# Inequalities from inequations

An inequality is a statement of the form $A\le B$ or $A < B$. (What's up with vertical alignment of formulas in WP?) An inequation is $A\ne B$. I can’t think of any adequate Russian word for “inequation”, but that’s besides the point. The literal analog “неравенство” is already used for “inequality”.

Suppose we want to prove that a map $f\colon \mathbb C \to \mathbb C$ is Lipschitz, that is, $\exists\ L$ such that $|f(z)-f(w)|\le L|z-w|$ for all $z,w\in\mathbb C$. All we know is that the map $f_{\lambda}(z):=f(x)+\lambda z$ is injective for all $\lambda\in \mathbb C$ with a large modulus, i.e., for $|\lambda|\ge C$. In the following we consider only such values of $\lambda$.

Fix distinct $z,w\in \mathbb C$ and record the inequation: $f(z)+\lambda z\ne f(w)+\lambda w$. Rearrange as $f(z)-f(w)\ne -\lambda (z-w)$. Note that we can multiply $\lambda$ by any unimodular complex number, since the inequation holds whenever $|\lambda|\ge C$. Thus, we have a stronger inequation: $|f(z)-f(w)|\ne |\lambda| |z-w|$. Yes, putting absolute values in an inequation makes it stronger, opposite to what happens with equations.

Still keeping $z$ and $w$ fixed, increase the modulus of $\lambda$ until the inequality $|f(z)-f(w)| < |\lambda| |z-w|$ holds. Continuity with respect to $\lambda$ tells us that $\ne$ was indeed $<$ all the way. So we bring $|\lambda|$ back down to $|\lambda|=C$, and happily record the inequality $|f(z)-f(w)| < C |z-w|$, the desired Lipschitz continuity.

Self-promotion: a paper in which the above trick was used.

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