7
$\begingroup$

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?

$\endgroup$
3
+25
$\begingroup$

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 $\alpha_2=0$.

exp-x

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: $$\Re\{\left(f(x_2) - f(x_1)\right)^T(x_2-x_1)\} \leq C||x_2-x_1||^2$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.