In my current lecture Numerical Analysis of Ordinary Differential Equations we introduced the concept of One-sided Lipschitz functions.

A function $f: D \rightarrow \mathbb{C}^d$ satisfies a one-sided Lipschitz condition on it's domain $D \subseteq \mathbb{R} \times \mathbb{C}^d$ if there exists $C \in \mathbb{R}$, such that the inequality

$$\Re((f(t,x) - f(t,y))^t(x-y)) \leq C|x-y|^2$$

holds for all $(t,x),(t,y) \in D$.

I always visualised the Lipschitz condition using this cone-intuition, but can't seem to find an analogue for the one-sided Lipschitz condition. Is there a visual interpretation of the one-sided version that, for example helps me understand why $f(x) = e^{-x}$ satisfies a one-sided Lipschitz condition?


As you have already mentioned, Lipschitz condition is sometimes illustrated by a cone condition:

Note that the Lipschitz constant $C$ is related to the (white) cone angle, i.e. if $\theta$ is the angle between two generating lines of the cone (which we call $L_1$ and $L_2$) then $$C=\cot\frac{\theta}2$$ The key to understand or visualize one-sided Lipschitz condition is the $\frac{\theta}2$ term. If a function is not Lipschitz at all, then there is no $C$ or in other words, $C$ tends to infinity and thus $\frac{\theta}2$ or $\theta$ are zero. This means that the cone vanishes.

Let $\alpha_1$ be the angle between $L_1$ and the vertical line. Also let $\alpha_2=\theta-\alpha_1$. For a Lipschits function, these are both equal to $\frac{\theta}2$. But if a function is Lipschitz only in one direction, it would mean either of $\cot\alpha_1$ or $\cot\alpha_2$ is undefined. So either $\alpha_1=0$ or $\alpha_2=0$. In other words, the cone is oblique.

For example, $f(x)=e^{-x}$ is one-sided Lipschitz with $C=0$. You can easily verify that in this case $\alpha_1=90^\circ$ and $\alpha_2=0$.


Edit: How to mathematically formulate this property?

For real functions of a single variable, the relationship between the intuition and mathematical formulation is straightforward. In this case, if a function is Lipschitz on a domain $D\subset\mathbb R$ then $$|f(x_1)-f(x_2)|\le C|x_1-x_2|,\quad \{x_1,x_2\}\subset D$$ And if it is one-sided Lipschitz, there would be no absolute values in this inequality. For example, if $f$ is Lipschitz on the right side then for every real values of $x_1$ and $x_2$ on its domain, there is $C\ge 0$ such that $$f(x_2)-f(x_1)\le C(x_2-x_1)$$ which can be re-written as $$\left(f(x_2)-f(x_1)\right)(x_2-x_1)\le C|x_2-x_1|^2\tag{*}$$ Now for multi-variable complex functions, the $(*)$ condition can be generalized to the form you mentioned: $$\Re\{\left(f(x_2) - f(x_1)\right)^T(x_2-x_1)\} \leq C||x_2-x_1||^2$$

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  • $\begingroup$ Thank you. Is there also a way to justify this intuition using the definition of the one sided Lipschitz condition? $\endgroup$ – Herickson Jul 12 '19 at 11:20
  • $\begingroup$ @Herickson see the edited answer $\endgroup$ – polfosol Jul 13 '19 at 13:16

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