Problem 7 - Basic Analysis: Introduction to Real Analysis

Question

Let \(f:(0,1) \rightarrow \mathbb{R}\) be a bounded continuous function. Show that the function \(g(x):=\) \(x(1-x) f(x)\) is uniformly continuous.

Step-by-Step Solution

Answer

The function \(g(x) = x(1-x)f(x)\) is uniformly continuous on \((0,1)\).

Step 1: Define Uniform Continuity

A function is uniformly continuous on a set if, for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y\) in the set, if \(|x-y| < \delta\) then \(|g(x) - g(y)| < \epsilon\). This definition will be our target to prove for the function \(g(x) = x(1-x)f(x)\).

Step 2: Analyze the Domain and Boundedness

Given that \(f\) is a bounded function, there exists a constant \(M > 0\) such that \(|f(x)| \leq M\) for all \(x \in (0,1)\). The function \(x(1-x)\) is continuous over \((0,1)\) and attains its maximum value at \(x = \frac{1}{2}\), which is \(\frac{1}{4}\). Thus, \(|x(1-x)| \leq \frac{1}{4}\).

Step 3: Express the Difference \(|g(x) - g(y)|\)

Consider two points \(x, y \in (0,1)\). The difference between the values of the function is:\[|g(x) - g(y)| = |x(1-x)f(x) - y(1-y)f(y)|.\]This can be bounded using the triangle inequality:\[|x(1-x)f(x) - y(1-y)f(y)| \leq |x(1-x)||f(x) - f(y)| + |f(y)||(x(1-x) - y(1-y))|\]

Step 4: Estimate Each Part of the Bound

As established, \(|x(1-x)| \leq \frac{1}{4}\) and \(|f(x) - f(y)|\) can be made arbitrarily small since \(f\) is continuous (hence uniformly continuous on the compact interval \([\delta,1-\delta]\) for any \(\delta > 0\)). Hence, for any \(\epsilon > 0\), there exists a \(\delta_1 > 0\) such that \(|f(x) - f(y)| < \frac{\epsilon}{2M}\) for \(|x-y| < \delta_1\).Similarly, by choosing \(\delta_2 > 0\) sufficiently small, \(|x(1-x) - y(1-y)| < \frac{\epsilon}{2M}\) if \(|x-y| < \delta_2\).

Step 5: Combine the Estimates to Show Uniform Continuity

Now, choose \(\delta = \min(\delta_1, \delta_2)\). Then for \(|x-y| < \delta\), both parts of the sum in Step 3 are less than \(\frac{\epsilon}{2}\). Therefore:\[|g(x) - g(y)| \leq \frac{1}{4}\frac{\epsilon}{2M} + M \frac{\epsilon}{2M} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.\]This proves that \(g(x)\) is uniformly continuous on \((0,1)\).

Step 6: Conclusion: Final Verification

By the construction above, for any \(\epsilon > 0\), we have found a \(\delta\) which satisfies the uniform continuity condition. Thus, the function \(g(x) = x(1-x)f(x)\) is uniformly continuous on \((0,1)\).

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that focuses on the study of rates at which quantities change. It is concerned with the concept of the derivative, which provides a way to measure how a function's output changes as its input changes. For the function \( g(x) = x(1-x)f(x) \), understanding its derivative can give insights into how it behaves as \( x \) varies within the interval \((0, 1)\). In differential calculus, derivatives are crucial for solving problems related to slope, rate of change, and optimization.

When dealing with functions like \( g(x) \), where one function is nested within another, we often use techniques such as the product rule to find the derivative. This involves differentiating each part of the function and applying relevant rules. While differential calculus might not directly prove uniform continuity, it helps us understand the behavior of the function around small changes in \( x \). This can indirectly support uniform continuity by ensuring that there aren't abrupt changes in the function's output.

Continuity

Continuity is a foundational concept in calculus and real analysis, describing a function that does not have any sudden jumps or breaks. A function \( f(x) \) is said to be continuous on a domain if, intuitively, you can draw its graph without lifting your pencil from the paper. For the function \( g(x) = x(1-x)f(x) \), we start with \( f(x) \) being continuous and bounded on \((0,1)\).

These properties ensure that there are no unexpected breaks in \( g(x) \) as \( x \) changes slightly. The composition and multiplication of continuous functions (\( x(1-x) \) and \( f(x) \) in this context) result is a function that's also continuous. Such properties are essential when moving towards uniform continuity, as they prevent large variations in \( g(x) \) and ensure stability over the interval.

Real Analysis

Real analysis is a significant area of mathematics focused on the study of real numbers and real-valued functions. It provides the rigorous foundation needed to comprehend and prove concepts like continuity and uniform continuity. In our context, the task was to prove that \( g(x) = x(1-x)f(x) \) is uniformly continuous over \((0, 1)\).

Uniform continuity is stronger than mere continuity. It requires that the function remains 'well-behaved' across its entire domain, even as the inputs get arbitrarily close. Real analysis provides the tools and definitions, such as the ε-δ (epsilon-delta) definition of uniform continuity, to address such problems with precision.

• The uniform continuity of \( g(x) \) involves ensuring that for every \( \epsilon > 0 \), we can find a \( \delta > 0 \) such that \(|g(x) - g(y)| < \epsilon\) whenever \(|x-y| < \delta\).
• The analysis leverages the boundedness and continuity of \( f(x) \) and the continuity of \( x(1-x) \) as a strategy to show this property holds throughout the interval.
• Beyond the intuition provided by continuity, real analysis formalizes these ideas and creates a framework where rigorous proofs can be constructed, ensuring the solutions are not only correct but also comprehensive and satisfying.