Problem 14 - Basic Analysis: Introduction to Real Analysis

Question

Suppose \(f:[-1,0] \rightarrow \mathbb{R}\) and \(g:[0,1] \rightarrow \mathbb{R}\) are continuous and \(f(0)=g(0) .\) Define \(h:[-1,1] \rightarrow \mathbb{R}\) by \(h(x):=f(x)\) if \(x \leq 0\) and \(h(x):=g(x)\) if \(x>0 .\) Show that h is continuous.

Step-by-Step Solution

Answer

Function \( h \) is continuous on the interval \([-1, 1] \).

Step 1: Understand Continuity at 0

Since we are given that \( f(0) = g(0) \) and both functions \( f \) and \( g \) are continuous on their respective domains, we need to show that \( h(x) \) is continuous at \( x = 0 \). Continuity at a point requires that the limit from the left equals the limit from the right and both equal the function value at that point.

Step 2: Check Left-Hand Limit at 0

For \( x \) approaching 0 from the left (\( x \to 0^- \)), the function \( h(x) = f(x) \). Since \( f \) is continuous on \([-1, 0] \) and \( f(0) = g(0) \), we have \( \lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} f(x) = f(0) \).

Step 3: Check Right-Hand Limit at 0

For \( x \) approaching 0 from the right (\( x \to 0^+ \)), the function \( h(x) = g(x) \). Since \( g \) is continuous on \([0, 1] \) and \( g(0) = f(0) \), we have \( \lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} g(x) = g(0) \).

Step 4: Verify Continuity at 0

We have \( \lim_{x \to 0^-} h(x) = f(0) \) and \( \lim_{x \to 0^+} h(x) = g(0) \). Given \( f(0) = g(0) \), both side limits equal \( h(0) \), confirming that \( h \) is continuous at 0.

Step 5: Verify Continuity on Intervals

Away from \( x = 0 \), the function \( h(x) \) is defined as \( f(x) \) on \([-1, 0) \) and \( g(x) \) on \((0, 1] \). Since \( f \) is continuous on \([-1, 0] \) and \( g \) is continuous on \([0, 1] \), \( h \) is continuous on each open interval.

Step 6: Conclusion on Continuity of h

Having shown that \( h \) is continuous at \( x = 0 \) and on the intervals \([-1, 0) \) and \((0, 1] \), we conclude that \( h \) is continuous on the entire domain \([-1, 1] \).

Key Concepts

Piecewise Functions

Piecewise functions are a fascinating concept in mathematics where a single function is defined by multiple sub-functions, each valid over different intervals of the domain. In essence, a piecewise function allows for different rules or expressions to prescribe the value of the function depending on the input's range.
For example, consider the function \( h(x) \) in our exercise: it combines two functions \( f(x) \) for \( x \leq 0 \) and \( g(x) \) for \( x > 0 \). This piecewise definition allows us to handle different conditions seamlessly within one continuous domain.

• Key advantage: Flexibility in defining complex systems.
• Common use: Modeling real-world scenarios with different phases.

Understanding piecewise functions is crucial as they appear in various fields, including engineering, physics, and economics. They help describe processes that have different behaviors in different situations, making them a powerful tool in analysis.

Limits

The concept of limits is foundational in calculus and real analysis, serving as the bedrock for defining continuity and derivatives. Limits help us understand the behavior of a function as the input approaches a particular value.
In our problem, the limit assesses whether \( h(x) \) approaches the same value from both sides of \( x = 0 \). Specifically, the limits as \( x \to 0^- \) and \( x \to 0^+ \) help ensure that a piecewise function like \( h(x) \) behaves continuously across its entire domain.

• Left-hand limit \( \lim_{x \to 0^-} h(x) \): Comes from the portion defined by \( f(x) \).
• Right-hand limit \( \lim_{x \to 0^+} h(x) \): Comes from the portion defined by \( g(x) \).

Computing these limits and ensuring they meet at the transition point (\( x = 0 \)) certifies the function does not "jump" or have breaks.

Function Continuity Proof

Continuity is a desirable property of functions, indicating that there are no abrupt changes or holes. To prove a function is continuous, we typically need to show that for every point in the domain, the limit of the function as we approach the point equals the function's value at that point.
For the piecewise function \( h(x) \) given in the exercise, proving continuity involves:

• Showing the left and right hand limits of \( h(x) \) at \( x = 0 \) are equal. Since both \( f(x) \) and \( g(x) \) meet at this point (\( f(0) = g(0) \)), \( h(0) = f(0) = g(0) \).
• Proving \( h(x) \) remains continuous within the sub-domains \([-1, 0]\) and \((0, 1] \), which are already covered by the continuity of \( f(x) \) and \( g(x) \).

This approach validates that there are no disruptions as \( x \) transitions from negative to positive, hence ensuring the entire function \( h(x) \) is seamless across its domain.