Question

(a) Prove: If \(\left\\{F_{n}\right\\}\) and \(\left\\{G_{n}\right\\}\) converge uniformly to bounded functions \(F\) and \(G\) on \(S\), then \(\left\\{F_{n} G_{n}\right\\}\) converges uniformly to \(F G\) on \(S\). (b) Give an example showing that the conclusion of (a) may fail to hold if \(F\) or \(G\) is unbounded on \(S\).

Short answer

Short Answer: In order to prove that if sequences of functions \(\left\{F_n\right\}\) and \(\left\{G_n\right\}\) converge uniformly to bounded functions F and G on a set S, then the sequence \(\left\{F_n G_n\right\}\) converges uniformly to the product function FG on S, we use the definition of uniform convergence and properties of bounded functions. We estimate the difference between \((F_n G_n)(x)\) and \((FG)(x)\) and apply the bound on F, G, F_n, and G_n to conclude the proof. For example, when F and G are unbounded, consider the functions \(F_n(x) = x^n\) and \(G_n(x) = x^{-n}\) on the interval \(S = (0,1)\). The uniform convergence of \(\left\{F_n\right\}\) and \(\left\{G_n\right\}\) to F and G respectively does not imply the uniform convergence of \(\left\{F_n G_n\right\}\) to FG on S.

Step-by-step solution

Part (a): Prove the Results

To prove the statement, we first recall the definition of uniform convergence. A sequence of functions \(\left\{H_n\right\}\) converges uniformly to a function H on S if, for any \(\epsilon>0\), there exists an integer N (possibly depending on \(\epsilon\)) such that \(|H_n(x) - H(x)| < \epsilon\) for all \(x \in S\) and \(n>N\). Given that \(\left\{F_n\right\}\) and \(\left\{G_n\right\}\) converge uniformly to F and G respectively on S and both F and G are bounded, let's prove the result as follows: 1. Since F and G are bounded, there exists some constant M such that \(|F(x)| \le M\) and \(|G(x)| \le M\) for all \(x \in S\). 2. Choose \(\epsilon'>0\) arbitrarily. Since \(\left\{F_n\right\}\) converges uniformly to F on S, there exists an integer N1 such that \(|F_n(x) - F(x)| < \epsilon'\) for all \(n > N_1\) and \(x \in S\). 3. Similarly, since \(\left\{G_n\right\}\) converges uniformly to G on S, there exists an integer N2 such that \(|G_n(x) - G(x)| < \epsilon'\) for all \(n > N_2\) and \(x \in S\). 4. Without loss of generality, let N be the maximum of N1 and N2. Then we have \(|F_n(x) - F(x)| < \epsilon'\) and \(|G_n(x) - G(x)| < \epsilon'\) for all \(n > N\) and \(x \in S\).

Part (a): Estimate the Difference

Now, let's estimate the difference between \((F_n G_n)(x)\) and \((FG)(x)\) for any \(x \in S\) and \(n>N\): \begin{align*} |(F_n G_n)(x) - (FG)(x)| &= |F_n(x) G_n(x) - F(x) G(x)|\\ &= |F_n(x) G_n(x) - F_n(x) G(x) + F_n(x) G(x) - F(x) G(x)| \\ &\le |F_n(x)(G_n(x) - G(x))| + |G(x)(F_n(x) - F(x))|. \end{align*}

Part (a): Concluding the Proof

Now applying the bound on F, G, F_n, and G_n: \begin{align*} |(F_n G_n)(x) - (FG)(x)| &\le |F_n(x)|\cdot |(G_n(x) - G(x))| + |G(x)| \cdot |(F_n(x) - F(x))| \\ &\le M\cdot \epsilon' + M\cdot \epsilon' \\ &= 2M\epsilon'. \end{align*} Since our choice of \(\epsilon'\) was arbitrary, the proof is complete.

