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Chapter 5: Problem 2

Show that the function \(f(x):=1 / x^{2}\) is uniformly continuous on \(A:=[1, \infty)\), but that it is not uniformly continuous on \(B:=(0, \infty)\).

Short Answer

Expert verified
The function \(f(x):=1 / x^{2}\) is uniformly continuous on \(A:=[1, \infty)\) and not uniformly continuous on \(B:=(0, \infty)\).

Step by step solution

01

Prove Uniform Continuity on A

We must show that, for all \(蔚 > 0\), there exists a \(未 > 0\) such that for all \(x, y \in A\) with \(|x-y| < 未\), we have \(|f(x) - f(y)| < 蔚\). To do this, note that \(|f(x) - f(y)| = |1/x^2 - 1/y^2| = |y^2 - x^2| / (x^2y^2). If \(|x-y| < 未\), then \(|y^2 - x^2| = |(y - x)(y + x)| 鈮� 2未\). So \(|f(x) - f(y)| 鈮� 2未 / y^4\) for \(a 鈮� 1\). To make this less than \(蔚\), we can choose \(未 = 蔚y^4 / 2\).
02

Prove Non-Uniform Continuity on B

We can prove the function \(f(x) \neq 1 / x^2\) is not uniformly continuous on \(B\) by finding a counterexample. Take \(蔚 = 1\). Let \(x = 1/n\) and \(y = 1/(n+1)\) for some integer \(n\). Then, \(|x-y| = 1/n^2 - 1/(n+1)^2\), which can be made less than any \(未 > 0\) by choosing a sufficiently large \(n\). However, \(|f(x) - f(y)| = n^4/(n+1)^2 - n^4 = n^2/(n+1)^2\), which is always greater than 1/4 for all \(n\), regardless of the choice of \(未\). So for these \(x, y\), \(|x-y| < 未\) but \(|f(x) - f(y)| 鈮� 蔚\), proving \(f(x)\) is not uniformly continuous on \(B\).

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Most popular questions from this chapter

If \(g(x):=\sqrt{x}\) for \(x \in[0,1]\), show that there does not exist a constant \(K\) such that \(|g(x)| \leq K|x|\) for all \(x \in[0,1]\). Conclude that the uniformiy continuous \(g\) is not a Lipschitz function on \([0,1]\).

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) and \(g: l \rightarrow \mathbb{R}\) be continuous functions on \(I\). Show that the set \(E:=\\{x \in I: f(x)=g(x)\\}\) has the property that if \(\left(x_{n}\right) \subseteq E\) and \(x_{n} \rightarrow x_{0}\), then \(x_{0} \in E\)

If \(f\) and \(g\) are continuous on \(\mathbb{R}\), let \(S:=\\{x \in \mathbb{R}: f(x) \geq g(x)\\} .\) If \(\left(s_{n}\right) \subseteq S\) and \(\lim \left(s_{n}\right)=s\) show that \(s \in S\).

Let \(I \subseteq \mathbb{R}\) be an interval and let \(f: I \rightarrow \mathbb{R}\) be increasing on \(I\). Suppose that \(c \in I\) is not an endpoint of \(I\). Show that \(f\) is continuous at \(c\) if and only if there exists a sequence \(\left(x_{n}\right)\) in \(I\) such that \(x_{n}c\) for \(n=2,4,6, \cdots\), and such that \(c=\lim \left(x_{n}\right)\) and \(f(c)=\lim \left(f\left(x_{n}\right)\right)\)

Let \(f\) be defined for all \(x \in \mathbb{R}, x \neq 2\), by \(f(x)=\left(x^{2}+x-6\right) /(x-2) .\) Can \(f\) be defined at \(x=2\) in such a way that \(f\) is continuous at this poiat?
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