Question

Prove that if the functions \(f_{n}\) and \(g_{n}\) are real or complex (or if the \(g_{n}\) are vector valued and the \(f_{n}\) are scalar valued), and if $$ f_{n} \rightarrow f \text { and } g_{n} \rightarrow g \text { (uniformly) on } B \text { , } $$ then $$ f_{n} g_{n} \rightarrow f g \text { (uniformly) on } B $$ provided that either \(f\) and \(g\) or the \(f_{n}\) and \(g_{n}\) are bounded on \(B\) (at least from some \(n\) onward \() ;\) cf. Problem 11 . Disprove it for the case where only one of \(f\) and \(g\) is bounded.

Short answer

The uniform convergence of products \(f_n g_n \rightarrow fg\) holds if either \(f_n, g_n\) or \(f, g\) are bounded. It may fail if only one is bounded.

Step-by-step solution

Understanding Uniform Convergence

Uniform convergence of the functions \(f_n\) to \(f\) means that for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n > N\), \(|f_n(x) - f(x)| < \epsilon\) for all \(x \in B\). Similarly, \(g_n\) converges to \(g\) uniformly if for every \(\epsilon > 0\), there exists an \(M\) such that for all \(n > M\), \(|g_n(x) - g(x)| < \epsilon\) for all \(x \in B\). This definition implies that the convergence does not depend on the specific choice of \(x\) but only on \(n\).

Establish Boundedness Conditions

If the sequences \(f_n\) and \(g_n\) or the limits \(f\) and \(g\) are bounded on \(B\), say by \(M\), then \(|f_n(x)| \leq M\) and \(|g_n(x)| \leq M\) for all \(n\) and all \(x \in B\). This ensures that the product \(f_n(x)g_n(x)\) remains bounded as well.

Using Uniform Convergence to Prove Product Convergence

By the uniform convergence of \(f_n\) and \(g_n\), let \(\epsilon > 0\). Then there exists an \(N\) such that for all \(n > N\), \(|f_n(x) - f(x)| < \frac{\epsilon}{2M}\) and \(|g_n(x) - g(x)| < \frac{\epsilon}{2M}\) for all \(x \in B\). Thus, the product can be analyzed as: \[|f_n(x)g_n(x) - f(x)g(x)| \leq |f_n(x)(g_n(x) - g(x))| + |g(x)(f_n(x) - f(x))|.\] Use boundedness, \(|f_n(x)| \leq M\) and \(|g(x)| \leq M\), to write: \[\leq M \cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M} = \epsilon.\]

Conclusion of Uniform Convergence for Product

Thus, by the triangle inequality and boundedness, we have \(|f_n(x)g_n(x) - f(x)g(x)| < \epsilon\) for all \(x \in B\) and for all \(n > N\), proving uniform convergence of the product \(f_n g_n \rightarrow fg\).

Disproving the Statement when only one function is bounded

Consider \(f_n = g_n = 1/n\) on the interval \([1, \infty)\). As \(n \rightarrow \infty\), both \(f_n\) and \(g_n\) go to 0 uniformly, but the product \(f_n g_n = 1/n^2\) does not converge uniformly to 0 on any unbounded \(B\). This shows that if only one of the functions, say \(f\), is bounded, the conclusion does not hold if \(g\) or the \(g_n\) are unbounded.

Key concepts

Bounded Functions

Bounded functions are an important concept in mathematical analysis. When we say a function is bounded on a set, it means there is a real number that limits the function values above or below. More formally, if a function \(f\) is bounded on a set \(B\), there exists a number \(M > 0\) such that for every \(x \in B\), the absolute value of \(f(x)\) is less than or equal to \(M\). This can be expressed as:

• \(|f(x)| \leq M\) for all \(x \in B\).

Boundedness is a helpful property because it restricts the behavior of the function, making analysis more manageable. It is especially useful when dealing with the convergence of sequences or series, as seen in steps of the original solution. In the context provided, either the sequences \((f_n)\) and \((g_n)\) or their limits \(f\) and \(g\) must be bounded to ensure uniform convergence of their product, \((f_n g_n) \to fg\). Without boundedness, the convergence properties could fail.

Uniform Limit Theorem

The Uniform Limit Theorem is a significant result in real analysis. It states that if a sequence of functions \(f_n\) converges uniformly to a function \(f\), and each \(f_n\) is continuous on a set \(B\), then \(f\) is also continuous on \(B\). Uniform convergence is stronger than pointwise convergence, as it implies convergence does not depend on the particular point \(x\), but instead the rate of convergence is uniform across all points in the domain. Here’s what uniform convergence entails:

• For every \(\epsilon > 0\), there exists an index \(N\) where for all \(n > N\), \(|f_n(x) - f(x)| < \epsilon\) for every \(x \in B\).

This definition is crucial for ensuring that certain operations on functions, like taking the product \(f_n g_n\), yield predictable results. It guarantees that if initial sequences \(f_n\) and \(g_n\) uniformly converge to \(f\) and \(g\), respectively, then their limit behaves well when considering their product, under the condition they are bounded.

Product of Functions

When dealing with the product of functions, especially in terms of convergence, understanding the behavior of each function factor is vital. The claim based on uniform convergence and boundedness is that if \((f_n)\) converges uniformly to \(f\), and \((g_n)\) converges uniformly to \(g\), then \(f_n g_n\) converges to \(fg\) uniformly on the set \(B\), provided these functions or their limits are bounded.To see why this is important, consider the inequality used in the solution:\[|f_n(x)g_n(x) - f(x)g(x)| \leq |f_n(x)(g_n(x) - g(x))| + |g(x)(f_n(x) - f(x))|.\]With uniform convergence, for large enough \(n\), both \(|f_n(x) - f(x)|\) and \(|g_n(x) - g(x)|\) can be made small enough, improving control over their products. Boundedness, such as \(|f_n(x)| \leq M\) and \(|g(x)| \leq M\), helps ensure these product differences remain small, thereby enforcing the uniform convergence of the product \(f_n g_n\) to \(fg\).Without boundedness, as illustrated in the solution with the counterexample of only one function being bounded, uniform convergence may fail if one sequence or function can grow without limit.