The ML Inequality

Complex Analysis · Axiom Academy

One ceiling, one length: why the size of a contour integral can never exceed M·L. Walk along the contour C and watch the size |f(z)| of the integrand. Suppose it never pokes above a ceiling M . Then the integral is trapped: its magnitude is at most that ceiling height M times the distance L you walked. The picture is a box — height M , width L — and the integral lives inside it. M = the ceiling: the largest |f(z)| reaches on C L = the length of the contour C A loose lasso, on real numbers Integrate f(z)=z^2 along the straight segment from 0 to 1+i . The exact value is , with magnitude . On that segment |z| tops out at , so , and the length is . The ceiling gives — and sure enough . The bound holds; it is just generous. The proof is three honest moves. Parametrize the contour by z(t) for t from a to b , so the integral becomes . Now: (1) the size of an integral is at most the integral of sizes; (2) replace each |f| by its ceiling M ; (3) recognize the leftover integral of |z'(t)| as the arc length L . : a sum (or integral) of arrows is never longer than the sum of their lengths. everywhere, so the integrand only grows when we swap it for M . is precisely the arc length L of the contour. Stack the three and the bound drops out. No cancellation, no luck. 3. Why We Care: Vanishing Arcs

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