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, http://www.flickr.com/photos/koalie/3520170352/

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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