Chapter 3: Problem 13
True/False, prove or find a counterexample. If \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a continuous function such that \(\left.f\right|_{\mathbb{Z}}\) is bounded, then \(f\) is bounded.
Short Answer
Expert verified
False, a counterexample shows the statement is incorrect.
Step by step solution
01
Understand the Problem
We need to determine if a continuous function \( f: \mathbb{R} \rightarrow \mathbb{R} \) that is bounded on the integers \( \mathbb{Z} \) is necessarily bounded on the entire set of real numbers \( \mathbb{R} \). A function is bounded if there exists a real number \( M \) such that \( |f(x)| \leq M \) for all \( x \) in its domain.
02
Analyze the Given Conditions
The condition states that the restriction of \( f \) to the integers, \( \left.f\right|_{\mathbb{Z}} \), is bounded. This means there exists a real number \( M \) such that for all integers \( n \), \( |f(n)| \leq M \). We need to explore whether this boundedness on \( \mathbb{Z} \) implies boundedness on all of \( \mathbb{R} \).
03
Consider Relevant Theorems
Consider the properties of continuous functions: If a function is continuous on a compact set, then it is bounded on that set. However, \( \mathbb{R} \) is not compact, so this theorem does not directly apply. Therefore, the boundedness on \( \mathbb{Z} \), a discrete set, may not extend to \( \mathbb{R} \).
04
Construct a Counterexample
A counterexample would show that even if \( f \) is bounded on \( \mathbb{Z} \), \( f \) might not be bounded on \( \mathbb{R} \). Consider the function \( f(x) = x^2 \sin(\frac{1}{x}) \) for \( x eq 0 \) and \( f(0) = 0 \). This function is continuous on \( \mathbb{R} \). For integer \( n \), \( f(n) = n^2 \sin(\frac{1}{n}) \), and since \( |\sin(\frac{1}{n})| \leq 1 \), \( |f(n)| \leq n^2 \). Thus, \( f \) is bounded on any finite subset of \( \mathbb{Z} \). But as \( x \rightarrow 0 \), \( f(x) \rightarrow 0 \) and is not necessarily bounded on all of \( \mathbb{R} \) because as \( x \) increases around the unit circle size, \( f(x) \) grows proportionally to \( x^2 \).
05
Conclude with the Solution
The function \( f(x) = x^2 \sin(\frac{1}{x}) \) shows that even if \( f \) is bounded on \( \mathbb{Z} \), it is not necessarily bounded on \( \mathbb{R} \). Therefore, the statement is false.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bounded Functions
A function is described as bounded if there exists a real number \( M \) such that for every input \( x \) within its domain, the absolute value of the function \( f(x) \) is less than or equal to \( M \). In simpler terms, a bounded function has all of its outputs kept within a fixed upper and lower limit.
Consider the function \( g(x) = \sin(x) \) for all real numbers \( x \). The sine function is bounded because its outputs are constrained within -1 and 1 no matter what real value \( x \) takes. This consistent limitation signifies that the function does not go to infinity as the inputs grow, making \( g(x) \) a bounded function.
An important thing to note about bounded functions is the domain under consideration. A function might be bounded on one subset of its domain but unbounded on another. For instance, the given problem discusses a function \( f \) that is bounded on the integers \( \mathbb{Z} \), but it does not necessarily imply that \( f \) will be bounded on the entire set of real numbers \( \mathbb{R} \).
Consider the function \( g(x) = \sin(x) \) for all real numbers \( x \). The sine function is bounded because its outputs are constrained within -1 and 1 no matter what real value \( x \) takes. This consistent limitation signifies that the function does not go to infinity as the inputs grow, making \( g(x) \) a bounded function.
An important thing to note about bounded functions is the domain under consideration. A function might be bounded on one subset of its domain but unbounded on another. For instance, the given problem discusses a function \( f \) that is bounded on the integers \( \mathbb{Z} \), but it does not necessarily imply that \( f \) will be bounded on the entire set of real numbers \( \mathbb{R} \).
Real Analysis
Real analysis is a branch of mathematics dealing with real numbers and real-valued functions. It involves rigorous study of sequences, series, limits, continuity, differentiation, and integration.
One core concept within real analysis is continuity. A function is continuous if small changes in the input result in small changes in the output, without any sudden jumps. For instance, \( f(x) = x^2 \) is continuous across all real numbers because adjusting \( x \) slightly will only cause a small change in \( f(x) \). This exploration allows mathematicians to understand behaviors of functions in a precise manner.
One core concept within real analysis is continuity. A function is continuous if small changes in the input result in small changes in the output, without any sudden jumps. For instance, \( f(x) = x^2 \) is continuous across all real numbers because adjusting \( x \) slightly will only cause a small change in \( f(x) \). This exploration allows mathematicians to understand behaviors of functions in a precise manner.
- Continuity on Different Sets: A function might behave very differently depending on the subset of the real numbers it's examined on. This relates to boundedness in the sense that continuity on a compact set implies boundedness, but \( \mathbb{R} \) itself is not compact.
- Importance of Real Analysis: It provides tools to tackle various mathematical problems by establishing limits and behaviors of functions, which aids in deeper comprehension of mathematical concepts.
Counterexample
A counterexample is a specific case for which a general statement, presumed true, fails. It is a crucial concept especially for proving or disproving propositions in mathematics.
Let’s examine the proposed counterexample in the exercise: the function \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) for \( x eq 0 \) and \( f(0) = 0 \) . This function is designed to demonstrate that being bounded on \( \mathbb{Z} \), the integers, does not imply being bounded on \( \mathbb{R} \), the real numbers.
Why does this serve as a counterexample?
Let’s examine the proposed counterexample in the exercise: the function \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) for \( x eq 0 \) and \( f(0) = 0 \) . This function is designed to demonstrate that being bounded on \( \mathbb{Z} \), the integers, does not imply being bounded on \( \mathbb{R} \), the real numbers.
Why does this serve as a counterexample?
- Bounded on \( \mathbb{Z} \): For integer values \( n \), \( f(n) = n^2 \sin\left(\frac{1}{n}\right) \) suggests \( |f(n)| \leq n^2 \), which means the function stays within bounds for each integer.
- Unbounded on \( \mathbb{R} \): As \( x \) gets very large, the function's behavior is dominated by the \( x^2 \) term, making it grow unbounded.