Uniform Continuity and Cauchy Sequences

[Enzipino]

Hello,

I've been attempting to do these problems from my textbook:

1. Suppose that \(\displaystyle f\) is a continuous function on a bounded set \(\displaystyle S\). Prove that the
following two conditions are equivalent:
(a) The function \(\displaystyle f\) is uniformly continuous on \(\displaystyle S\).
(b) It is possible to extend \(\displaystyle f\) to a continuous function on the set \(\displaystyle S\).

2. Let \(\displaystyle f:[0, \infty)\to\Bbb{R}\) be defined by \(\displaystyle f(x) = \sqrt{x}.\) Prove that \(\displaystyle f\) is uniformly continuous on \(\displaystyle [0, \infty)\)

For #1, I don't know what I'm supposed to do. I know the definition of uniform continuity but I just don't know how to go about using it for this.

For #2, I did some extra scratch work and what I did was:
\(\displaystyle \left| f(t)-f(x) \right| = \left| \sqrt{t}-\sqrt{x} \right| \le {\left| \sqrt{t}-\sqrt{x} \right|}^{2} \le \left| \sqrt{t}-\sqrt{x} \right|\left| \sqrt{t}+\sqrt{x} \right| = \left| t-x \right|<\delta\) but now I don't know what to let my \(\displaystyle \delta\) be so that it works out.

 

[Opalg]

1. Suppose that \(\displaystyle f\) is a continuous function on a bounded set \(\displaystyle S\). Prove that the following two conditions are equivalent:
(a) The function \(\displaystyle f\) is uniformly continuous on \(\displaystyle S\).
(b) It is possible to extend \(\displaystyle f\) to a continuous function on the set \(\displaystyle S\).

Hi Enzipino and welcome to MHB!

I think there must be something missing from condition (b). If $f$ is already defined on $S$, you don't need any extension to define it on $S$! I suspect that (b) should say

(b) It is possible to extend \(\displaystyle f\) to a continuous function on the closure $\overline{S}$ of \(\displaystyle S\).​

Given that revised form of (b), it is then easy to see that (a) follows. In fact, $\overline{S}$ is compact (in other words, closed and bounded), and a continuous function on a compact set is always uniformly continuous.

To go from (a) to the revised (b), use the fact that a uniformly continuous function takes Cauchy sequences to Cauchy sequences. Then, given $x \in \overline{S},$ there exists a sequence $\{x_n\}$ in $S$ with $x_n\to x$ as $n\to\infty$. Deduce that $\{f(x_n)\}$ is a Cauchy sequence and must therefore be convergent. Define $f(x)$ to be the limit of that sequence.

2. Let \(\displaystyle f:[0, \infty)\to\Bbb{R}\) be defined by \(\displaystyle f(x) = \sqrt{x}.\) Prove that \(\displaystyle f\) is uniformly continuous on \(\displaystyle [0, \infty)\)

The easiest way to do this is to split the interval $[0, \infty)$ into two pieces, $[0,1]$ and $[1, \infty)$. Show that $f$ is uniformly continuous on each of these (for different reasons!).

 

[Evgeny.Makarov]

\(\displaystyle \left| f(t)-f(x) \right| = \left| \sqrt{t}-\sqrt{x} \right| \le {\left| \sqrt{t}-\sqrt{x} \right|}^{2}\)

The last inequality does not necessarily hold.

Another way to prove statement 2 is the following.

Let $n$ be a positive integer and $p,q$ positive real numbers. Then $p^n+q^n\le (p+q)^n$. Substituting $p=x^{1/n}$ and $q=(y-x)^{1/n}$ where $0\le x\le y$, we get
\begin{align}
&y=x+(y-x)\le \left(x^{1/n}+(y-x)^{1/n}\right)^n\\
&\implies y^{1/n}\le x^{1/n}+(y-x)^{1/n}&&\text{raising both sides to }1/n\\
&\implies\quad0\le y^{1/n}-x^{1/n}\le (y-x)^{1/n}
\end{align}

 

[Enzipino]

Hi Enzipino and welcome to MHB!

I think there must be something missing from condition (b). If $f$ is already defined on $S$, you don't need any extension to define it on $S$! I suspect that (b) should say

(b) It is possible to extend \(\displaystyle f\) to a continuous function on the closure $\overline{S}$ of \(\displaystyle S\).​

Given that revised form of (b), it is then easy to see that (a) follows. In fact, $\overline{S}$ is compact (in other words, closed and bounded), and a continuous function on a compact set is always uniformly continuous.

To go from (a) to the revised (b), use the fact that a uniformly continuous function takes Cauchy sequences to Cauchy sequences. Then, given $x \in \overline{S},$ there exists a sequence $\{x_n\}$ in $S$ with $x_n\to x$ as $n\to\infty$. Deduce that $\{f(x_n)\}$ is a Cauchy sequence and must therefore be convergent. Define $f(x)$ to be the limit of that sequence.

