# Extreme value theorem, proof question

• fishturtle1
In summary, the conversation discusses the contradiction between the limits of a function ##f## and its absolute value when approaching infinity. The proof being discussed is from Ross page 133 and involves the Bolzano-Weierstrass Theorem and the Theorem on Continuous Functions on Closed Intervals. The conversation delves into the reasoning behind the contradiction and the use of natural numbers and sub-sequences to prove it. The domain is clarified to be in ##[a,b]## and the discussion also touches on the notation used in the proof.
fishturtle1

## Homework Statement

Why does ##\lim_{n \rightarrow \infty} f(x_n) = f(c)## contradict ##\lim_{n \rightarrow \infty} \vert f(x_n) \vert = +\infty##?

edit: where ##c## is in ##[a,b]##

## Homework Equations

Here's the proof I'm reading from Ross page 133.

18.1 Theorem
Let ##f## be a continuous real valued function on a closed interval [a,b]. Then ##f## is a bounded function. Moreover, f assumes its maximum and minimum values on [a,b]; that is there exist ##x_0, y_0## in [a,b] such that ##f(x_0) \le f(x) \le f(y_0)## for all ##x \epsilon [a,b]##.

Proof: Assume ##f## is not bounded on ##[a,b]##. Then to each ##n \epsilon \mathbb{N}## there corresponds an ##x_n \epsilon [a,b]## such that ##\vert f(x_n) > n##. By Bolzano-Weierstrass Theorem 11.5, ##(x_n)## has a subsequence ##(x_{n_k})## that converges to some real number ##x_0## in ##[a,b]##. Since ##f## is continuous at ##x_0##, we have ##\lim_{k \rightarrow \infty} f(x_{n_k}) = f(x_0)##, But we also have ##\lim_{k\rightarrow\infty} \vert f(x_{n_k}) \vert = +\infty##. which is a contradiction. It follows that ##f## is bounded.

//And then another paragraph I haven't really read yet that finishes the proof

## The Attempt at a Solution

1) From def. of limit: For all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##k > N(\varepsilon)## implies ##\vert f(x_{n_k}) - f(x_0) \vert < \varepsilon##. This implies ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon##.

2) Since ##f## is not bounded we have for all ##\varepsilon > 0##, there exists some ##k## such that ##\vert f(x_{n_k}) \vert > \vert f(x_0) \vert + \varepsilon##.

So there is only a contradiction between 1) and 2), if the ##k## in 2) is greater than ##N(\varepsilon)## in 1). But how do we guarantee this?

Last edited:
Just pick a greater k. Any greater k works as well because you have infinitely many ##n_k##'s.

verty said:
Just pick a greater k. Any greater k works as well because you have infinitely many ##n_k##'s.

I understand there are infinitely many ##n_k##'s such that ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon##, but i don't see how there are infinitely many k's such that ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon##?

The definition of a sequence ##x_n## being unbounded is for all ##M \epsilon \mathbb{R}##, there exists some ##n## such that ##\vert x_n \vert > M##. Since this is an exists statement, I don't see how we can pick a greater ##k## in 2) since it doesn't seem one is guaranteed to exist.

The k's are natural numbers: 1,2,3,4,... . There are infinitely many. There is a one-one mapping between the k's and the n's. It doesn't pick all the n's, there are gaps, but there are enough n's for this to be possible. So pick a later k, you just throw away a few n_k's from the front of the sequence but who cares about them, the desired property still holds for the tail of the sequence.

PS. Hmm, let me ponder this. I think I have not answered your question.

fishturtle1 said:
...
The definition of a sequence ##x_n## being unbounded is for all ##M \epsilon \mathbb{R}##, there exists [an] ##n## such that ##\vert x_n \vert > M##. Since this is an exists statement, I don't see how we can pick a greater ##k## in 2) since it doesn't seem one is guaranteed to exist.

I think ##|x_n|## means the tail of that sequence.

is ##c \in [a,b]## ?

