Uniform Continuity of functions

[Bling Fizikst]

Homework Statement

below

Relevant Equations

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Discuss uniform continuity of the following functions:
$\tan x$ in $[0,\frac{\pi}{2})$
$\frac{1}{x}\sin^2 x$ in $(0,\pi]$
$\frac{1}{x-3}$ in $(0,3),(4,\infty),(3,\infty)$

I am completely new to this uniform continuity and couldn't find a lot of examples to learn the solving pattern .
For the first problem , i tried to use the definition ,
for $x,y$ in the given interval , $|x-y|<\delta\implies |\tan x-\tan y|=\frac{|\sin(x-y)|}{|\cos x\cos y|}\leq \frac{1}{|\cos x\cos y|}$
Just stuck in this and the other problems .

 

[Hill]

If you are stuck trying to show uniform continuity of the function, try to show that the function is NOT uniformly continuous.

 

[fresh_42]

Continuity is a local phenomenon. It always means continuous at a point $x_0.$ The usual definition is that for every $\varepsilon >0$ there is a $\delta >0$ such that $|f(x)-f(x_0)|<\varepsilon$ whenever $|x-x_0|<\delta.$

If you look closer to that definition, then you will find out that $x_0$ is chosen prior to $\delta$ as is $\varepsilon.$ This means that it actually should have been written as:
$$
\forall\;\varepsilon>0 \;\;\exists \;\delta(x_0,\varepsilon) \, : \, |x-x_0|<\delta(x_0,\varepsilon) \Longrightarrow |f(x)-f(x_0)|<\varepsilon\;.
$$
The difference is, that I noted that $\delta=\delta(x_0,\varepsilon)$ depends on location $x_0$ as well as on the margin $\varepsilon.$ We only say $x\to f(x)$ is continuous without mentioning the location if $f$ is continuous for all $x_0$ in the domain of $f.$ But $\delta$ may depend on the various locations $x_0.$

A third definition is uniform continuity. It simply expresses the fact that we can chose $\delta = \delta(\varepsilon)$ without using the location $x_0.$ We have
$$
\forall\;\varepsilon>0 \;\;\exists \;\delta(\varepsilon) \, : \, |x-x_0|<\delta(\varepsilon) \Longrightarrow |f(x)-f(x_0)|<\varepsilon\;.
$$
for any two $x,x_0$ in the domain of $f.$

The question in the exercise is thus: Can $\delta$ be chosen independently from the location $x_0$? Can we choose $\delta=\delta(\varepsilon)$ only depending on our choice of the margin $\varepsilon$ anywhere in the domain or not?

The typical example is $f(x)=\dfrac{1}{x}$ on $(0,1].$ It is continuous but not uniformly continuous because our $\delta$ would be
$$
\delta=\delta(x_0,\varepsilon)= \min\left\{\dfrac{x_0}{2}\, , \,\dfrac{x_0^2\varepsilon}{2}\right\}
$$
as
$$\left|f(x)-f(x_0)\right|=\left|\dfrac{1}{x}-\dfrac{1}{x_0}\right|=\left|\dfrac{x-x_0}{xx_0}\right|\leq \dfrac{2\left|x-x_0\right|}{x_0^2}<\dfrac{2\delta}{x_0^2}<\varepsilon\;.$$

 

[FactChecker]

The problem asks you to discuss those functions, not necessarily to mathematically prove what you discuss. Can you recognize an uniformly continuous function when you see its graph? Here are the three functions.
Discuss. :-)

 

[Bling Fizikst]

I actually wanted mathematical proof . That was the intention of the problem .

 

[PeroK]

I actually wanted mathematical proof . That was the intention of the problem .

Proofs don't appear out of thin air. You need to look at the functions in this case and decide why they are or are not uniformly continuous. Then, the proof is a matter of mathematical technique in turning an idea into a formal epsilon-delta argument.

To determine uniform continuity, what sort of thing are you looking for in a function? This is the other side of mathematics: taking a formal definition and understanding what sort of behaviour it is defining or encapsulating.

Post #4 is a great start to solving this problem.

 

[Hill]

I actually wanted mathematical proof . That was the intention of the problem .

Then you should go one step back in $$|x-y|<\delta\implies |\tan x-\tan y|=\frac{|\sin(x-y)|}{|\cos x\cos y|}\leq \frac{1}{|\cos x\cos y|}$$ and rather look for what pushes $\frac{|\sin(x-y)|}{|\cos x\cos y|}$ from below.

