KindiakMath

An Inequality Puzzle

Problem 1. Suppose f : \mathbb R \to \mathbb R satisfies the inequality

\displaystyle |f(x) - f(y)| \leq (x-y)^2, \quad x ,y \in \mathbb R,

and f(0) = 1. Given x \in \mathbb R, evaluate f(x).

(Click for Solution)

Solution. Fix c \in \mathbb R. For any h \in \mathbb R,

\displaystyle |f(c+h) - f(c)| \leq ((c+h)-c)^2 = h^2.

Dividing by |h| on both sides,

\displaystyle \left| \frac{f(c+h) - f(c)}{h} \right| \leq |h|.

Taking h \to 0, by the squeeze theorem,

\displaystyle f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = 0.

Since c \in \mathbb R is arbitrary, f' = 0. Thus, f is differentiable on \mathbb R with derivative f' = 0. Fix x \in \mathbb R. By the mean value theorem, there exists \xi between 0 and x such that

f(x) = f(0) + f'(\xi) \cdot (x-0) = 1 + 0 \cdot x = 1.

—Joel Kindiak, 23 Aug 25, 2237H

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joelkindiak

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