Generalized Directional Derivative of $-|x| $ (minus absolute $x$) (Clarke Derivative)

Question

I am having hard time calculating the Generalized Directional Derivative (Clarke Derivative) of $f(x)=-|x|$ at $x=0$.
The answer is $f^{\circ}(0;v)=|v|$.

The Generalized Directional Derivative is defined as

$f^{\circ}(x;v)=\displaystyle{\lim_{x \to \infty}} \displaystyle{\sup_{x \downarrow 0}} \frac{f(y+tv)-f(y)}{t}$

On this website, https://mathematix.wordpress.com/2021/10/13/generalized-directional-derivatives-some-examples/, the calculation of $f^{\circ}(0;v)$ when $f(x)=|x|$ and $f(x)=-|x|$ are shown and they have the same answer which is $|v|$.

I got stuck with the calculation of $f^{\circ}(0;v)=|v|$ when $f(x)=-|x|$. How come $f^{\circ}(0;v)=|v|$ for $f(x)=-|x|$ and $f(x)=|x|$ are the same? Can someone show me how to calculate $f^{\circ}(0;v)$ for $f(x)=-|x|$ in detail?

Answer 1

I am the author of the post you referenced. Have a look at Proposition 3, which states that $$ (-f)^{\circ}(x; h) = f^\circ(x; -h). $$ If we apply this to $f(x) = -|x|$ at $x=0$ we have $$ (-|{}\cdot{}|)^{\circ}(0; h) = |{}\cdot{}|^{\circ}(0; -h) = |-h| = |h|. $$

Answer 2

You always have $h \mapsto f^{\circ}(x; h)$ is positive homogenous you only need to compute for $h=\pm 1$. Since $| \cdot|$ is Lipschitz with rank one you always have $|f^{\circ}(x;\pm 1)| \le 1$.

$f^{\circ}(0; +1) = \limsup_{t \downarrow 0, y \to 0} {-|y+t| +|y| \over t} \ge \limsup_{t \downarrow 0, y<-t} {t \over t} = 1$ and similarly $f^{\circ}(0; -1) = \limsup_{t \downarrow 0, y \to 0} {-|y-t| +|y| \over t} \ge \limsup_{t \downarrow 0, y>t} {t \over t} = 1$.

Hence $f^{\circ}(x; h) = |h|$.

A similar argument shows that for the Euclidean norm we have $(-\|\cdot\|)^{\circ}(0; h) = \|h\|$.