/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Let \(D\) and \(E\) be subsets o... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 21

Let \(D\) and \(E\) be subsets of \(\mathbb{R}\), and let \(f: D \rightarrow \mathbb{R}\) and \(g: E \rightarrow \mathbb{R}\) be functions such that the range of \(f\) is contained in \(E .\) If \(f\) is uniformly continuous on \(D\) and \(g\) is uniformly continuous on \(E\), then show that \(g \circ f\) is uniformly continuous on \(D\).

Short Answer

Expert verified
Let \(\varepsilon > 0\). Since both \(f\) and \(g\) are uniformly continuous, there exists \(\delta_f\) and \(\delta_g\) that satisfy their respective continuity conditions. We take \(\delta = \min\{\delta_f, \delta_g\}\) and let \(x, y \in D\) with \(|x - y| < \delta\). Using the uniform continuity of both functions and the inclusion of the range of \(f\) in \(E\), we can show that \(|g(f(x)) - g(f(y))| < \varepsilon\), proving that \(g \circ f\) is uniformly continuous on \(D\).

Step by step solution

01

Recall the definitions

Recall that a function \(f:D\rightarrow\mathbb{R}\) is uniformly continuous on \(D\) if for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for any \(x, y \in D\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \varepsilon\).
02

Use the uniform continuity of \(g\)

Let \(\varepsilon > 0\). Since \(g\) is uniformly continuous on \(E\), there exists a \(\delta_g > 0\) such that for any \(u, v \in E\) with \(|u - v| < \delta_g\), we have \(|g(u) - g(v)| < \varepsilon\).
03

Use the uniform continuity of \(f\)

Now, we know that the range of \(f\) is contained in \(E\) and \(f\) is uniformly continuous on \(D\). So, there exists a \(\delta_f > 0\) such that for any \(x, y \in D\) with \(|x - y| < \delta_f\), we have \(|f(x) - f(y)| < \delta_g\).
04

Combine the uniform continuity of \(f\) and \(g\)

Let \(\delta = \min\{\delta_f, \delta_g\}\). Now, we will show that \(g\circ f\) is uniformly continuous on \(D\) using this \(\delta\). Let \(x,y\in D\) with \(|x - y| < \delta\). Since \(\delta \leq \delta_f\), we have \(|f(x) - f(y)|< \delta_g\). Also, because the range of \(f\) is contained in \(E\), both \(f(x)\) and \(f(y)\) belong to \(E\). Since \(|f(x) - f(y)| < \delta_g\), by the uniform continuity of \(g\) on \(E\), we obtain \(|g(f(x)) - g(f(y))| < \varepsilon\).
05

Conclude the proof

As we showed, for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that \(|x - y| < \delta\) for \(x,y \in D\) implies \(|g(f(x)) - g(f(y))| < \varepsilon\). Thus, the function \(g\circ f\) is uniformly continuous on \(D\).

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Most popular questions from this chapter

Let \(D \subseteq \mathbb{R}, c \in \mathbb{R}\), and let \(f: D \rightarrow \mathbb{R}\) be a function. (i) If \(c\) is a limit point of \(D\), then we say that a limit of \(f\) as \(x\) tends to \(c\) exists if there is a real number \(\ell\) such that \(\left(x_{n}\right)\) any sequence in \(D \backslash\\{c\\}\) and \(x_{n} \rightarrow c \Longrightarrow f\left(x_{n}\right) \rightarrow \ell\). Show that if a limit of \(f\) as \(x\) tends to \(c\) exists, then it is unique. (ii) If \(c\) is not a limit point of \(D\), then show that for any \(\ell \in \mathbb{R}\), the condition \(\left(x_{n}\right)\) any sequence in \(D \backslash\\{c\\}\) and \(x_{n} \rightarrow c \Longrightarrow f\left(x_{n}\right) \rightarrow \ell\) holds vacuously.

Show that \(f(x) \rightarrow \infty\) as \(x \rightarrow \infty\), if \(f:[0, \infty) \rightarrow \mathbb{R}\) is given by (i) \(f(x)=\frac{3 x^{2}+1}{2 x+1}\), (ii) \(f(x)=[x]\).

Analyze the following functions for uniform continuity: (i) \(f(x)=x, x \in \mathbb{R}\) (ii) \(f(x)=1 / x, x \in(0,1]\) (iii) \(f(x)=x^{2}, x \in(0,1)\) (iv) \(f(x)=\sqrt{1-x^{2}}, x \in[-1,1]\)

Let \(f:[0,1] \rightarrow \mathbb{R}\) be given by $$ f(x)=\left\\{\begin{array}{ll} 3 x / 2 & \text { if } 0 \leq x<\frac{1}{2} \\ (3 x-1) / 2 & \text { if } \frac{1}{2} \leq x \leq 1 \end{array}\right. $$ Show that \(f([0,1])=[0,1] .\) Is \(f\) continuous on \([0,1] ?\) Does \(f\) have the IVP on \([0,1] ?\)

Let \(k \in \mathbb{N}\) and \(f(x)=x^{1 / k}\) for \(x \in[0, \infty)\). If \(\epsilon \in \mathbb{R}\) is such that \(0<\epsilon \leq 1\), define \(\delta:=\min \left\\{(1+\epsilon)^{n}-1,1-(1-\epsilon)^{n}\right\\} .\) Show that \(\delta>0\) and $$ x \in[0, \infty) \text { and }|x-1|<\delta \Longrightarrow|f(x)-1|<\epsilon $$ Also, show that \(\delta\) is the greatest real number for which this holds.
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