In the previous post on the Rayleigh quotient we defined the directional derivative of a function \(f: \mathbb{R}^n \to \overline{\mathbb{R}}\) as

$$\begin{aligned}f'(x; h) = \lim_{t \downarrow 0} \frac{f(x+th)-f(x)}{t},\end{aligned}$$

provided that the limit exists.

It turns out that all convex functions are directionally differentiable on the interior (actually, the core) of their domains and \(f'(x; \cdot)\) is sublinear. However, the sublinearity property may fail when working with nonconvex functions. This motivates the definition of generalised directional derivatives which will hopefully be accompanied by some good calculus rules.

## 1. Generalised Directional Derivative

### 1.1. Dini Directional Derivative

The Dini directional derivative of a function \(f: \mathbb{R}^n \to \overline{\mathbb{R}}\) at a point \(x\) along a direction \(h\) is a generalised derivative given by

$$\begin{aligned}f^-(x; h) = \liminf_{t \downarrow 0} \frac{f(x+th)-f(x)}{t}.\end{aligned}$$

However, as we will see later, the Dini directional derivative is not always sublinear.

### 1.2. Clarke Directional Derivative

We also define the Clarke directional derivative of \(f\) at \(x\) for \(h\) is given by

$$\begin{aligned}f^\circ(x; h) {}={}& \limsup_{y \to x, t \downarrow 0} \frac{f(y+th)-f(y)}{t} \\ {}={}& \inf_{\delta>0}\, \sup_{\substack{\|y-x\|\leq \delta \\ t \in (0,\delta)}} \frac{f(y+th)-f(y)}{t}.\end{aligned}$$

We can show that

**Proposition 1.** The Clarke directional derivative is positive homogeneous and *sublinear*, i.e.,

$$\begin{aligned}f^\circ(x; u + v) \leq f^\circ(x; u) + f^\circ(x; v),\end{aligned}$$

for all \(x, u, v\).

*Proof.* To show that \(f^\circ(x; {}\cdot{})\) is positively homogeneous, take \(\lambda > 0\). We have

$$\begin{aligned}f^\circ(x; \lambda h) {}={}& \limsup_{y\to x, t\downarrow 0}\frac{f(y+\lambda t h) - f(y)}{t}\\{}\overset{\eta = \lambda t}{=}{}& \limsup_{y\to x, \eta\downarrow 0}\frac{f(y+\eta h) - f(y)}{\tfrac{1}{\lambda}\eta} = \lambda f^\circ(x; h).\end{aligned}$$

We need to show that \(f^\circ(x; h+h') \leq f^\circ(x; h) + f^\circ(x; h')\). We have

$$\begin{aligned}f^\circ(x; h+h') {}={}& \limsup_{y\to x, t\downarrow 0}\frac{f(y+th+th') - f(y)}{t}\\{}={}& \limsup_{y\to x, t\downarrow 0}\frac{f(y+th+th') - f(y+th') + f(y+th') - f(y)}{t} \\ {}\leq{}& \limsup_{y\to x, t\downarrow 0}\frac{f(y+th+th') - f(y+th') }{t} \\ &\qquad+ \limsup_{y\to x, t\downarrow 0}\frac{f(y+th') - f(y)}{t}\end{aligned}$$

Now in the first limsup let us define \(\tilde{y} = y+th\) and see that \(\tilde{y} \to x\) as \(y \to x\) and \(t \downarrow 0\), therefore the first limsup is equal to \(f^\circ(x; h)\). \(\blacksquare\)

Observe that the Clarke directional derivative majorises the Dini derivative \(f^-(x; h) \leq f^\circ(x; h)\). This is easy to prove - if not, please write a comment here below and I will elaborate.

Moreover, in \(f\) is Lipschitz, we can show the following

**Proposition 2.** Let \(f\) be locally Lipschitz with modulus \(K\) around \(x\). Then,

$$\begin{aligned}f^\circ(x; u) \leq K\|u\|,\end{aligned}$$

for all \(u\).

