# Extreme value theorem, proof question

## 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?

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verty
Homework Helper
Just pick a greater k. Any greater k works as well because you have infinitely many $n_k$'s.

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.

verty
Homework Helper
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.

verty
Homework Helper
...
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.

StoneTemplePython
Gold Member
2019 Award
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...

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.. gonna write this out on paper..

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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..

StoneTemplePython
Gold Member
2019 Award
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)

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...

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?

StoneTemplePython
Gold Member
2019 Award
How do you get this?
I may have mis-read your original post, but I was focusing on

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

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

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.

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?

- - - -
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