = \set{\nabla f(\bar x)}$. If $f$ is convex, both sets equal the subdifferential of convex analysis. If $\bar x$ is a local minimum of $f$, then $0 \in \hat\partial f(\bar x) \subseteq \partial f(\bar x)$. Other names are the Mordukhovich subdifferential and the basic subdifferential.

Example: $f(x) = -\abs{x}$ on $\R$ and $\bar x = 0$. For $x > 0$ the function is differentiable with derivative $-1$, so $\hat\partial f(x) = \set{-1}$; for $x < 0$, $\hat\partial f(x) = \set{1}$. Hence $\set{-1, 1} \subseteq \partial f(0)$, and in fact $\partial f(0) = \set{-1, 1}$.

Non-example: the vector $v = 0$ for the same function. It is not a regular subgradient at $0$, since $(f(y) - f(0) - 0)/\abs{y} = -1$ for all $y \ne 0$. It is not a limit of regular subgradients at nearby points either, since these are all $\pm 1$. So $0 \notin \partial f(0)$, although $0$ lies in the Clarke generalized gradient $[-1, 1]$.