*Proof.* We have

$$\begin{aligned}f^\circ(x; h) {}={}& \inf_{\delta>0}\, \sup_{\substack{\|y-x\|\leq \delta \\ t \in (0,\delta)}} \frac{f(y+th)-f(y)}{t} \\ {}\leq{}& \inf_{\delta>0}\, \sup_{\substack{\|y-x\|\leq \delta \\ t \in (0,\delta)}} \frac{|f(y+th)-f(y)|}{t} \\ {}\leq{}& \inf_{\delta>0}\, \sup_{\substack{\|y-x\|\leq \delta \\ t \in (0,\delta)}} \frac{K\|y+th -y\|}{t} = K\|h\|, \end{aligned}$$

which completes the proof. \(\blacksquare\)

### 1.3. Michel-Penot Directional Derivative

The Michel-Penot directional derivative is defined as

$$\begin{aligned}f^\#(x; h) {}={}& \sup_{u\in\mathbb{R}^n} \limsup_{t \downarrow 0} \frac{f(x+t(h+u))-f(x+th)}{t}.\end{aligned}$$

**Proposition 3.** The Michel-Penot directional derivative is sublinear, i.e.,

$$\begin{aligned}f^\#(x; h_1 + h_2) \leq f^\#(x; h_1) + f^\#(x; h_2),\end{aligned}$$

for all \(x, u, v\).

*Proof.* Let us define the function

$$\begin{aligned}\hat{f}(x, h, u){}={}\limsup_{t \downarrow 0} \frac{f(x+th+tu))-f(x+tu)}{t}.\end{aligned}$$

Note that \(f^\#(x; h) = \sup_u \hat{f}(x, h, u)\). We shall prove that \(\hat{f}\) is sublinear in its second argument; for fixed \(x\) and \(u\) we have

$$\begin{aligned}\hat{f}(x, h_1+h_2, u) {}={}& \limsup_{t \downarrow 0} \frac{1}{t}\left[f(x+t(h_1+h_2+u))-f(x+tu)\right] \\ {}={}& \limsup_{t \downarrow 0} \frac{1}{t}\big[ f(x+t(h_1+h_2+u)) -f(x+th_2+tu) \\ &\qquad\quad+f(x+th_2+tu)-f(x+tu)\big] \\ {}\leq{}& \limsup_{t \downarrow 0} \frac{1}{t}\left[ f(x+t(h_1+h_2+u)) -f(x+th_2+tu)\right] \\ &\qquad\quad+\limsup_{t \downarrow 0} \frac{1}{t}\left[ f(x+th_2+tu)-f(x+tu)\right] \\ {}\leq{}& \sup_u\limsup_{t \downarrow 0} \frac{1}{t}\left[ f(x+th_1+t(h_2+u))) -f(x+t(h_2+u))\right] \\ &\quad\qquad{}+{} f^{\#}(x, h_2) \\ {}={}& \sup_{v}\limsup_{t \downarrow 0} \frac{1}{t}\left[ f(x+th_1+t v) - f(x+tv)\right] + f^{\#}(x, h_2) \\ {}={}& f^{\#}(x, h_1) + f^{\#}(x, h_2). \end{aligned}$$

Now taking a supremum with respect to \(u\), the assertion follows. \(\blacksquare\)

A notable property is stated below.

**Proposition 4.** We have

$$\begin{aligned}f^{-}(x; h) \leq f^\#(x; h) \leq f^{\circ}(x; h),\end{aligned}$$

for all \(x, h\).

*Proof.* This is Proposition 6.1.1 in [1] and the proof can be found there. \(\blacksquare\)

Observe that

$$\begin{aligned}f^-(x; h) \leq f^\#(x; h) \leq f^\circ(x; h).\end{aligned}$$

To prove the second inequality, \(f^\#(x; h) \leq f^\circ(x; h)\), fix \(u\) and observe that

$$\begin{aligned}f^\circ(x; h) = \limsup_{y\to x, t\downarrow 0} \frac{f(y+th) - f(y)}{t} \geq \limsup_{\substack{y=x+tu,\\ t\downarrow 0}} \frac{f(x+tu+th) - f(x+tu)}{t},\end{aligned}$$

and take the supremum with respect to \(u\) on both sides.

## 2. Examples

### 2.1. Absolute value

Let us first see how the above three derivatives behave on a nonsmooth convex function. Consider the function \(f(x) = |x|\), with \(x \in \mathbb{R}\). The Dini derivative of \(f\) at \(x=0\) for \(h\) is

$$\begin{aligned}f^-(x; h) = \liminf_{t\downarrow 0} \frac{|th|}{t} = |h|.\end{aligned}$$

