Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Prove the sequential criterion for uniform continuity: A function \(f: A \rightarrow T\) is uniformly continuous on a set \(B \subseteq A\) iff for any two (not necessarily convergent) sequences \(\left\\{x_{m}\right\\}\) and \(\left\\{y_{m}\right\\}\) in \(B,\) with \(\rho\left(x_{m}, y_{m}\right) \rightarrow 0,\) we have \(\rho^{\prime}\left(f\left(x_{m}\right), f\left(y_{m}\right)\right) \rightarrow 0\) (i.e., \(f\) preserves concurrent pairs of sequences; see Problem 4 in Chapter \(3, \S 17 ).\)

Short Answer

Expert verified
A function \( f: A \to T \) is uniformly continuous on \( B \subseteq A \) if and only if for sequences \( \{x_m\} \) and \( \{y_m\} \) in \( B \) with \( \rho(x_m, y_m) \to 0 \), we have \( \rho'(f(x_m), f(y_m)) \to 0 \).

Step by step solution

01

Understand Uniform Continuity

A function \( f: A \to T \) is uniformly continuous on \(B \subseteq A\) if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in B \) with \( \rho(x, y) < \delta \), we have \( \rho'(f(x), f(y)) < \epsilon \).
02

Sequence Convergence Condition

We are given two sequences \( \{x_m\} \) and \( \{y_m\} \) in \( B \) such that \( \rho(x_m, y_m) \to 0 \). This means that for every \( \epsilon > 0 \), there exists an \( M \) such that for all \( m \geq M \), \( \rho(x_m, y_m) < \epsilon \).
03

Apply Uniform Continuity to Sequences

Since \( f \) is uniformly continuous on \( B \), there exists a \( \delta > 0 \) corresponding to any \( \epsilon' > 0 \) such that if \( \rho(x, y) < \delta \), then \( \rho'(f(x), f(y)) < \epsilon' \). For our sequences, choose \( \epsilon' > 0 \) and find the corresponding \( \delta > 0 \).
04

Prove the Sequence Limit Condition

For the sequences \( \{x_m\} \) and \( \{y_m\} \), since \( \rho(x_m, y_m) \to 0 \), there exists an \( M \) such that for all \( m \geq M \), \( \rho(x_m, y_m) < \delta \). By uniform continuity, this means \( \rho'(f(x_m), f(y_m)) < \epsilon' \) for all \( m \geq M \). Thus, \( \rho'(f(x_m), f(y_m)) \to 0 \).
05

Establish Equivalence of Conditions

The converse is also true: if \( \rho(x_m, y_m) \rightarrow 0 \) implies \( \rho^{\prime}(f(x_m), f(y_m)) \rightarrow 0 \) for any sequences \( \{x_m\} \) and \( \{y_m\} \), this ensures that for any given \( \epsilon > 0 \), choosing sequences \( \{x_m = x\} \) and \( \{y_m = y\} \) achieves \( \rho(x, y) < \delta \). This shows uniform continuity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Sequential Criterion
The Sequential Criterion is a fascinating and useful mathematical concept when dealing with functions and their continuity properties. It provides a sequence-based method to determine uniform continuity. Instead of directly dealing with open intervals or neighborhoods, the Sequential Criterion allows us to express uniform continuity using sequences.

To apply it, consider sequences \( \{x_m\} \) and \( \{y_m\} \) coming from a set \( B \subseteq A \). When these sequences get closer—meaning the distance between them \( \rho(x_m, y_m) \) approaches zero—the corresponding outputs, according to the function \( f \), also ought to get closer. This is quantified by the condition \( \rho'(f(x_m), f(y_m)) \rightarrow 0 \).

If this criterion holds true for all possible pairs of sequences in \( B \), then \( f \) is uniformly continuous on set \( B \). Thus, this Sequential Criterion wraps the essence of uniform continuity into a sequence-oriented package.
Sequences in Analysis
In mathematical analysis, sequences are fundamental constructs used to explore the limits and behavior of functions. They are ordered lists of elements, typically numbers, and can offer critical insight into the nature of functions when individual elements of the sequence approximate arbitrary closeness.

Similar to the progression of numbers, sequences can exhibit various behaviors such as convergence, divergence, or oscillation. A common application in analysis is observing how sequences behave under functions. By examining the function's output sequences, one can infer continuity properties.

Engineers and mathematicians often use sequences to handle abstract problems or create simulations of real-world phenomena. Sequences thus serve as invaluable tools for both theoretical and applied mathematics, particularly when examining properties like uniform continuity.
Convergence of Sequences
A crucial concept in calculus and analysis is sequence convergence, which deals with the tendency of sequences to approach some limit. A sequence \( \{a_m\} \) is said to converge to a limit \( L \) if, for every positive number \( \epsilon \), however small, there exists a point in the sequence beyond which all terms are within \( \epsilon \) of \( L \).