From what I can tell it shows up in the very first line of the original post and nowhere else? It would be nice to know what the domain is...

verty said:
The k's are natural numbers: 1,2,3,4,... . There are infinitely many. There is a one-one mapping between the k's and the n's. It doesn't pick all the n's, there are gaps, but there are enough n's for this to be possible. So pick a later k, you just throw away a few n_k's from the front of the sequence but who cares about them, the desired property still holds for the tail of the sequence.

PS. Hmm, let me ponder this. I think I have not answered your question.

I think I get it. We have ##f(x_{n_k})## is unbounded. From this we have for all ##M > 0##, there exists ##m \epsilon \mathbb{N}## such that ##f(x_{n_{k_m}})##. From this we have ##f(x_{n_{k_m}}) < f(x_{n_{k_{m+1}}})##. (maybe this should be ##\le##) .. So we can keep picking more and more positive ##m## until we have a ##k_m## that contradicts 1) from the OP. But this, I think, relies on ##f(x_{n_{k_m}}) < f(x_{n_{k_{m+1}}})## being a strictly less than inequality and not ##\le##..

edit: just to expand on this,
we have for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##k > N(\varepsilon)## implies ##\vert f(x_{n_k}) - f(x_0) \vert < \varepsilon##.

But it seems we can keep picking a more and more positive ##k_m## until ##k_m > N(\varepsilon)## which would implies ##\vert f(x_{n_k}) \vert > \varepsilon + \vert f(x_0)\vert ## which would give us a contradiction. edit2: actually I don't think this will give a contradiction.. going to write this out on paper..

Last edited:
StoneTemplePython said:
is ##c \in [a,b]## ?

From what I can tell it shows up in the very first line of the original post and nowhere else? It would be nice to know what the domain is...
Sorry i was trying to write a general statement, but yes I intended for ##c## to be in ##[a,b]##. ill edit it

maybe i should have just wrote the same limits from the proof..

fishturtle1 said:
Sorry i was trying to write a general statement, but yes I intended for ##c## to be in ##[a,b]##. ill edit it

ok, so you know

##\big \vert f(x_n) \big \vert \leq \text{max}\Big( \big \vert f(y_0) \big \vert, \big \vert f(x_0)\big \vert\Big)##

for all ##x_n##

(notationally I used your ##x_0## but it seems to overload ##x_n## which isn't great... I probably would have used some greek letters here)

fishturtle1 said:
I think I get it. We have ##f(x_{n_k})## is unbounded. From this we have for all ##M > 0##, there exists ##m \epsilon \mathbb{N}## such that ##f(x_{n_{k_m}})##. From this we have ##f(x_{n_{k_m}}) < f(x_{n_{k_{m+1}}})##. (maybe this should be ##\le##) .. So we can keep picking more and more positive ##m## until we have a ##k_m## that contradicts 1) from the OP. But this, I think, relies on ##f(x_{n_{k_m}}) < f(x_{n_{k_{m+1}}})## being a strictly less than inequality and not ##\le##..

I wouldn't worry about this strictness of the above inequality here. This is the gist of it.

The idea is if the limiting magnitude tends to infinity, then for any ##\lambda \in (0, \infty)## I can find ##\big \vert f(x_n) \big \vert \gt \lambda## by selecting large enough ##n##. But that doesn't jive with the above inequalities...

StoneTemplePython said:
ok, so you know

##\big \vert f(x_n) \big \vert \leq \text{max}\Big( \big \vert f(y_0) \big \vert, \big \vert f(x_0)\big \vert\Big)##

for all ##x_n##

How do you get this?

fishturtle1 said:
How do you get this?

I may have mis-read your original post, but I was focusing on

fishturtle1 said:
Then ##f## is a bounded function. Moreover, f assumes its maximum and minimum values on [a,b]; that is there exist ##x_0, y_0## in [a,b] such that ##f(x_0) \le f(x) \le f(y_0)## for all ##x \epsilon [a,b]##.

it's possible I mis-read the difference between ##x## and ##x_n##?

