Problem 13 - Basic Analysis: Introduction to Real Analysis

Question

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.

Step-by-Step Solution

Answer

False, a counterexample shows the statement is incorrect.

Step 1: 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.

Step 2: 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 \( \mathbb{R} \).

Step 3: 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 \( \mathbb{R} \).

Step 4: 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 \( x^2 \).

Step 5: 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.

Key Concepts

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 \( \mathbb{R} \).

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.

• 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.

Thus, real analysis helps distinguish which properties, like boundedness, can or cannot be extended from certain sets (like \( \mathbb{Z} \)) to others (like \( \mathbb{R} \)).

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?

• 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.

The counterexample solidifies the understanding that boundedness on a discrete set like the integers doesn’t imply the same on the continuous set of real numbers, thus proving the original statement false.