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Chapter 5: Problem 5

Show that if \(f\) and \(g\) are uniformly continuous on a subset \(A\) of \(\mathbb{R}\), then \(f+g\) is uniformly continuous on \(A\).

Short Answer

Expert verified
The sum of two uniformly continuous functions \(f\) and \(g\) is also uniformly continuous. This result can be obtained by using the definition of uniform continuity and the triangle inequality on absolute differences. The main idea of the proof is to show that if \(f\) and \(g\) are uniformly continuous on \(A\), then for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever the absolute difference in inputs is less than \(\delta\), the absolute difference in the function values of \(f+g\) is less than \(\varepsilon\).

Step by step solution

01

- Understand Uniform Continuity

Before proving the statement, it is crucial to know what uniform continuity entails. A function \(f: A \rightarrow \mathbb{R}\) on a set \(A \subseteq \mathbb{R}\) is uniformly continuous if for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(x, y \in A\) and \(|x - y| < \delta\), then \(|f(x) - f(y)| < \varepsilon\). That means the difference in function values is small whenever the difference in inputs is small, for all point in the domain \(A\). The task is to show that this holds for \(f+g\) if it holds for \(f\) and \(g\) individually.
02

- Setup of the Proof

Now a straightforward way to this proof is to start by assuming that \(f\) and \(g\) are uniformly continuous on \(A\). So, for a given \(\varepsilon > 0\), there exists \(\delta_1, \delta_2 > 0\) such that for any \(x, y \in A\) if \(|x - y| < \delta_1\), then \(|f(x) - f(y)| < \frac{\varepsilon}{2}\) and if \(|x - y| < \delta_2\), then \(|g(x) - g(y)| < \frac{\varepsilon}{2}\). The objective is now to construct \(\varepsilon-\delta\) proof for the function \(f+g\).
03

- Applying the Definitions

To find corresponding \(\delta\) for \(f+g\), choose \(\delta = \min(\delta_1, \delta_2)\). Then for any \(x, y \in A\), if \(|x - y| < \delta\), it means both \(|x - y| < \delta_1\) and \(|x - y| < \delta_2\). From the assumption of uniform continuity of functions \(f\) and \(g\), this implies \(|f(x) - f(y)| < \frac{\varepsilon}{2}\) and \(|g(x) - g(y)| < \frac{\varepsilon}{2}\).
04

- Constructing the Proof

Now consider the absolute difference \(|f(x) + g(x) - f(y) - g(y)|\). By triangle inequality, \(|f(x) + g(x) - f(y) - g(y)| \leq |f(x) - f(y)| + |g(x) - g(y)|\), which by step 3 is less than \(\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\). Thus, whenever \(|x - y| < \delta\), \(|f(x) + g(x) - f(y) - g(y)| < \varepsilon\). Showing that \(f+g\) is uniformly continuous on \(A\).

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