- - - -
Another way of getting at this desired results is:

If the sequence converges to the image of ##c## under ##f## whose magnitude is bounded then it cannot converge to to something that gets mapped to infinite magnitude under ##f##. This reads as

##\lim_{n \rightarrow \infty} f(x_n) = f(c) \leq \text{max}\Big( \big \vert f(y_0) \big \vert, \big \vert f(x_0)\big \vert\Big) \lt \infty##

fishturtle1 said:
How do you get this?
Ok so using your definition of unbounded (which i agree with and I realized mine had a typo) and picking up from the end of Ross's paragraph: We have
1) ##\lim_{k\rightarrow\infty} f(x_{n_k}) = f(x_0)## and
2) ##\lim_{k\rightarrow\infty} \vert f(x_{n_k}) \vert = +\infty##.

Using definition of limit we get

1) for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that ##k > N(\varepsilon)## implies ##\vert f(x_{n_k}) - f(x_0) \vert < \varepsilon## which implies ##\vert f(x_{n_k}) \vert - \vert f(x_0) \vert < \varepsilon## which is equivalent to ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon##. Fix this ##\varepsilon## for the following sentences.

2) for ##\vert f(x_0) \vert + \varepsilon##, there exists infinitely many ##k##'s ##\vert f(x_{n_k} \vert > \vert f(x_0) \vert + \varepsilon##.

Thus we need only choose a positive enough ##k## such that from from 1) we have ##\vert f(x_{n_k}) \vert < \vert f(x_0) \vert + \varepsilon## and from 2) we have ##\vert f(x_{n_k} \vert > \vert f(x_0) \vert + \varepsilon##, a contradiction.

StoneTemplePython said:
I may have mis-read your original post, but I was focusing on
it's possible I mis-read the difference between ##x## and ##x_n##?
I think you might have assumed what we are trying to prove?

StoneTemplePython said:
- - - -
Another way of getting at this desired results is:

If the sequence converges to the image of ##c## under ##f## whose magnitude is bounded then it cannot converge to to something that gets mapped to infinite magnitude under ##f##. This reads as

##\lim_{n \rightarrow \infty} f(x_n) = f(c) \leq \text{max}\Big( \big \vert f(y_0) \big \vert, \big \vert f(x_0)\big \vert\Big) \lt \infty##
Doesn't this assume ##f## is bounded which we are trying to prove?

I think I should have written my OP clearer.. My confusion is reading Ross's proof of Extreme value theorem. In the first paragraph which I wrote in relevant equations, he proves f is bounded by first assuming it is not bounded. And the contradiction he gets is what I was confused on... but I think I understand it in Post #12?

I just realized putting what I was trying to prove in "Relevant equations" is really misleading.. SORRY for the confusion. I should have put that in the first or third part of the template

## 1. What is the Extreme Value Theorem?

The Extreme Value Theorem is a mathematical theorem that states that for a continuous function on a closed and bounded interval, there exists at least one point where the function reaches its maximum value and at least one point where it reaches its minimum value.

## 2. What is the significance of the Extreme Value Theorem?

The Extreme Value Theorem is significant because it guarantees the existence of maximum and minimum values for continuous functions on specific intervals. This allows us to make conclusions about the behavior of the function and use it to solve real-world problems.

## 3. What is the proof for the Extreme Value Theorem?

The proof for the Extreme Value Theorem involves the use of the Bolzano-Weierstrass Theorem and the Intermediate Value Theorem. It also utilizes the concepts of supremum and infimum to show that a continuous function on a closed and bounded interval must have a maximum and minimum value.

## 4. Can the Extreme Value Theorem be applied to all functions?

No, the Extreme Value Theorem can only be applied to continuous functions on closed and bounded intervals. It cannot be applied to discontinuous functions or functions on open intervals.

## 5. How is the Extreme Value Theorem used in real-world applications?

The Extreme Value Theorem is used in various fields such as economics, physics, and engineering to solve optimization problems. It helps us find the maximum or minimum values of a function, which can be used to make decisions and improve efficiency in various systems and processes.

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