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Chapter 5: Problem 9

If \(f\) is uniformly continuous on \(A \subseteq \mathbb{R}\), and \(|f(x)| \geq k>0\) for all \(x \in A\), show that \(1 / f\) is uniformly continuous on \(A\).

Short Answer

Expert verified
The function \(1/f\) is uniformly continuous on \(A\).

Step by step solution

01

Definition of Uniform Continuity

The definition of uniform continuity states that a function \( f \) is uniformly continuous on \( A \) if, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \) in \( A \), if \( |x-y| < \delta \), then \( |f(x)-f(y)| < \epsilon \). This needs to be displayed for the function \( 1/f \).
02

Setup for Proving Uniform Continuity

Given \( |f(x)| \geq k > 0 \), it follows that \( 1/|f(x)| \leq 1/k \). This shows that 1/f is bounded. Moreover, since f is uniformly continuous, there exists a \( \delta > 0 \) such that for all \( x, y \) in \( A \), if \( |x-y| < \delta \), then \( |f(x)-f(y)| < \epsilon k^2 \).
03

Proving Uniform Continuity

Applying the triangle inequality, it's found that \( |1/f(x) - 1/f(y)| = |f(y)-f(x)|/(|f(x)||f(y)|) \leq |f(y)-f(x)|/(k^2) < \epsilon \). Thus, for all \( x, y \) in \( A \), if \( |x-y| < \delta \), then \( |1/f(x)-1/f(y)| < \epsilon \). It's demonstrated that 1/f is uniformly continuous on A.

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