Ah yes, I apologize. It is $\overline{S}$. Thank you very much for the hint! I don't recall covering Cauchy Sequences in my intro class so I will have to read up on it but I do see where I can go with it now.

 

[Enzipino]

I was just looking through my textbook to brush up on Cauchy sequences since I was planning on using it for this problem but my book utterly leaves it out. But I did come across this Theorem:
Suppose that \(\displaystyle f\) is a function and \(\displaystyle f:S\to\Bbb{R}\). If \(\displaystyle f\) is uniformly continuous on \(\displaystyle S\) then, given any two sequences \(\displaystyle ({x}_{n})\) and \(\displaystyle ({t}_{n})\) in $S$ such that ${t}_{n}-{x}_{n}\to 0$ as $n\to \infty$, we have $f({t}_{n})-f({x}_{n})\to 0$ as $n\to \infty$

Could I use this instead of a Cauchy Sequence? Or use it alongside Cauchy sequences?

 

[Opalg]

I was just looking through my textbook to brush up on Cauchy sequences since I was planning on using it for this problem but my book utterly leaves it out. But I did come across this Theorem:
Suppose that \(\displaystyle f\) is a function and \(\displaystyle f:S\to\Bbb{R}\). If \(\displaystyle f\) is uniformly continuous on \(\displaystyle S\) then, given any two sequences \(\displaystyle ({x}_{n})\) and \(\displaystyle ({t}_{n})\) in $S$ such that ${t}_{n}-{x}_{n}\to 0$ as $n\to \infty$, we have $f({t}_{n})-f({x}_{n})\to 0$ as $n\to \infty$

Could I use this instead of a Cauchy Sequence? Or use it alongside Cauchy sequences?

That Theorem is needed in my approach to the problem, to show that the extension of $f$ to $\overline S$ is well-defined. In other words, if you take two different sequences $\{x_n\}$ and $\{t_n\}$ converging to the same point $x\in\overline S$ then $\lim_{n\to\infty}\{f(x_n)\} = \lim_{n\to\infty}\{f(t_n)\}$.

But you still need a result about Cauchy sequences in order to deduce that those limits exist at all.

 

[Krizalid1]

2. Let \(\displaystyle f:[0, \infty)\to\Bbb{R}\) be defined by \(\displaystyle f(x) = \sqrt{x}.\) Prove that \(\displaystyle f\) is uniformly continuous on \(\displaystyle [0, \infty)\)

Following Opalg's hint, you have \(\displaystyle [0,1]\) is a compact set and \(\displaystyle f\) is clearly continuous on \(\displaystyle [0,1].\) Does that ring any bell?

Besides, proving uniform continuity on \(\displaystyle [1,\infty)\) follows easily by definition.

 

[Enzipino]

That Theorem is needed in my approach to the problem, to show that the extension of $f$ to $\overline S$ is well-defined. In other words, if you take two different sequences $\{x_n\}$ and $\{t_n\}$ converging to the same point $x\in\overline S$ then $\lim_{n\to\infty}\{f(x_n)\} = \lim_{n\to\infty}\{f(t_n)\}$.

But you still need a result about Cauchy sequences in order to deduce that those limits exist at all.

So I was pulling my hair out at Barnes and Noble doing this problem because I was completely unsure if I could just use Cauchy Sequences to prove this problem since my textbook is mean and likes to not name certain things. I decided to go back and look at some of the previous problems in the book and whadda you know, I found these two example problems:

a) Prove that if $f$ is uniformly continuous on a set $S$ and $\{x_n\}$ is a convergent sequence in $S$, then the sequence $(f({x}_{n}))$ is also convergent. (AKA the Cauchy Sequence)​

b) Suppose that $f$ is uniformly continuous on a set $S$, that $x$ is a real number and that $\{x_n\}$ and $\{t_n\}$ are sequences in S that converge to the number $x$. Prove that $\lim_{n\to\infty}\{f(x_n)\} = \lim_{n\to\infty}\{f(t_n)\}$ (AKA the freaking result that I needed).​

Of course, knowing my professor in order for me to use this, I have to first prove it but that doesn't seem to hard because if I can prove the first one then the second one pretty much follows and then with that I can just do:

Suppose $\{x_n\}$,$\{t_n\}$$\in S$ such that ${x}_{n}-{t}_{n}\to0$ and let $\varepsilon>0$. From the theorem I posted before, we have that $f({x}_{n})-f({t}_{n})\to0$ as $n\to\infty$. Now, let $x\in\overline S$ and suppose $\{x_n\}$,$\{t_n\}$ both converge to $x$. By (a), their functions must also be convergent to $f(x)$. Then by (b), $\lim_{n\to\infty}f(x_n) = \lim_{n\to\infty}f(t_n)$.​

P.S. I really appreciate all the help Opalg.