Converging sequences are pivotal when discussing the continuity of functions, including uniform continuity. Uniform continuity can be considered a condition that is well-expressed through sequence behavior.

Use these properties to show a function \( f \) is uniformly continuous: if any two sequences \( \{x_m\} \) and \( \{y_m\} \) in a domain \( B \) converge to the same point, then the images under \( f \) converge to the same value—showing that a function preserves the convergence attribute of sequences.
  • This can help ensure reliable function behavior over sets and is valuable in mathematical proofs.
  • Ultimately, the convergence of sequences is a bridge to understanding deeper properties of functions, like uniform continuity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\bar{p}_{0}, \bar{p}_{1}, \ldots, \bar{p}_{m}\) be fixed points in \(E^{n}\left({ }^{*}\right.\) or in another normed space \()\). Let $$f(t)=\bar{p}_{k}+(t-k)\left(\bar{p}_{k+1}-\bar{p}_{k}\right)$$ whenever \(k \leq t \leq k+1, t \in E^{1}, k=0,1, \ldots, m-1\) Show that this defines a uniformly continuous mapping \(f\) of the interval \([0, m] \subseteq E^{1}\) onto the "polygon" $$\bigcup_{k=0}^{m-1} L\left[p_{k}, p_{k+1}\right]$$ In what case is \(f\) one to one? Is \(f^{-1}\) uniformly continuous on each \(L\left[p_{k}, p_{k+1}\right] ?\) On the entire polygon?

Give explicit definitions for the following "unsigned infinity" limit statements: (a) \(\lim _{x \rightarrow p} f(x)=\infty\) (b) \(\lim _{x \rightarrow p^{+}} f(x)=\infty\); (c) \(\lim _{x \rightarrow \infty} f(x)=\infty\)

Prove that the convergence or divergence (pointwise or uniformly) of a sequence \(\left\\{f_{m}\right\\}\), or a series \(\sum f_{m}\), of functions is not affected by deleting or adding a finite number of terms. Prove also that \(\lim _{m \rightarrow \infty} f_{m}\) (if any) remains the same, but \(\sum_{m=1}^{\infty} f_{m}\) is altered by the difference between the added and deleted terms.

Let \(f_{n} \rightarrow f\) (uniformly) on \(B\). Prove the equivalence of the following statements: (i) Each \(f_{n}\), from a certain \(n\) onward, is bounded on \(B\). (ii) \(f\) is bounded on \(B\). (iii) The \(f_{n}\) are ultimately uniformly bounded on \(B ;\) that is, all function values \(f_{n}(x), x \in B,\) from a certain \(n=n_{0}\) onward, are in one and the same globe \(G_{q}(K)\) in the range space. For real, complex, and vector-valued functions, this means that $$ \left(\exists K \in E^{1}\right)\left(\forall n \geq n_{0}\right)(\forall x \in B) \quad\left|f_{n}(x)\right|

Prove, independently, the principle of nested intervals in \(E^{n},\) i.e., Theorem 5 with $$ F_{m}=\left[\bar{a}_{m}, \bar{b}_{m}\right] \subseteq E^{n}; $$ where $$ \bar{a}_{m}=\left(a_{m 1}, \ldots, a_{m n}\right) \text { and } \bar{b}_{m}=\left(b_{m 1}, \ldots, b_{m n}\right) $$ [Hint: As \(F_{m+1} \subseteq F_{m}, \bar{a}_{m+1}\) and \(\bar{b}_{m+1}\) are in \(F_{m} ;\) hence by properties of closed intervals, \(a_{m k} \leq a_{m+1, k} \leq b_{m+1, k} \leq b_{m k}, \quad k=1,2, \ldots, n\) Fixing \(k\), let \(A_{k}\) be the set of all \(a_{m k}, m=1,2, \ldots\) Show that \(A_{k}\) is bounded above by each \(b_{m k},\) so let \(p_{k}=\sup A_{k}\) in \(E^{1}\). Then $$ (\forall m) \quad a_{m k} \leq p_{k} \leq b_{m k} . \text { (Why?) } $$ Unfixing \(k\), obtain such inequalities for \(k=1,2, \ldots, n\). Let \(\bar{p}=\left(p_{1}, \ldots, p_{k}\right)\). Then $$ (\forall m) \quad \bar{p} \in\left[\bar{a}_{m}, \bar{b}_{m}\right], \text { i.e., } \bar{p} \in \bigcap F_{m}, \text { as required. } $$ Note that the theorem fails for nonclosed intervals, even in \(E^{1} ;\) e.g., take \(F_{m}=\) \((0,1 / m]\) and show that \(\bigcap_{m} F_{m}=\emptyset\).]

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free