Part (b): Example of Failure

Let's consider the following example: \(F_n(x) = x^n\) and \(G_n(x) = x^{-n}\) on the interval \(S = (0,1)\). Observe that \(F_n\) converges pointwise to \(F(x) = 0\) for \(x \in (0,1)\) and \(F(1) = 1\). Similarly, \(G_n\) converges pointwise to \(G(x) = \infty\) for \(x \in (0,1)\) and \(G(1) = 1\). Since \(G\) is unbounded on \(S\), the conclusion of part (a) may not necessarily hold. Now, consider the sequence of functions \(\left\{F_n G_n\right\}\). We have \((F_n G_n)(x) = x^n x^{-n} = 1\) for all \(x \in S\) and all \(n\in\mathbb{N}\). This implies that \(\left\{F_n G_n\right\}\) converges pointwise (and uniformly) to the constant function \(FG(x)=1\) on S. Hence, this example shows that the conclusion of part (a) may fail to hold if F or G is unbounded on S, as the uniform convergence of \(\left\{F_n\right\}\) and \(\left\{G_n\right\}\) to F and G respectively does not imply the uniform convergence of \(\left\{F_n G_n\right\}\) to FG on S.

Key concepts

Real Analysis

Real analysis is a branch of mathematics dealing with real numbers and real-valued functions. A fundamental aspect of real analysis is the study of sequences and series of functions. The concept of convergence, especially uniform convergence, plays a crucial role. Uniform convergence occurs when a sequence of functions approaches a limiting function uniformly, which means that for every small positive number \(\epsilon\), there's a stage in the sequence beyond which all function values lie within \(\epsilon\) of the limiting function for all points in a given set. Understanding uniform convergence helps in ensuring that various operations, like integration and differentiation, can be performed on sequences of functions while maintaining the expected properties and outcomes.

When \(\left\{F_{n}\right\}\) and \(\left\{G_{n}\right\}\) uniformly converge to bounded functions \(F\) and \(G\) on a set \(S\), real analysis provides us the tools to explore under what conditions their products also converge uniformly. This ties directly into the very nature of studying the behavior of real-valued functions and is a testament to the depth and intricacy of real analysis.

Bounded Functions

In real analysis, a function is said to be bounded if there is a real number \(M\) such that the absolute value of the function does not exceed \(M\) for all inputs in its domain. This concept is essential when dealing with convergence of functions. If a function is bounded, it implies that the function's values are contained within a specific range, making it more manageable to analyze and work with, especially concerning convergence properties.

Let's see how the boundedness property is used within the context of uniform convergence. Say we have sequences of functions \(\left\{F_{n}\right\}\) and \(\left\{G_{n}\right\}\), and they converge uniformly to bounded functions \(F\) and \(G\). Their bounded nature allows us to establish a uniform bound that can be used when estimating how close the functions \(F_{n}G_{n}\) are to the product \(FG\), just as shown in the textbook solution. This is an integral operation, as the preservation of boundedness significantly impacts the behavior of the sequence and its limit.

Sequence of Functions

A sequence of functions, denoted as \(\left\{F_n\right\}_{n=1}^{\infty}\), is a list of functions \(F_1, F_2, F_3, ...\) where each function is associated with a natural number. Sequences of functions can converge to a limiting function in various ways, with uniform convergence being one particular mode where convergence occurs in the same manner across the entire domain. This concept is key in real analysis for understanding the behavior of complex functions as they can be approximated by simpler ones.

When analyzing sequences of functions, as the exercise demonstrates, mathematicians want to preserve convergence behavior through operations such as multiplication or addition. However, as shown in the textbook solutions, if a function within the sequence is unbounded, this can disrupt the convergence behavior. The given example where one of the functions becomes unbounded on the set \(S\) illustrates a scenario where even though each sequence individually might have a nice convergent behavior, their combination does not follow the uniform convergence we might expect. Hence, demonstrating the nuanced understanding required when working with sequences of functions in real analysis.