TITLE: Is there a nonstandard characterization of Lipschitz continuity?
QUESTION [2 upvotes]: 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.

REPLY [1 votes]: 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. 
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