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Chapter 9: Problem 9

Prove that if \(\left\\{f_{n}\right\\}\) and \(\left\\{g_{n}\right\\}\) both converge uniformly on a set \(D\), then so too does the sequence \(\left\\{f_{n}+g_{n}\right\\}\).

Short Answer

Expert verified
The sequence \( \{f_n + g_n\} \) converges uniformly to \( f + g \) on \( D \).

Step by step solution

01

Understand Uniform Convergence

First, recall the definition of uniform convergence. A sequence of functions \( \{f_n\} \) converges uniformly to a function \( f \) on a set \( D \) if for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n > N \) and all \( x \in D \), \( |f_n(x) - f(x)| < \epsilon \). This means that after some term in the sequence, all subsequent functions stay within \( \epsilon \) of the limit function for every point in \( D \).
02

Uniform Convergence of \( \{f_n\} \) and \( \{g_n\} \)

Since \( \{f_n\} \) converges uniformly to \( f \) on \( D \), for every \( \epsilon > 0 \), there exists \( N_1 \) such that for all \( n > N_1 \), \( |f_n(x) - f(x)| < \epsilon/2 \) for all \( x \in D \). Similarly, since \( \{g_n\} \) converges uniformly to \( g \) on \( D \), there exists \( N_2 \) such that for all \( n > N_2 \), \( |g_n(x) - g(x)| < \epsilon/2 \) for all \( x \in D \).
03

Establish Uniform Convergence of \( \{f_n + g_n\} \)

Consider the sequence \( \{f_n + g_n\} \). We need to show that \( f_n + g_n \) converges uniformly to \( f + g \) on \( D \). For any \( \epsilon > 0 \), choose \( N_3 = \max(N_1, N_2) \). Then, for all \( n > N_3 \) and all \( x \in D \), we have: \[ |f_n(x) + g_n(x) - (f(x) + g(x))| = |(f_n(x) - f(x)) + (g_n(x) - g(x))| \]Using the triangle inequality, this becomes:\[ \leq |f_n(x) - f(x)| + |g_n(x) - g(x)| < \epsilon/2 + \epsilon/2 = \epsilon \]
04

Conclusion on Uniform Convergence

Since for every \( \epsilon > 0 \), there exists an \( N_3 \) such that for all \( n > N_3 \) and all \( x \in D \), \( |f_n(x) + g_n(x) - (f(x) + g(x))| < \epsilon \), we conclude that \( \{f_n + g_n\} \) converges uniformly to \( f + g \) on \( D \). This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sequence of Functions
A sequence of functions,
  • is simply a set of functions aligned in a specific order.
  • Think of each function as a step in the sequence that approaches a final outcome or limit function.
In our context, we have sequences \( \{f_n\} \) and \( \{g_n\} \), each of which consists of an infinite number of functions ordered from \( f_1, f_2, f_3, \ldots \) and \( g_1, g_2, g_3, \ldots \).
Each function in these sequences possesses unique characteristics and collectively form the movement from one function towards a certain end value or behavior (called the limit function).
The concept of a sequence of functions is crucial to understand, as it sets the foundation for analyzing how these functions behave as they "move" closer and closer to a specified function (the limit) as the sequence progresses.
This convergence in behavior is what we explore further with uniform convergence, a particular way these sequences approach their limit functions together as seen in uniform convergence analysis.
Triangle Inequality
The triangle inequality is a fundamental concept within the realm of mathematical analysis,
  • it's a property of distance in space.
  • Mathematically, it states that the sum of the lengths of any two sides of a triangle is greater than or equal to the length of the remaining side.
In terms of absolute values, this property is described using the equation \[ |a + b| \leq |a| + |b| \]In the context of proving uniform convergence,
  • the triangle inequality helps us handle the bounds on the sum \((f_n + g_n) - (f + g)\).
  • By applying the inequality, we can split the journey from \(f_n + g_n\) to \(f + g\) into the manageable parts: \(f_n - f\) and \(g_n - g\).
This allows us to pair each variation to its separate limit function, providing a means to prove that the combined difference is still constrained within the
  • combined acceptable limit \(\epsilon\).
In practice, using the triangle inequality simplifies the convergence proof to make sure each sequence approaches its respective limit function sufficiently close.
Limit Function
The concept of a limit function is integral in the study of sequences and series in calculus and analysis.
  • It serves as the ultimate target or endpoint of a sequence of functions.
  • As each function in a sequence modifies slightly to become more like a specific end function, that endpoint is called the limit function.
For context, if \( \{f_n\} \) is a sequence of functions and it converges, it approaches a limit function \( f \).
Uniform convergence ensures that as a whole, the sequence \( f_n(x) \) converges to \( f(x) \) uniformly all over some set \( D \).
It's not just each individual point \( x \) in \( D \) approaching \( f(x) \),
  • every single point approaches in a manner that eventually doesn't stray more than \( \epsilon \) away.
In combining sequences \( f_n \) and \( g_n \),
  • the limit functions form a new limit \( f + g \),
  • providing a specific instance of a combined sequence tending toward a target end function uniformly.
The harmony of the sequences' approach is what permits this uniform targeting, accentuating the importance of each sequence achieving a shared limiting behavior.

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Most popular questions from this chapter

Show that the series $$ \sum_{k=0}^{\infty} \frac{x^{k}}{k !} $$ converges uniformly on \([-a, a]\) for every \(a \in \mathbb{R}\) but does not converge uniformly on all of the real line. (Does it converge pointwise on the real line?) Obtain a series representation for $$ \int_{-a}^{a} \sum_{k=0}^{\infty} \frac{x^{k}}{k !} d x $$

Show that if \(f_{n} \rightarrow f\) uniformly on \([a, b]\) and each \(f_{n}\) is continuous then the sequence of functions $$ F_{n}(x)=\int_{a}^{x} f_{n}(t) d t $$ also converges uniformly on \([a, b]\).

Suppose that \(\left\\{f_{n}\right\\}\) is a sequence of continuous functions on an interval \([a, b]\) converging uniformly to a function \(f\) on the open interval \((a, b)\). If \(f\) is also continuous on \([a, b]\), show that the convergence is uniform on \([a, b]\).

Prove that if \(f_{n} \rightarrow f\) pointwise on a finite set \(D\), then the convergence is uniform.

Give an example of a sequence of continuous functions \(\left\\{f_{n}\right\\}\) on the interval \([0, \infty)\) that is monotonic decreasing and converges pointwise to a continuous function \(f\) on \([0, \infty)\) but for which the convergence is not uniform. Why does this not contradict Theorem \(9.24 ?\)
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