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Chapter 5: Problem 11

Let \(K>0\) and let \(f: \mathbb{R} \rightarrow \mathbb{R}\) satisfy the condition \(|f(x)-f(y)| \leq K|x-y|\) for all \(x, y \in \mathbb{R}\). Show that \(f\) is continuous at every point \(c \in \mathbb{R}\).

Short Answer

Expert verified
The function \(f\) is continuous at every point \(c \in \mathbb{R}\) as the function satisfies the definition of continuity using the given inequality and the chosen \(\delta\).

Step by step solution

01

Define the Problem

We are given \(|f(x)-f(y)| \leq K|x-y|\) for all \(x, y \in \mathbb{R}\) and a constant \(K > 0\). We need to show that the function \(f\) is continuous everywhere in \(\mathbb{R}\). Therefore we should demonstrate that for every point \(c \in \mathbb{R}\), \(f\) is continuous at \(c\).
02

Using the Definition of Continuity

Let's denote \(\epsilon > 0\). In order for \(f\) to be continuous at \(c\), for any \(\epsilon > 0\) we must find a \(\delta > 0\) such that if \(|x - c| < \delta\), then \(|f(x) - f(c)| < \epsilon\). This is the definition of continuity that we will use to prove that \(f\) is continuous everywhere within \(\mathbb{R}\).
03

Apply the Given Inequality to The Continuity Definition

By using the given inequality, we can make \(|f(x) - f(c)| \leq K * |x - c|\). Comparing this with the definition of continuity we obtained in step 2, we see that we can choose \(\delta = \epsilon / K\). Then, for \(|x - c| < \delta\), we have that \(|f(x) - f(c)| \leq K * |x - c| = K * \delta = \epsilon\). This satisfies the definition of continuity at point \(c\). Therefore, the function \(f\) is continuous at every point in \(\mathbb{R}\).

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