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Extreme value theorem, proof question

  • #1
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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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Answers and Replies

  • #2
verty
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Just pick a greater k. Any greater k works as well because you have infinitely many ##n_k##'s.
 
  • #3
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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.
 
  • #4
verty
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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.
 
  • #5
verty
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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.
 
  • #6
StoneTemplePython
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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...
 
  • #7
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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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  • #8
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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..
 
  • #9
StoneTemplePython
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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...
 
  • #10
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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?
 
  • #11
StoneTemplePython
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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##
 
  • #12
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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.
 
  • #13
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28
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
 

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