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Chapter 3: Problem 4

Let \(f: D \rightarrow \mathbb{R}\) and let \(x_{0} \in D\). Prove that if \(f\) is continuous at \(x_{0},\) then \(|f|\) is continuous at this point. Is the converse true in general?

Short Answer

Expert verified
Yes, if \(f\) is continuous at \(x_0\), then \(|f|\) is continuous at \(x_0\). The converse is not true generally.

Step by step solution

01

Understand Continuity Definition

Recall the definition of continuity. A function \(f\) is continuous at \(x_0\) if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\) in the domain \(D\), whenever \(|x - x_0| < \delta\), it follows that \(|f(x) - f(x_0)| < \epsilon\).
02

Express Absolute Continuity Condition

To prove that \(|f|\) is continuous at \(x_0\), we need to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(|x - x_0| < \delta\), it follows that \(| |f(x)| - |f(x_0)| | < \epsilon\).
03

Use Triangle Inequality

Utilize the triangle inequality which states that \(| |f(x)| - |f(x_0)| | \leq |f(x) - f(x_0)|\). This inequality is always true, simplifying our proof.
04

Apply Continuity of f

Given that \(f\) is continuous at \(x_0\), for the \(\epsilon\) given, there exists a \(\delta > 0\) such that \(|f(x) - f(x_0)| < \epsilon\) whenever \(|x-x_0| < \delta\).
05

Conclude |f| is Continuous

Using the triangle inequality from Step 3, since \(| |f(x)| - |f(x_0)| | \leq |f(x) - f(x_0)| < \epsilon\), it follows that \(|f|\) is continuous at \(x_0\). Therefore, the continuity of \(f\) implies the continuity of \(|f|\) at \(x_0\).
06

Analyze the Converse Statement

Consider the converse: if \(|f|\) is continuous at \(x_0\), is \(f\) continuous at \(x_0\)? This is not true generally. For a counterexample, consider \(f(x) = x\) if \(x eq 0\) and \(f(0) = 1\). Here \(|f(x)| = |x|\), which is continuous at 0, but \(f\) itself is not continuous at 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous functions
A function is continuous if it behaves predictably as inputs change gradually. To be specific, consider a function \( f: D \rightarrow \mathbb{R} \) defined on a domain \( D \). The function \( f \) is continuous at a point \( x_0 \in D \) if for every small number \( \epsilon > 0 \), there exists a small number \( \delta > 0 \) such that whenever \( |x - x_0| < \delta \), it results in \( |f(x) - f(x_0)| < \epsilon \).
  • This definition ensures that as \( x \) gets arbitrarily close to \( x_0 \), \( f(x) \) stays close to \( f(x_0) \).
  • A function without discontinuity is consistent, meaning no jumps or breaks.
  • The behavior is smooth because small changes in input shouldn't cause large changes in output.
Understanding this definition helps to appreciate how continuity is not just about absence of breaks but about predictable and smooth behavior under tiny perturbations.
Absolute value and continuity
The absolute value function affects continuity by ensuring non-negative outputs. If \( f \) is continuous at a point, its absolute value \( |f| \) function must also be continuous. Let's explore why:
  • Starting from the definition of continuity, for \(|f|\) to be continuous at \(x_0\),
    • For every \(\epsilon > 0\), there should exist a \(\delta > 0\) such that whenever \(|x - x_0| < \delta\), it follows that \(| |f(x)| - |f(x_0)| | < \epsilon\).
  • The absolute value \(|f(x)|\) just alters the value to be non-negative but maintains the close proximity of \( f(x) \) and \( f(x_0) \).
  • Since \(|f|\) simply takes the non-negative part of \( f \), it doesn't break continuity.
Triangle inequality in continuity
The triangle inequality is a powerful tool that aids in verifying continuity, especially when dealing with absolute values. In essence, the triangle inequality states that for any real numbers \( a \) and \( b \), \(|a + b| \leq |a| + |b|\).When applying this to our context:
  • The inequality \(| |f(x)| - |f(x_0)| | \leq |f(x) - f(x_0)|\) shows that the fluctuation in \(|f|\) is always limited by the fluctuation in \(f\).
    • This means that if \(f\) is continuous at \(x_0\), then \(|f|\) must also be continuous there, because the inequality ensures \(| |f(x)| - |f(x_0)| | < \epsilon\).
  • Since \(|f(x) - f(x_0)| < \epsilon\) can be achieved with a suitable \(\delta\), so can \(| |f(x)| - |f(x_0)| | < \epsilon\).
The triangle inequality thus serves as a bridge from the continuity of \(f\) to the continuity of \(|f|\). This method simplifies proofs because it allows us to break down the problem of proving continuity into something manageable and intuitive.

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Most popular questions from this chapter

Prove that each of the following functions is uniformly continuous on the given domain: (a) \(f(x)=a x+b, a, b \in \mathbb{R},\) on \(\mathbb{R}\). (b) \(f(x)=1 / x\) on \([a, \infty),\) where \(a>0\).

Let \(f: D \rightarrow \mathbb{R}\) be continuous at \(c \in D\) and let \(\gamma \in \mathbb{R}\). Suppose \(f(c)>\gamma\). Prove that there exists \(\delta>0\) such that $$ f(x)>\gamma \text { for every } x \in B(c ; \delta) \cap D. $$

Prove that the equation \(x^{2}-2=\cos (x+1)\) has at least two real solutions. (Assume known that the function \(\cos x\) is continuous.)

Find the following limits: (a) \(\lim _{x \rightarrow 2} \frac{3 x^{2}-2 x+5}{x-3}\), (b) \(\lim _{x \rightarrow-3} \frac{x^{2}+4 x+3}{x^{2}-9}\)

Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x}\) be a limit point of \(D .\) Prove that if \(\lim _{x \rightarrow \bar{x}} f(x)=\ell,\) then \(\lim _{x \rightarrow \bar{x}}|f(x)|=|\ell|\) Give an example to show that the converse is not true in general.
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