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Chapter 3: Problem 37

Let \(D \subseteq \mathbb{R}\) and \(f: D \rightarrow \mathbb{R}\). Prove the following. (i) If \(D\) is bounded and \(f\) is uniformly continuous on \(D\), then \(f\) is bounded on \(D .\) Is this true if \(f\) is merely continuous on \(D ?\) (ii) Let \(\left(x_{n}\right)\) be a Cauchy sequence in \(D .\) If \(f\) is uniformly continuous on \(D\), then \(\left(f\left(x_{n}\right)\right)\) is also a Cauchy sequence. Is this true if \(f\) is merely continuous on \(D ?\)

Short Answer

Expert verified
(i) If $D$ is bounded and $f$ is uniformly continuous on $D$, then $f$ is bounded on $D$. No, this does not hold if $f$ is merely continuous on $D$. Counterexample: $f(x) = \frac{1}{x}$ on $D = [0, 1]$. (ii) If $f$ is uniformly continuous on $D$ and $(x_n)$ is a Cauchy sequence in $D$, then $(f(x_n))$ is also a Cauchy sequence. No, this does not hold if $f$ is merely continuous on $D$. Counterexample: $f(x) = \frac{1}{x}$ on $D = \mathbb{R}$ and $(x_n) = \frac{1}{n}$.

Step by step solution

01

Recall the definition of uniform continuity

We know that a function f is uniformly continuous on D if for every 蔚 > 0, there exists a 未 > 0 such that for all x, y in D, if |x - y| < 未, then |f(x) - f(y)| < 蔚.
02

Use this definition to show that f is bounded on D

We are given that D is bounded. Let's cover D with a finite amount of intervals of length 未, which is given by the definition of uniform continuity. For each interval, pick a point x_i. Then, for any x in D, there exists an i such that |x - x_i| < 未. By uniform continuity, we have: |f(x) - f(x_i)| < 蔚 So, f(x) < f(x_i) + 蔚 Since there are only a finite number of x_i, we can take M = max{|f(x_i)|} + 蔚. Then, for all x in D, |f(x)| 鈮� M, so f is bounded on D.
03

Counterexample for mere continuity

Now, let's consider the case when f is merely continuous on D. Consider D = [0, 1] and f(x) = 1/x for x in (0, 1] and f(0) = 0. Since f is continuous on D, we can see that f(x) is not bounded on D as f(x) goes to infinity as x approaches 0 from the right. #Proof of Part (ii)#
04

Recall the definition of a Cauchy sequence

A sequence (x_n) is a Cauchy sequence if for every 蔚 > 0, there exists an N 鈭� 鈩� such that for all m, n 鈮� N, |x_m - x_n| < 蔚.
05

Prove that f(x_n) is a Cauchy sequence

Given that f is uniformly continuous on D and (x_n) is a Cauchy sequence in D, we want to show that (f(x_n)) is also a Cauchy sequence. Let 蔚 > 0, and since f is uniformly continuous, there exists a 未 > 0 such that for all x, y in D, if |x - y| < 未, then |f(x) - f(y)| < 蔚. Now, since (x_n) is a Cauchy sequence, there exists N 鈭� 鈩� such that for all m, n 鈮� N, |x_m - x_n| < 未. Since x_m and x_n are in D, we can use the definition of uniform continuity: |f(x_m) - f(x_n)| < 蔚 Thus, (f(x_n)) satisfies the definition of a Cauchy sequence.
06

Counterexample for mere continuity

Now, let's consider the case when f is merely continuous on D. Let D = 鈩�. For the sequence (x_n) = 1/n, this sequence is a Cauchy sequence in D. However, if we consider the function f(x) = 1/x on D, f is continuous but not uniformly continuous. Then for x_n = 1/n, we have f(x_n) = n, which is not a Cauchy sequence as the terms increase without limit as n goes to infinity. Thus, we can't guarantee that continuous functions maintain the Cauchy property.

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Key Concepts

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

Bounded Functions
Understanding the concept of bounded functions is key to grasping other fundamental topics in calculus and analysis. A function is said to be bounded on a set if there is a real number that serves as an upper limit to the absolute value of the function on that set. In simpler terms, imagine having an invisible ceiling and floor that the graph of the function can't break through. This ceiling and floor are what mathematicians refer to as bounds.

When we say that a domain, let's call it 鈥楧鈥�, is bounded, we mean that there is a real number beyond which the elements of 鈥楧鈥� don鈥檛 exist. For instance, the set of all numbers between 0 and 10 is bounded because there isn鈥檛 any number in the set greater than 10 or less than 0. Now, if we have a function 鈥榝鈥� that maps every element from this set to some real number, and if there's a particular number 鈥楳鈥� so large that no output of the function 鈥榝(x)鈥� exceeds 鈥楳鈥� in absolute value, then function 鈥榝鈥� is bounded.