 

[Bling Fizikst]

Then you should go one step back in $$|x-y|<\delta\implies |\tan x-\tan y|=\frac{|\sin(x-y)|}{|\cos x\cos y|}\leq \frac{1}{|\cos x\cos y|}$$ and rather look for what pushes $\frac{|\sin(x-y)|}{|\cos x\cos y|}$ from below.

intuitively , i get that it shouldn't be uniformly continuous as the denominator tends to $0$ as $x,y$ tends to $\frac{\pi}{2}$ making it unbounded . But how do i prove it with rigor?

 

[PeroK]

intuitively , i get that it shouldn't be uniformly continuous as the denominator tends to $0$ as $x,y$ tends to $\frac{\pi}{2}$ making it unbounded . But how do i prove it with rigor?

Are you able to state the condition/definition that a function is not uniformly continuous?

The first step is to state what you are trying to prove.

PS This is not as easy as you might think! You have to negate the condition for uniform continuity.

 

[fresh_42]

intuitively , i get that it shouldn't be uniformly continuous as the denominator tends to $0$ as $x,y$ tends to $\frac{\pi}{2}$ making it unbounded . But how do i prove it with rigor?

The function is obviously continuous in its domain. Assuming it was uniformly continuous means that there is a $\delta(\varepsilon)$ such that $|\tan(x)-\tan(y)|<\varepsilon$ whenever $|x-y|<\delta(\varepsilon).$

You have to show that $\delta(\varepsilon)$ cannot be chosen independently from $y$ as $y$ approaches $\pi/2.$ Take a value $\delta(\varepsilon)$ that does not depend on $y$ and find a contradiction by choosing two different values $y$ close to $\pi/2,$ one that satisfies the condition and one that does not.

Look at my example $x \longmapsto 1/x$ in post #3. It becomes quite evident if you use the series expansion of $\tan x.$

 

[Hill]

intuitively , i get that it shouldn't be uniformly continuous as the denominator tends to $0$ as $x,y$ tends to $\frac{\pi}{2}$ making it unbounded . But how do i prove it with rigor?

You can make $$\frac{|\sin(x-y)|}{|\cos x\cos y|}>\frac{const.}{|\cos x|}$$.

 

[FactChecker]

intuitively , i get that it shouldn't be uniformly continuous as the denominator tends to $0$ as $x,y$ tends to $\frac{\pi}{2}$ making it unbounded .

That is not enough. You could also say that about the second function, $\frac {\sin^2(x)}{x}$, but it is different.

But how do i prove it with rigor?

That depends on what class you are in. Have you had calculus? Look at the slope or the derivative definitions. Do you know the relationship between uniform continuity and the slope or derivative?
I wrote a basic hint because it seemed like you were trying to prove the wrong thing in your work.

 

[FactChecker]

I actually wanted mathematical proof . That was the intention of the problem .

Are you sure they expect a rigorous proof? Did they say "discuss" or "prove"? Proofs can get hard, even for simple functions.
I still think that your first step should be to say what you think the answer is for each function and why. You should talk about the slope or derivative.

 

[Mark44]

I actually wanted mathematical proof . That was the intention of the problem .

No it wasn't if your problem statement means anything. Emphasis added.

Discuss uniform continuity of the following functions:

 

[Bling Fizikst]

I am unsure about the epsilon delta . The only thing i could think of was consider two sequences in $(0,\frac{\pi}{2}) : x_n=\frac{\pi}{2}-\frac{1}{n}, y_n=\frac{\pi}{2}-\frac{1}{n+1}$
Observing that $\mid x_n-y_n\mid\to 0$ and $\mid f(x_n)-f(y_n)\mid = \mid\cot\frac{1}{n}-\cot\frac{1}{n+1}\mid$ is of the form $\infty-\infty$ which can get arbitrarily large . But unsure if this is alright

 

[Bling Fizikst]

Are you sure they expect a rigorous proof? Did they say "discuss" or "prove"? Proofs can get hard, even for simple functions.
I still think that your first step should be to say what you think the answer is for each function and why. You should talk about the slope or derivative.

source: elementary analysis by Ross

 

[PeroK]

Exercise 19.4:

Prove that if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on $S$.

That's the key to it!

 

[FactChecker]

Exercise 19.4:

Prove that if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on $S$.

That's the key to it!

I'm not sure that it will apply on the unbounded domains in the problem.

