3. To show that f is not uniformly continuous on A, you should demonstrate that there is some e> 0 such that no matter what &> 0 you pick, you can find points a, y E A such that r y < 8 and f(r)- f (u) 2 e. Consider the proof below demonstrating that f(a) is not uniformly continuous on (0, 1). Proof: Consider e = 1/2. Let &> 0 be given. Then, there exists n E N such that <6. Moreover, we can choosen 2 3. If r and y which are in (0, 1), then 0 < |a - yl = and IS(*)- ()-/(2/n)- f(1/n)| n/22 1/2 e. Using a similar proof, show that f(a) = is not uniformly continuous on (0, 1).

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Chapter2: Second-order Linear Odes
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3. To show that \( f \) is not uniformly continuous on \( A \), you should demonstrate that there is some \( \varepsilon > 0 \) such that no matter what \( \delta > 0 \) you pick, you can find points \( x, y \in A \) such that \(|x - y| < \delta\) and \(|f(x) - f(y)| \geq \varepsilon\). Consider the proof below demonstrating that \( f(x) = \frac{1}{x} \) is not uniformly continuous on \( (0, 1) \).

**Proof:** Consider \( \varepsilon = 1/2 \). Let \( \delta > 0 \) be given. Then, there exists \( n \in \mathbb{N} \) such that \( \frac{1}{n} < \delta \). Moreover, we can choose \( n \geq 3 \). If \( x = \frac{2}{n} \) and \( y = \frac{1}{n} \) which are in \( (0, 1) \), then

\[
0 < |x - y| = \left| \frac{2}{n} - \frac{1}{n} \right| = \frac{1}{n} < \delta
\]

and

\[
|f(x) - f(y)| = \left|\frac{1}{2/n} - \frac{1}{1/n}\right| = \left|n/2 - n\right| = n/2 \geq 1/2 = \varepsilon.
\]

Using a similar proof, show that \( f(x) = \frac{1}{x} \) is not uniformly continuous on \( (0, 1) \).
Transcribed Image Text:3. To show that \( f \) is not uniformly continuous on \( A \), you should demonstrate that there is some \( \varepsilon > 0 \) such that no matter what \( \delta > 0 \) you pick, you can find points \( x, y \in A \) such that \(|x - y| < \delta\) and \(|f(x) - f(y)| \geq \varepsilon\). Consider the proof below demonstrating that \( f(x) = \frac{1}{x} \) is not uniformly continuous on \( (0, 1) \). **Proof:** Consider \( \varepsilon = 1/2 \). Let \( \delta > 0 \) be given. Then, there exists \( n \in \mathbb{N} \) such that \( \frac{1}{n} < \delta \). Moreover, we can choose \( n \geq 3 \). If \( x = \frac{2}{n} \) and \( y = \frac{1}{n} \) which are in \( (0, 1) \), then \[ 0 < |x - y| = \left| \frac{2}{n} - \frac{1}{n} \right| = \frac{1}{n} < \delta \] and \[ |f(x) - f(y)| = \left|\frac{1}{2/n} - \frac{1}{1/n}\right| = \left|n/2 - n\right| = n/2 \geq 1/2 = \varepsilon. \] Using a similar proof, show that \( f(x) = \frac{1}{x} \) is not uniformly continuous on \( (0, 1) \).
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