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Chapter 5: Problem 12

Show that if \(f\) is continuous on \([0, \infty)\) and uniformly continuous on \([a, \infty)\) for some positive constant \(a\), then \(f\) is uniformly continuous on \([0, \infty)\).

Short Answer

Expert verified
The function \(f\) is uniformly continuous on [0, \(\infty\)] because it's uniformly continuous on [a, \(\infty\)] and on [0, a].

Step by step solution

01

State the definition of uniform continuity

A function \(f\) is called uniformly continuous on a set \(S\) if for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for every pair of points \(x, y \in S\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \epsilon\).
02

Apply uniform continuity on [a, \(\infty\)]

Given that \(f\) is uniformly continuous on [a, \(\infty\)], for each \(\epsilon > 0\) there exists a \(\delta_1 > 0\) so that for all \(x, y \in [a, \infty)\) with \(|x - y| < \delta_1\), it holds that \(|f(x) - f(y)| < \epsilon\).
03

Apply continuity on [0, \(\infty\)]

Given that \(f\) is continuous on [0, \(\infty\)], it is specifically continuous on the closed interval [0, a], and hence uniformly continuous there because any continuous function on a closed interval is uniformly continuous. Thus, for each \(\epsilon > 0\) there exists a \(\delta_2 > 0\) so that for all \(x, y \in [0, a]\) with \(|x - y| < \delta_2\), it holds that \(|f(x) - f(y)| < \epsilon\).
04

Combine the intervals

Let \(\delta = \min(\delta_1, \delta_2)\). Then for any \(x, y \in [0, \infty)\) with \(|x - y| < \delta\), if both points are in [a, \(\infty\)] or both points are in [0, a], we know \(|f(x) - f(y)| < \epsilon\) based on steps 2 and 3. If one point is in [0, a] and the other is in [a, \(\infty\)], we know \(|f(x) - f(y)| < \epsilon\) as well because \(|x - y| < \delta\) must be smaller than a, and hence both x and y must be in [0, a]. This implies \(f\) is uniformly continuous on [0, \(\infty\)].

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Let \(f:(0,1) \rightarrow \mathbb{R}\) be bounded but such that \(\lim _{x \rightarrow 0} f\) does not exist. Show that there are two sequences \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) in \((0,1)\) with \(\lim \left(x_{n}\right)=0=\lim \left(y_{n}\right)\), but such that \(\lim \left(f\left(x_{n}\right)\right)\) and \(\lim \left(f\left(y_{n}\right)\right)\) exist but are not equal.

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