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”.

Moon in inequation, by Coralie Mercier,

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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