For the Clarke derivative we have that

$$\begin{aligned}f^\circ(0; h) = \limsup_{y\to 0, t\downarrow 0} \frac{|y+th|-|y|}{t} \leq \limsup_{y\to 0, t\downarrow 0} \frac{|y| + |th| - |y|}{t} = |h|,\end{aligned}$$

using the triangle inequality. By choosing the sequences \(y_\nu = \tfrac{1}{\nu^2}\) and \(t_\nu = \tfrac{1}{\nu}\) we have that

$$\begin{aligned}f^\circ(0; h) {}={}& \limsup_{y\to 0, t\downarrow 0} \frac{|y+th|-|y|}{t} \\ {}\geq{}& \lim_{\nu} \frac{|y_\nu + t_\nu h|-|y_\nu|}{t_\nu} = |h|,\end{aligned}$$

therefore, \(f^\circ(0; h) = |h|\). Since the Michel-Penot directional derivative is between the Dini and Clarke derivatives, we conclude that \(f^\#(0; h) = |h|\) (however, it can be easily determined using the definition).

### 2.2. Negative Absolute value

Consider the function \(f(x) = -|x|\) with \(x \in \mathbb{R}\). This is a nonsmooth concave function. The Dini derivative of \(f\) at \(x=0\) along the direction \(h\) is

$$\begin{aligned}f^-(0; h) {}={}& \liminf_{t\downarrow 0} \frac{-|th|}{t} = -|h|.\end{aligned}$$

The Michel-Penot derivative is

$$\begin{aligned}f^\#(0; h) {}={}& \sup_{u\in\mathbb{R}}\limsup_{t\downarrow 0} \frac{-|th + tu|+|tu|}{t} = \sup_u |u| - |h+u| = -|h|.\end{aligned}$$

The last equality is because (i) \(\sup_u |u| - |h+u| \geq |0| - |h+0| = -|h|\) and (ii) by the triangle inequality \(\sup_u |u| - |h+u| \leq -|h|\).

Lastly, the Clarke derivative of \(f(x) = -|x|\) at 0 is

$$\begin{aligned}f^\circ(0; h) {}={} \limsup_{y \to 0, t\downarrow 0} \frac{|y|-|y+th|}{t} {}\geq{} |h|,\end{aligned}$$

by using the triangle inequality. By following a similar procedure as in \(|x|\) in Example 1, we conclude that \(f^\circ(0; h) = |h|\).We can arrive at the same conclusion using Proposition 5 below.

We can state the following result that gives the Clarke derivative of \(-f\) in terms of that of \(f\).

**Proposition 5.** It is \((-f)^\circ(x; h) = f^\circ(x; -h)\).

*Proof.* We have

$$\begin{aligned}f^\circ(x; -h) {}={} \limsup_{y \to x, t\downarrow 0} \frac{f(y-th)-f(y)}{t},\end{aligned}$$

Define \(z = y + th\) and observe that as \(t \downarrow 0\) and \(y \to x\), \(z \to x\). Additionally, \(y = z + th\). Therefore we have

$$\begin{aligned}f^\circ(x; -h) {}={} \limsup_{y \to x, t\downarrow 0} \frac{f(y-th)-f(y)}{t} = \limsup_{z\to x, t \downarrow 0} \frac{f(z) - f(z+th)}{t} = -f^\circ(x; h),\end{aligned}$$

which completes the proof. \(\blacksquare\)

Note that the property of Proposition 5 does not hold for the Dini derivative, but the Michel-Penot derivative satisfies \(f^\#(x; -h) = (-f)^\#(x; h)\).

It is also easy to prove that

**Proposition 6.** It is \((\lambda f)^\circ(x; h) = \lambda f^\circ(x; h)\), for \(\lambda \geq 0\).

Proposition 6 holds for the Michel-Penot derivative too.

### 2.3. An interesting case

Consider the following function

$$\begin{aligned}f(x) = \begin{cases}x^2 \sin\tfrac{1}{x}, & \text{ for } x \neq 0 \\ 0, &\text{ for } x=0\end{cases}\end{aligned}$$

This is how the graph of this function looks like:

and then if we zoom in even more, this is what we see

This function is not differentiable at zero; this is its derivative:

We are interested in the (generalised) derivatives at \(x=0\). The Dini derivative is

$$\begin{aligned}f^-(x; h) {}={} \liminf_{t \downarrow 0}\frac{t^2h^2 \sin\tfrac{1}{th}}{t}=\liminf_{t \downarrow 0} th^2 \sin\tfrac{1}{th}=0.\end{aligned}$$

