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Chapter 5: Problem 7

If \(f(x):=x\) and \(g(x):=\sin x\), show that both \(f\) and \(g\) are uniformly continuous on \(\mathbb{R}\), but that their product \(f g\) is not uniformly continuous on \(\mathbb{R}\).

Short Answer

Expert verified
\(f(x) = x\) and \(g(x) = \sin(x)\) are uniformly continuous on \(\mathbb{R}\), but their product, \(f(x)g(x)\) is not uniformly continuous on \(\mathbb{R}\).

Step by step solution

01

Demonstrate Uniform Continuity of \(f(x) = x\)

A function \(f(x)\) is said to be uniformly continuous on \(\mathbb{R}\) if for every number \(\epsilon > 0\), there exists a number \(\delta > 0\) such that whenever \(|x - y| < \delta\), it follows that \(|f(x) - f(y)| < \epsilon\). In the case of \(f(x) = x\), we see that if \(|x - y| < \delta\) then \(|x - y| < \epsilon\). Therefore, setting \(\delta = \epsilon\) always creates the necessary condition for uniform continuity, which shows that \(f(x) = x\) is uniformly continuous.
02

Demonstrate Uniform Continuity of \(g(x) = \sin(x)\)

For \(g(x) = \sin(x)\), for every two points \(x, y \in \mathbb{R}\) we have \(|\sin(x) - \sin(y)| \leq |x - y|\). So, if \(|x - y| < \delta\) then \(|\sin(x) - \sin(y)| \leq |x - y| < \delta\). If we set \(\delta = \epsilon\), we ensure \(|\sin(x) - \sin(y)| < \epsilon\), showing uniform continuity of \(g(x) = \sin(x)\).
03

Show The Product \(f(x)g(x)\) is Not Uniformly Continuous

\(f(x)g(x) = x \sin(x)\). For all \(n \in \mathbb{N}\) let us choose \(x_n = 2n\pi\) and \(y_n = 2n\pi + \pi /2n\). As \(n\) goes to infinity, \(x_n - y_n = \pi /2n\) tends to zero, but \(|(x_n \sin x_n) - (y_n \sin y_n)| = |\pi /2|\). Since it's not tending to zero as \(n\) tends to infinity, \(f(x)g(x)\) is not uniformly continuous.

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Most popular questions from this chapter

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) be continuous on \(I\). If \(f\) has an absolute maximum (respectively, minimum \(]\) at an interior point \(c\) of \(I\), show that \(f\) is noc injective on \(I\).

Let \(J:=(a, b)\) and let \(g: J \rightarrow \mathbb{R}\) be a continuous function with the property that for every \(x \in J\), the function \(g\) is bounded on a neighborhood \(V_{\delta}(x)\) of \(x .\) Show by example that \(g\) is not necessarily bounded on \(J\).

Let \(A \subseteq B \subseteq \mathbb{R}\), let \(f: B \rightarrow \mathbb{R}\) and let \(g\) be the restriction of \(f\) to \(A\) (that is, \(g(x)=f(x)\) for \(x \in A)\) (a) If \(f\) is continuous at \(c \in A\), show that \(g\) is continuous at \(c\). (b) Show by example that if \(g\) is continuous at \(c\), it need not follow that \(f\) is continuous at \(c\).

Let \(I:=[a, b]\), let \(f: I \rightarrow \mathbb{R}\) be continuous on \(I\), and assume that \(f(a)<0, f(b)>0\). Let \(W:=\\{x \in I: f(x)<0\\}\). and let \(w:=\sup W\). Prove that \(f(w)=0\). (This provides an alternative proof of Theorem \(5.3 .5 .)\)

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