Extreme value theorem, proof question

[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?

 

[verty]

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

 

[fishturtle1]

Thanks for the reply.

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]

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]

Thanks for the reply.
...
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]

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...

 

[fishturtle1]

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..

 

[fishturtle1]

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]

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...

 

[fishturtle1]

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]

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

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

 

[fishturtle1]

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.

 

[fishturtle1]

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