In order to determine the Clarke derivative we use the fact that the function is continuously differentiable at all \(y \neq 0\), therefore, by Taylor's approximation theorem,

$$ \begin{aligned}f(y + th) {}={} y(t) + f'(y)th + o(|th|),\end{aligned}$$

for adequately small \(t\), and for \(y \neq 0\),

$$\begin{aligned}f'(y) {}={} 2y\sin\tfrac{1}{y} - \cos\tfrac{1}{y}.\end{aligned}$$

Then, for any fixed \(h\),

$$\begin{aligned}f^\circ(0; h) {}={}& \limsup_{y\to 0, t \downarrow 0}\frac{f'(y)th + o(|th|)}{t} \\ {}={}& \limsup_{y\to 0, t \downarrow 0} f'(y)|h| + \frac{o(|th|)}{t} = |h|.\end{aligned}$$

The Michel-Penot derivative is

$$\begin{aligned}f^\#(0; h) {}={}& \sup_u\limsup_{t \downarrow 0}\frac{(th+tu)^2\sin\frac{1}{th+tu} - t^2u^2\sin\frac{1}{tu}}{t} \\ {}={}& \sup_u \limsup_{t \downarrow 0} t \left[ (h+u)^2\sin\frac{1}{th+tu} - u^2\sin\frac{1}{tu}\right] = 0.\end{aligned}$$

### 2.4. An even more interesting example

Consider the function

$$\begin{aligned}f(x) {}={} \begin{cases}3^n, & \text{ if } 3^n \leq x \leq 2(3^n), n \in \mathbb{Z} \\ 2x - 3^{n+1}, &\text{ if } 2(3^n) \leq x \leq 3^{n+1} \\ 0, & \text{ if } x \leq 0\end{cases}\end{aligned}$$

The Dini derivative of \(f\) at 0 for a negative direction \( h < 0\) is clearly equal to 0. For \( h>0\), by definition

$$\begin{aligned}f'(0; h) = \liminf_{t \downarrow 0}\frac{th}{t} = \lim_{\epsilon \downarrow 0} \inf_{0 < t < \epsilon}\frac{f(th)}{t} = \frac{h}{2},\end{aligned}$$

so, overall \( f'(0; h) = \max\{0, \tfrac{h}{2}\}.\)

The liminf of the Dini derivative with \( h>0\) is illustrated below:

For the Clarke derivative with \( h>0\) it is convenient to use the following expression:

$$\begin{aligned}f^\circ(0; h) ={}& \limsup_{y\to 0, t \downarrow 0}\frac{f(y+th) - f(y)}{t} \\ {}={}& \lim_{\epsilon \downarrow 0} \underbrace{\sup_{\substack{|y| < \epsilon \\ 0 < t < \epsilon}} \frac{f(y+th) - f(y)}{t}}_{2} = 2h,\end{aligned}$$

whereas for \( h < 0\) we have \( f^\circ(0; h) = 0\), so overall \( f^\circ(0; h) = \max\{0, 2h\}\).

Lastly, the Michel-Penot derivative at 0 along a direction \( h<0\) is equal to zero because it is bounded between the Dini and Clarke derivatives, which vanish for \( h<0\). For \( h > 0\) the Michel-Penot derivative is \( f^\#(0; h) = 2h\) and the optimal u in the supremum is \( u = 2\). Overall \( f^\#(0; h) = \max\{0, 2h\}\).

Below you can see an animation showing the convergence of the limsup in the definition of the Michel-Penot derivative for \( h=1\) and for different values of \( u\) (\( u=1,\tfrac{1}{2},2\)).

### 2.5. Summary

Function | Dini, \(f'(0; h)\) | Michel-Penot \(f^\#(0; h)\) | Clarke \(f^\circ(0; h)\) |
---|---|---|---|

\(f(x) = |x|\) | \(|h|\) | \(|h|\) | \(|h|\) |

\(f(x) = -|x|\) | \(-|h|\) | \(-|h|\) | \(|h|\) |

Example 2.3 | \(0\) | \(0\) | \(|h|\) |

Example 2.4 | \(\max\{0, \tfrac{h}{2}\}\) | \(\max\{0, 2h\}\) | \(\max\{0, 2h\}\) |

## Bibliographic notes

[1] The examples given here are the solutions of Exercise 1 in Section 6.1 (Generalized derivatives) of the book: JM Borwein and AS Lewis, Convex Analysis and Nonlinear Optimization, Canadian Mathematical Society, Second Edition, Springer, 2010 (p. 127-128).

[2] FH Clarke, Optimization and Nonsmooth Analysis, SIAM Classics in Applied Mathematics, Wiley, 1983: A great book to understand the concept of Clarke’s generalised derivative.