Problem 5 - Basic Analysis: Introduction to Real Analysis

Question

Let \(A, B\) be intervals. Let \(f: A \rightarrow \mathbb{R}\) and \(g: B \rightarrow \mathbb{R}\) be uniformly continuous functions such that \(f(x)=g(x)\) for \(x \in A \cap B\). Define the function \(h: A \cup B \rightarrow \mathbb{R}\) by \(h(x):=f(x)\) if \(x \in A\) and \(h(x):=g(x)\) if \(x \in B \backslash A\) a) Prove that if \(A \cap B \neq \emptyset,\) then \(h\) is uniformly continuous. b) Find an example where \(A \cap B=\emptyset\) and \(h\) is not even continuous.

Step-by-Step Solution

Answer

a) Proven uniformly continuous if intersection is non-empty. b) Example: \(A=(0,1), B=(2,3)\).

Step 1: Understand the Problem

We need to show under what conditions the function \(h\) is uniformly continuous and provide an example where it isn't even continuous when certain conditions do not hold.

Step 2: Given Information and Definitions

We have \(f: A \rightarrow \mathbb{R}\) and \(g: B \rightarrow \mathbb{R}\) as uniformly continuous functions, and we define \(h(x)\) as a piecewise function using \(f\) and \(g\) based on the domain.

Step 3: Prove uniform continuity when \(A \cap B \neq \emptyset\)

Since \(f\) and \(g\) are uniformly continuous, for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in A\), \(|f(x) - f(y)| < \epsilon\) if \(|x-y|<\delta\) and similarly for \(g\) on \(B\). Since \(f(x) = g(x)\) for \(x \in A \cap B\), we have a consistent value on the overlap ensuring \(h(x)\) is uniformly continuous.

Step 4: Example for \(A \cap B = \emptyset\)

Consider intervals \(A = (0, 1)\) and \(B = (2, 3)\). Define \(f(x) = x\) for \(x \in A\) and \(g(x) = x^2\) for \(x \in B\). Define \(h: A \cup B \rightarrow \mathbb{R}\). Since \(A\) and \(B\) do not intersect, \(h\) will not be continuous across the union.

Key Concepts

Piecewise Function

A piecewise function is a function that is defined by different expressions or rules depending on the input value. For the exercise, the function \( h \) is a piecewise function defined as:

• \( h(x) = f(x) \) when \( x \) is in interval \( A \)
• \( h(x) = g(x) \) when \( x \) is in interval \( B \setminus A \)

This means that \( h(x) \) applies the rule of \( f \) when inputs are within \( A \) and applies \( g \) elsewhere in \( B \).
Piecewise functions like \( h \) are commonplace in real-world problems where different formulas govern certain ranges or conditions.
Always consider how the different pieces interact, especially where domains overlap, to fully understand such functions.

Intervals in Real Analysis

Intervals are crucial in understanding real-world constraints and behaviors in mathematics, especially in real analysis. In the provided problem, two intervals, \( A \) and \( B \), define the domains of the functions \( f \) and \( g \) respectively.
Intervals can be of several types:

• Open Intervals (e.g., \((a, b)\)): Does not include endpoint values.
• Closed Intervals (e.g., \([a, b]\)): Includes both endpoint values.
• Half-Open or Half-Closed Intervals (e.g., \((a, b]\) or \([a, b)\)): Includes one endpoint value but not the other.

The intersection \( A \cap B \) is the set of all points that the two intervals share. If they overlap, \( h \) may maintain uniform continuity as \( f \) and \( g \) align over this range.

Continuity Conditions

Continuity conditions often ensure smooth behaviors of functions over different domains. A function is continuous at a point if, intuitively, there is no "jump" or sudden change in value as the input approaches that point.
For a piecewise function like \( h \), continuity hinges on these key conditions:

• At points within a single defined interval, \( f \) and \( g \) must be continuous, as they are specified in the problem.
• At the intersection of \( A \) and \( B \), \( f(x) = g(x) \) ensures that \( h \) remains continuous across the overlap.

When \( A \) and \( B \) do not intersect, such as in the provided example with \( A = (0, 1) \) and \( B = (2, 3) \), there is no region of overlap, causing possible discontinuities.

Uniformly Continuous Functions

Uniform continuity is a stronger form of continuity that guarantees consistent behavior across the entire domain. A function \( f \) is uniformly continuous on a domain if, for every \( \epsilon > 0 \), there exists a single \( \delta > 0 \) such that for all \( x, y \) in that domain, if \(|x-y|<\delta\), then \(|f(x) - f(y)| < \epsilon\).
This definition ensures that small changes in input lead to small changes in output, and the specific \( \delta \) does not depend on the point in the domain, contrasting with ordinary pointwise continuity.
In the problem, both \( f \) and \( g \) are uniformly continuous on their respective domains. This property is critical in ensuring that the unified focus function \( h \) remains uniformly continuous when \( A \cap B eq \emptyset \). The intersection ensures a seamless transition, maintaining the uniform continuity property. Without overlap, as in the \( A \cap B = \emptyset \) case in the exercise, \( h \) may lose this property entirely, resulting in potential discontinuities.