Is there a nonstandard characterization of Lipschitz continuity?

Question

Let $f: \mathbb R \to \mathbb R$ be Lipschitz continuous with finite constant $L$. Then $$ |f(x) - f(y) \le L |x-y|, \tag{1} $$ and, by direct transfer, this property holds for $^*\!f$.
For continuity, there is the infinitesimal characterization $$ x \approx y \to f(x) \approx f(y). \tag{2} $$

I'm wondering if there exists an infinitesimal characterization of (1).
I ask because (1) lacks the infinitesimal feeling that (2) gives.

Answer 1

Note the definition you give is not for continuity (of standard functions), but for uniform continuity. $f(x) = x^2$, for example, doesn't satisfy that property: e.g. take $x = H+1/H$ and $y = H$, where $H$ is a transfinite number.

I'm not entirely sure there's a need for an "infinitesimal" version of Lipschitz continuity, since it is already free from the usual annoyances of standard analysis, but here goes.

For any positive constant $\delta$, you can restrict the standard definition to require that inequality only for $|x - y| < \delta$. For a a nonstandard version, it should be enough to prove the inequality only for $x \approx y$.

I assert the following:

A standard function $f$ is Lipschitz continuous if and only if $(f(x) - f(y))/(x-y)$ is limited for all $x,y$ with $x \approx y$ and $x \neq y$.

(equivalently, you could replace $x \approx y$ with "$x-y$ limited", or even remove that condition entirely).

Proof: $(\Rightarrow)$ is straightforward. To prove $(\Leftarrow)$, suppose that $q(x,y) = (f(x) - f(y))/(x-y)$ is limited for all $x,y$ with $x \approx y$, $x \neq y$. Then $f$ is continuous.

Now let $x<y$ be any nonstandard numbers, and let $x = x_0 < x_1 < \cdots < x_n = y$ be a (hyperfinite) sequence such that $x_i \approx x_{i+1}$. The set of values $|q(x_{i+1}, x_i)|$ is hyperfinite (and internal), and thus has a maximum value $M$, which is limited. Thus $|f(x_{i+1}) - f(x_i)| \leq M |x_{i+1} - x_i|$, and it follows that $|q(x,y)| \leq M$.

Thus, $q(x,y)$ is limited for all $x \neq y$.

Let $H$ be positive transfinite, and $\epsilon$ be positive infinitesimal. $|q(x,y)|$ is a continuous function on the (internally) closed and bounded set $|x| \leq H \wedge |x-y| \in [\epsilon, H]$, and thus attains a maximum value $M$, which is limited. Let $L$ be any standard number larger than $M$.

We now have

$$ |f(x) - f(y)| \leq L |x-y| $$

for every standard $x,y$, and therefore $f$ is Lipschitz continuous.

Answer 2

Technically the answer is Yes, but morally, No.

The Lipschitz condition with exponent $\alpha < 1$ has very little content for infinitesimally close (bounded) $x,y$. Any differentiable function on a bounded set will satisfy $|f(x)-f(y)| \leq L|x-y|^\alpha$ for infinitesimally near points, and an arbitrary value of $L$. Thus, it is a large-scale constraint on the function and there is no way to localize the condition to small neighborhoods.

For $\alpha > 1$ there are no locally nonconstant functions satisfying the condition. Any constraints created by the Lipschitz condition are large-scale, controlling the variation of function values on separated components of the domain. Within components it is constant.

For $\alpha=1$, the inequality (for any metric space where distance can be estimated arbitrarily closely by paths of "short jumps") is equivalent, with the same constant $L$, to its restriction to infinitesimally close $x$ and $y$, with a one line proof "apply triangle inequality to an estimating path". So the restatement will look the same as the original with unnecessary extra words added, like infinitely close and standard.

Related Q&A at $|f(x)-f(y)|\le(x-y)^2$ without gaplessness