 

[FactChecker]

I am unsure about the epsilon delta . The only thing i could think of was consider two sequences in $(0,\frac{\pi}{2})$ : $x_n=\frac{\pi}{2}-\frac{1}{n}$ $y_n=\frac{\pi}{2}-\frac{1}{n+1}$
Observing that $\mid x_n-y_n\mid\to 0$ and $\mid f(x_n)-f(y_n)\mid = \mid\cot\frac{1}{n}-\cot\frac{1}{n+1}\mid$ is of the form $\infty-\infty$ which can get arbitrarily large . But unsure if this is alright

That is a good try. A couple of comments:
1) First, you should state whether you are trying to prove or disprove that it is uniformly continuous.
2) I had to edit your Latex to read your post.

Now for the question of $\tan(x), x \in [0, \frac{\pi}{2})$. To prove it is not uniformly continuous, consider $x_n = \tan^{-1}(10^n)$ where $x_n \in [0, \frac{\pi}{2})$. Clearly, $x_n$ is a Cauchy sequence (it is monotonically increasing and bounded) but $\tan(x_n)=10^n$ is not a Cauchy sequence.
EDIT: (See Theorem 19.4 and Example 6 of Ross, Elementary Analysis)

 

[Hill]

I am unsure about the epsilon delta

To show that $f(x)$ is not uniformly continuous, you show that for some $\epsilon$ and any $\delta$ there exist such $x$ and $y$ with $|x-y| \lt \delta$ that $|f(x)-f(y)| \gt \epsilon$.

 

[PeroK]

To show that $f(x)$ is not uniformly continuous, you show that for some $\epsilon$ and any $\delta$ there exist such $x$ and $y$ with $|x-y| \lt \delta$ that $|f(x)-f(y)| \gt \epsilon$.

It's sometimes useful to note that this is equivalent to finding sequences $x_n$ and $y_n$ with the property that, for some (fixed) $\epsilon$:
$$|x_n - y_n| < \frac 1 n, \ \text{and} \ |f(x_n) - f(y_n)| > \epsilon$$

 

[WWGD]

Note the result that uniform continuity is necessary to preserve the Cauchyness of sequence.

 

[fresh_42]

Note the result that uniform continuity is necessary to preserve the Cauchyness of sequence.

I think continuity is sufficient. Cauchy sequences converge to a real number $x_0$ and continuity is a local phenomenon, too. Uniform continuity is a global property.

 

[WWGD]

I think continuity is sufficient. Cauchy sequences converge to a real number $x_0$ and continuity is a local phenomenon, too. Uniform continuity is a global property.

Try $\frac{1}{x-\sqrt 2}$ on $\mathbb Q$.

 

[fresh_42]

Try $\frac{1}{x-\sqrt 2}$ on $\mathbb Q$.

We are not in $\mathbb{Q}.$ And what does $\sqrt{2}$ even mean in $\mathbb{Q}$, or continuity? We have real numbers in the domains and limits available. Hence
\begin{align*}
|f(x_n)-f(x_m)|&=|f(x_n)-f(x_0)+f(x_0)-f(x_m)|\\
&\leq \underbrace{|f(x_n)-f(x_0)|}_{<\varepsilon/2}+\underbrace{|f(x_0)-f(x_m)|}_{<\varepsilon/2}< \varepsilon
\end{align*}
for all $|x_m-x_0|,|x_n-x_0|<\delta(x_0,\varepsilon)$ and sufficiently large $n,m\in \mathbb{N}.$

Continuity is equivalent to $\displaystyle{ \lim_{n \to \infty}f(x_n)=f(\lim_{n \to \infty}x_n)}.$

 

[WWGD]

We are not in $\mathbb{Q}.$ And what does $\sqrt{2}$ even mean in $\mathbb{Q}$? We have real numbers in the domains and limits available. Hence
\begin{align*}
|f(x_n)-f(x_m)|&=|f(x_n)-f(x_0)+f(x_0)-f(x_m)|\\
&\leq \underbrace{|f(x_n)-f(x_0)|}_{<\varepsilon/2}+\underbrace{|f(x_0)-f(x_m)|}_{<\varepsilon/2}< \varepsilon
\end{align*}
for all $|x_m-x_0|,|x_n-x_0|<\delta(x_0,\varepsilon)$ and sufficiently large $n,m\in \mathbb{N}.$

When you restrict to Rationals and consider a sequence that should converge to $\sqrt 2$, the image of the sequence will blow up.

