Clarke generalized gradient

Remarks

If $K$ is a Lipschitz constant of $f$ near $x$, then $\abs{f^\circ(x; v)} \le K \norm{v}$, and $v \mapsto f^\circ(x; v)$ is sublinear. So $\partial_C f(x)$ is convex, nonempty by the Hahn–Banach theorem, and weak-* compact by the Banach–Alaoglu theorem, since it is weak-* closed and bounded by $K$. If $f$ is continuously differentiable near $x$, then $\partial_C f(x) = \set{f'(x)}$. If $f$ is convex, then $\partial_C f(x)$ is the subdifferential of convex analysis. For $X = \R^n$, Clarke's formula describes it through gradients, which exist almost everywhere by Rademacher's theorem. It is the convex hull of the limits $\lim_k \nabla f(x_k)$ over sequences $x_k \to x$ at which $f$ is differentiable. This formula still holds when the $x_k$ must avoid a given set of measure zero.

Example: $f(x) = -\abs{x}$ on $\R$ at $x = 0$. For $y > 0$ small the difference quotient in direction $v$ is $-v$, and for $y < 0$ small it is $v$, so $f^\circ(0; v) = \abs{v}$. Hence $\partial_C f(0) = \set{\zeta : \zeta v \le \abs{v} \text{ for all } v} = [-1, 1]$.

Non-example: $f(x) = \sqrt{\abs{x}}$ on $\R$ is not Lipschitz near $0$. The quotient $(f(t) - f(0))/t = t^{-1/2}$ is unbounded as $t \downarrow 0$, so $f^\circ(0; 1) = +\infty$, and the construction gives no useful information.