B Another proof of the existence of extreme values on open intervals

[mcastillo356]

TL;DR Summary

I've been given another and easier proof, but there is something I don't understand.

Hello, PF

This is Theorem 8 of Chapter 4 of the ninth edition of Calculus, by Robert A. Adams: "Existence of extreme values on open intervals". I have an alternative and easier proof, based on epsilon-delta arguments, but it's not mine, and I don't understand it completely.

The fact is that the original doubt was formulated in a Spanish forum. I quote the paragraph origin of the doubt:

"Since $\displaystyle\lim_{x \to{a^{+}}}{f(x)}=L$ there must exist a number $x_1$ in $(a,u)$ such that
$f(x)<f(u)$ whenever $a<x<x_1$
Similarly, there must exist a number $x_2$ in $(u,b)$ such that
$f(x)<f(u)$ whenever $x_2<x<b$"

And this is the reply
"Actually what's in the book I don't see very well. Focusing on doubt, if definition of limit is simply applied appropriately for $a,b\in{R}$ i.e. for real numbers:

$\exists{\epsilon} |~ 0<\epsilon<f(u)-L$. Definition of limit implies $\exists{\delta}>0 |~ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |<\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}$

Similarly,

$\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M$ , definition of limit implies $\exists{\delta'}>0 |~ if \ x\in{(b-\delta',b)}\Rightarrow{\left |{f(x)-M}\right |<\epsilon'}\Rightarrow{f(x)<M+\epsilon'<f(u)}$

Specifically, for your doubts you can take $x_1=a+\delta \wedge x_2=b-\delta$'

Therefore it is known that the maximum of $f$ in the interval $[a+\delta,b-\delta']$. And I think you can continue"

Question:
Let's place the function in the first quadrant, for simplifying the matter:

Why does $\exists{\epsilon} |~ 0<\epsilon<f(u)-L$ ?. Similarly, why does $\exists{\epsilon'} |~ 0<\epsilon'<f(u)-M$?

My attempt is not an attempt:

1- If limit exists, it must be $\forall{\epsilon>0}$
2- $f(u)-L>0$
3- $f(u)-M>0$

Hence, 2 and 3 are fixed numbers, but I can choose any $\epsilon$. So I can state

$\exists{\epsilon} |~ 0<\epsilon<f(u)-L$
$\exists{\epsilon'} |~0<\epsilon'<f(u)-M$

the same way I could state

$\exists{\epsilon} |~ 0<f(u)-L<\epsilon$
$\exists{\epsilon'} |~ 0<f(u)-M<\epsilon'$

Greetings!

 

[pbuk]

This is Theorem 8 of Chapter 4 of the ninth edition of Calculus, by Robert A. Adams: "Existence of extreme values on open intervals".

What is this theorem? Not everyone owns the ninth edition of Calculus, by Robert A. Adams.

 

[mcastillo356]

What is this theorem? Not everyone owns the ninth edition of Calculus, by Robert A. Adams.

Can I write it down in this thread?

 

[Office_Shredder]

I think it's fine to write down what the theorem is if you're worried about copyright stuff.

 

[pbuk]

Mathematical theorems and their proofs are not subject to copyright.

 

[mcastillo356]

Hello, PF

This is Theorem of "Existence of extreme values on open intervals". I have an alternative and easier proof, based on epsilon-delta arguments, but it's not mine, and I don't understand it completely.

I quote the theorem:

If $f$ is continuous on the open interval $(a,b)$, and if

$\displaystyle\lim_{x \to{a^{+}}}{f(x)}=L\quad{\mbox{and}}\quad{\displaystyle\lim_{x \to{b^{-}}}{f(x)}=M}$

Then the following conclusions hold:

(i) If $f(u)>L$ and $f(u)>M$ for some $u$ in $(a,b)$, then $f$ has an absolute maximum value on $(a,b)$
(ii) If $f(v)<L$ and $f(v)<M$ for some $v$ in $(a,b)$, then $f$ has an absolute minimum value on $(a,b)$

In this theorem $a$ may be $-\infty$, in which case $\lim_{x \to{a^{+}}}{}$ should be replaced with $\lim_{x \to{{-\infty}}}{}$, and $b$ may be $\infty$, in which case $\lim_{x \to{\infty}}{}$.

Also, either or both $L$ and $M$ may be either $\infty$ or $-\infty$

PROOF We prove part (i); the proof of (ii) is similar. We are given that there is a number $u$ in $(a,b)$ such that $f(u)>L$ and $f(u)>M$. Here $L$ and $M$ may be finite numbers or $-\infty$. Since $\lim_{x \to{a+}}{f(x)}=L$ there must exist a number $x_1$ in $(a,u)$ such that
$f(x)<f(u)$ whenever $a<x<x_1$
Similarly, there must exist a number $x_2$ in $(u,b)$ such that
$f(x)<f(u)$ whenever $x_2<x<b$
(See attached Figure)
Thus, $f(x)<f(u)$ at all points of $(a,b)$ that are not in the closed, finite subinterval $[x_1,x_2]$. By Theorem of Existence of extreme values, the function $f$, being continuous on $[x_1,x_2]$, must have an absolute maximum value on that interval, say at the point $w$. Since $u$ belongs to $[x_1,x_2]$, we must have $f(w)\geq{f(u)}$, so $f(w)$ is the maximum value of $f(x)$ for all of $(a,b)$.