If the domain is bounded, doesn't mean the function must be bounded too. A classic counterexample is the function 鈥榝(x) = 1/x鈥� near zero; it grows infinitely large as 鈥榵鈥� approaches zero despite the domain being bounded. This illustrates the importance of specifying conditions like uniform continuity to ensure a function is bounded.
Cauchy Sequences
Delving into the realm of Cauchy sequences opens the door to understanding concepts of convergence and stability within sequences. In essence, a sequence is a Cauchy sequence if its terms get arbitrarily close to each other as the sequence progresses. This concept doesn鈥檛 even require the actual limit of the sequence to be known or exist within the same space! The formal definition states that for any arbitrarily small positive number 鈥樜碘��, there is a stage in the sequence after which all the terms are less than 鈥樜碘�� apart.

This concept is so robust that it forms the basis for defining what a complete metric space is. A space is complete if every Cauchy sequence in the space converges to a limit within the space. The twist comes when you mix Cauchy sequences with continuous functions. A uniformly continuous function has a wonderful property鈥攊t takes Cauchy sequences to Cauchy sequences. However, if a function lacks uniform continuity, it might disrupt the delicate balance of a Cauchy sequence, sending its image sprawling instead of converging.

Using an example, if we had the Cauchy sequence (1/n), and a function 鈥榝鈥� defined by 鈥榝(x) = sin(1/x)鈥�, in the absence of uniform continuity, the sequence 鈥榝(1/n)鈥� would start behaving erratically as 鈥榥鈥� increases, failing to be Cauchy. Remember, the transformative ability to maintain the Cauchy nature of a sequence is a specialty of uniformly continuous functions and not a general trait of all continuous functions.
Continuous Functions
A continuous function is one of the most celebrated creatures in the mathematical wilderness. At its heart, continuity is about predictability and smoothness. When we鈥檙e dealing with a continuous function on a domain, we can expect no sudden jumps, breaks, or infinite escalations in its values. In other words, as you move along the domain of the function, the output changes at a pace that doesn't shock you.

Mathematically, a function 鈥榝鈥� is continuous at a point if for every tiny increment 鈥樜碘�� that you choose, you can find a range 鈥樜粹�� such that when you take a step less than 鈥樜粹�� away from that point, the function's value changes by no more than 鈥樜碘��. It鈥檚 like knowing that no matter how small of a step you take, you won鈥檛 fall off a cliff because the function has no cliffs鈥攜ou can always anticipate its behavior.

However, continuous functions are diverse. Some are wild and unruly, presenting challenges when we try to extend the peace-of-mind guarantee they provide at each point to the entire domain. This is where uniform continuity raises the bar鈥攔equiring the 鈥樜粹�� you select to work uniformly over the entire domain. It requires the function to behave responsibly not just locally, but across all its domain. Literally, with uniform continuity, predictability is not just a local affair but a global guarantee.

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Most popular questions from this chapter

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) satisfy \(f(x+y)=f(x)+f(y)\) for all \(x, y \in \mathbb{R}\). If \(f\) is continuous at 0, then show that (i) \(f\) is continuous at every \(c \in \mathbb{R}\) and (ii) \(f(s x)=s f(x)\) for all \(s, x \in \mathbb{R} .\) Deduce that there exists \(r \in \mathbb{R}\) such that \(f(x)=r x\) for all \(x \in \mathbb{R}\).

Given \(\epsilon>0\), find \(\delta>0\) such that \(|f(x)-\ell|<\epsilon\) whenever \(0<|x-c|<\delta\), if (i) \(f(x)=x^{2}+1, c=1, \ell=2\), (ii) \(f(x)=\frac{1}{x}, c \neq 0, \ell=\frac{1}{c}\), (iii) \(f(x)=\frac{3 x^{2}+7 x+2}{2 x+4}, c=-2, \ell=-5 / 2\).

Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous. Use the IVP to show that for any \(c_{1}, \ldots, c_{n} \in[a, b]\), there is \(c \in[a, b]\) such that $$ f(c)=\frac{f\left(c_{1}\right)+\cdots+f\left(c_{n}\right)}{n} $$

Let \(f:[0,1] \rightarrow \mathbb{R}\) be given by $$ f(x)=\left\\{\begin{array}{ll} 3 x / 2 & \text { if } 0 \leq x<\frac{1}{2} \\ (3 x-1) / 2 & \text { if } \frac{1}{2} \leq x \leq 1 \end{array}\right. $$ Show that \(f([0,1])=[0,1] .\) Is \(f\) continuous on \([0,1] ?\) Does \(f\) have the IVP on \([0,1] ?\)

Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be given by \(f(x)=\left\\{\begin{array}{ll}1 & \text { if } x=0, \\ 1 / q & \text { if } x=p / q \text { where } p, q \in \mathbb{N} \text { and } p, q \text { have no common factor, } \\ 0 & \text { if } x \text { is irrational. }\end{array}\right.\) Show that \(f\) is discontinuous at each rational in \([0, \infty)\) and it is continuous at each irrational in \([0, \infty)\).
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