 

[fresh_42]

When you restrict to Rationals and consider a sequence that should converge to $\sqrt 2$, the image of the sequence will blow up.

How do you define continuity on $\mathbb{Q}$? If you have a location with a discontinuity, then uniformity won't save you.

 

[FactChecker]

I think continuity is sufficient. Cauchy sequences converge to a real number $x_0$ and continuity is a local phenomenon, too. Uniform continuity is a global property.

Good point, but not when the domain is an open set. Several of the examples in the OP are on open sets with a singularity at an open end of the domain.

 

[WWGD]

How do you define continuity on $\mathbb{Q}$? If you have a location with a discontinuity, then uniformity won't save you.

It's standard definition for a metric space. At any rate, just use the map $f(x)=1/x$, from $(0,1)\rightarrow \mathbb R$ and consider the Cauchy sequence $\{1/n \}$ in $(0,1)$. It's mapped into the non-Cauchy sequence $\{1,2,3 ...\}$.

 

[mathwonk]

1) a uniformly continuous function is bounded on a bounded domain.
2) a function whose derivative is bounded on an interval is also Lipschitz continuous on that interval, hence also uniformly continuous.

that does it for all parts. (note that a restriction of a uniformly continuous function from an infinite interval to a finite subinterval is still uniformly continuous.)

 

[FactChecker]

1) a uniformly continuous function is bounded on a bounded domain.
2) a function whose derivative is bounded on an interval is also Lipschitz continuous on that interval, hence also uniformly continuous.

that does it for all parts. (note that a restriction of a uniformly continuous function from an infinite interval to a finite subinterval is still uniformly continuous.)

Those are great proofs. But then I took a look at the text book this problem was in and those facts are not available at that point in the text. The book does have a theorem using Cauchy sequences and a nice example proof using it.

 

[mathwonk]

those facts follow from the definitions and the mean value theorem. of course perhaps the MVT was not available, but it usually occurs a few pages after the definition of a derivative.

even without that criterion, the problems follow with a little work.
the derivative criterion is only used in problems 2 and 4. 2 already follows if one knows the given function has a finite limit at 0, plus the fact that a continuous function on a closed bound interval is uniformly so. 4 follows from the same fact plus the fact that the given function is bounded and decreasing, (by separating the infinite interval into a finite one, plus an infinite one on which delta(f) is everywhere less than epsilon).
but maybe they also do not know continuous functions are uniformly cont. on compact domains?

 

[FactChecker]

those facts follow from the definitions and the mean value theorem. of course perhaps the MVT was not available, but it usually occurs a few pages after the definition of a derivative.

Good point. In that text, they haven't gotten to derivatives. The Cauchy sequence theorem and an associated example proof that a function is not uniformly continuous precedes the problem in the OP. That was the approach @Bling Fizikst was attempting to use in post #15. I tried to help him use that approach in post #19.

 

[fresh_42]

Good point, but not when the domain is an open set. Several of the examples in the OP are on open sets with a singularity at an open end of the domain.

This statement has nothing to do with what I said. Cauchy sequences and converging series are the same thing for real numbers, and convergence compatibility of a function is equivalent to simple continuity. There is no such thing as a "Cauchyness".

 

[fresh_42]

It's standard definition for a metric space. At any rate, just use the map $f(x)=1/x$, from $(0,1)\rightarrow \mathbb R$ and consider the Cauchy sequence $\{1/n \}$ in $(0,1)$. It's mapped into the non-Cauchy sequence $\{1,2,3 ...\}$.

The map $x\longmapsto 1/x$ is not uniformly continuous on $(0,1]$, as by the way, I already mentioned in my first reply in post #3. But that has nothing to do with Cauchy sequences.

Uniform continuity implies continuity which is equivalent to
$$
\displaystyle{\lim_{n \to \infty}f(x_n)=f(\lim_{n \to \infty}x_n)}
$$
that is equivalent to the use of Cauchy sequences instead of converging sequences over the reals. Your example just says that the first implication isn't reversible, as by the way, I already mentioned in my first reply in post #3. It is the non-existing limit at $0$ that matters.

The use of Cauchy sequences in this thread is simply off-topic. Yes, it makes things more convenient as you automatically have $|x_n-x_m|<\delta$ without bothering the limit $x_0.$ However, the triangle inequality solves this problem, as by the way, I already mentioned in my reply in post #25.