And this is the reply
"Actually what's in the book I don't see very well. Focusing on doubt, if definition of limit is simply applied appropriately for $a,b\in{R}$ i.e. for real numbers:

$\exists{\epsilon} \ / \ 0<\epsilon<f(u)-L$. Definition of limit implies $\exists{\delta}>0 \ / \ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}$

Similarly,

$\exists{\epsilon'} \ / \ 0<\epsilon'<f(u)-M$ , definition of limit implies $\exists{\delta'}>0 \ / \ if \ x\in{(b-\delta',b)}\Rightarrow{\left |{f(x)-M}\right |<\epsilon'}\Rightarrow{f(x)<M+\epsilon'<f(u)}$

Specifically, for your doubts you can take $x_1=a+\delta \wedge x_2=b-\delta'$

Therefore it is known that exists the maximum of $f$ in the interval $[a+\delta,b-\delta']$. And I think you can continue"

Question:
Let's place the function in the first quadrant, for simplifying the matter:

Why does $\exists{\epsilon} \ / \ 0<\epsilon<f(u)-L$?. Similarly, why does $\exists{\epsilon'} \ / \ 0<\epsilon'<f(u)-M$?

1-Limits exist. Therefore exist $\forall{\epsilon>0}$.
2-$f(u)-L$, as well as $f(u)-M$, are fixed numbers.
3-Therefore, I can state
$\exists{\;0<\epsilon<f(u)-L}$, or $\exists{0<f(u)-L<\epsilon}$. Similarly with $\epsilon'$

Greetings! If this is a mess, or am I overthinking, ignore. Thanks, love.

P.S. Don't know if done right the LaTeX

 

[pbuk]

$\exists{\epsilon} \ / \ 0<\epsilon<f(u)-L$. Definition of limit implies $\exists{\delta}>0 \ / \ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |}\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}$

Similarly,

$\exists{\epsilon'} \ / \ 0<\epsilon'<f(u)-M$ , definition of limit implies $\exists{\delta'}>0 \ / \ if \ x\in{(b-\delta',b)}\Rightarrow{\left |{f(x)-M}\right |<\epsilon'}\Rightarrow{f(x)<M+\epsilon'<f(u)}$

Specifically, for your doubts you can take $x_1=a+\delta \wedge x_2=b-\delta'$

Therefore it is known that exists the maximum of $f$ in the interval $[a+\delta,b-\delta']$. And I think you can continue"

That is not an alternative proof of the theorem, it is a proof of the statement

Since $\lim_{x \to{a+}}{f(x)}=L$ there must exist a number $x_1$ in $(a,u)$ such that $f(x)<f(u)$ whenever $a<x<x_1$

which appears as part of the proof in the book. I am not sure why the poster in the Spanish forum felt that expanding this part of the proof was needed, the statement in Adams' proof appears trivially true to me, and in case you don't see this Figure 4.16 provides a clear visualisation.

If this is a mess, or am I overthinking, ignore. Thanks, love.

You are definitely overthinking. You have provided an expanded proof of a trivial statement which is itself part of an expanded proof of a trivial statement which is part of a proof in Adams' book.

If there is an assumption or a step in an author's proof that you don't understand then of course you should go back and think it through (in this case by reference to Figure 4.16), and of course it is always possible that your end-of-course assessment may ask you to prove such a step.

However if you are going to learn anything from a book at some point you need to rise above this focus on the trivial.

 

[Mark44]

I have some comments on your (@mcastillo356) LaTeX formatting.

In the first one, I have left in the start of the quote so that you can easily find this quote. I have also modified the LaTeX delimiters to prevent the browser from rendering as it ordinarily would.

"Actually what's in the book I don't see very well. Focusing on doubt, if definition of limit is simply applied appropriately for $a,b\in{R}$ i.e. for real numbers:
$\exists{\epsilon} \ / \ 0<\epsilon<f(u)-L$. Definition of limit implies $\exists{\delta}>0 \ / \ if \ x\in{(a,a+\delta)}\Rightarrow{\left |{f(x)-L}\right |}\epsilon}\Rightarrow{f(x)<L+\epsilon<f(u)}$

1. Don't use \ / \ when you mean | ("such that"}.
2. The last LaTeX expression was so broken that I couldn't fix it in three tries, so I left it. It was more difficult to fix, mostly because it is so complex, with unmatched braces, extra braces, and everything in between being so tightly packed.
In mathematics, we use parentheses ( ), brackets [ ], and braces { }, to enclose expressions, but braces are special characters in LaTeX, so don't use them for the same purpose as they would be used mathematically.

Here's another chunk:

Question:
Let's place the function in the first quadrant, for simplifying the matter:

Why does $\exists{\epsilon} \ / \ 0<\epsilon<f(u)-L$?. Similarly, why does $\exists{\epsilon'} \ / \0<\epsilon'<f(u)-M$?

I would write the first LaTeX expression like this:
$\exists \epsilon > 0~|~ 0< \epsilon <f(u) - L$ -- raw LaTeX
$\exists \epsilon > 0 ~|~ 0< \epsilon <f(u) - L$ -- fully rendered

And similar for the second expression. Notice that I have removed several of the braces you used, and have used a "pipe" character -- | -- to represent "such that."

 

[mcastillo356]

Hi PF

Slash is never used the way my posts do.

https://en.wikipedia.org/wiki/Slash_(punctuation)#Combining_slash

Sorry, PF, and thanks at the same time for your good job

 

[Mark44]

Slash is never used the way my posts do.
https://en.wikipedia.org/wiki/Slash_(punctuation)#Combining_slash

There's a slash character (/) and a backslash character (\).
The wiki article is not relevant to how LaTeX does things.