Prove that an even degree poly has either a absolute max or min.

[╔(σ_σ)╝]

Homework Statement

My brain hasn't been working lately so if you see something weird in my proof pardon me and advice me against it.

So the problem states:
Show that an even degree polynomial has either an absolute max or min.

Homework Equations

The Attempt at a Solution

Let f(x) be an even degree poly.
Without loss of generality suppose, $\displaystyle\lim_{x \to\infty} f(x) = \infty$.It is obvious that f(x) is not bounded above, so we have no absolute max in this case.

Let [itex]S = $$\left\{x : f(x) \leq 0\right\}$$<br /> <br /> S is bounded below since f(x) does not go to $-\infty$.<br /> <br /> Let $A = \left\{f(x) :x \in S\right\}$<br /> <br /> A is bouded below since f(x) does not go to $-\infty$.<br /> <br /> LEt $\beta = infA$.<br /> <br /> For every n there exist $x_n \in [a,b]$ such that $\beta \leq f(x_n) < B + \frac{1}{n}$<br /> <br /> (Note a and b exist because of the f(x) does not go to $-\infty$ and f(x) goes to $\infty$ respectively. )<br /> <br /> $$f(x_n) \rightarrow \beta$$<br /> <br /> There exist [itex]x_{n_k} \rightarrow x[/tex] since [itex]x_n[/tex] is bounded.<br /> <br /> Also [itex]x \in [a,b] [/tex]By uniqueness of limits and continuity $f(x_{n_k}) \rightarrow f(x) = \beta$<br /> <br /> For the case where f(x) > 0 for all x and f(x) still goes to infinity as x goes to infinity I used the set $S = \left\{x : f(x) \leq f(0)= a_0 \right\}$ then the proof is analogous.<br /> <br /> Also the proof is the same when $\displaystyle\lim_{x \to\infty} f(x) = -\infty$ except we find a max instead.[/itex][/itex][/itex][/itex]

 

[micromass]

All seems well, save for one little thing. Where did you use that the polynomial has even degree?

 

[╔(σ_σ)╝]

I used it in my limits as x goes to +/- infinity. I used the property that the limits are the same as x goes both ways.

 

[micromass]

Yes. It seems your proof is completely correct then...

 

[╔(σ_σ)╝]

Thanks. Do you know of a better or easier proof?
Mine seems like overkill. It uses too many things.
I used Bolzano Weierstrass, completeness (sup and inf), and at least 2 limit theorems. :-(

 

[Dick]

Really you just need that f is continuous and lim f(x)->infinity as |x|->infinity. That means there is an M such that for all x>M f(x)>f(0). And there is an m such that for all x<m f(x)>f(0). There's nothing special about f(0), I just picked 0. Now you need the extreme value theorem on [m,M]. I don't think that's overkill. It's what you need.

 

[╔(σ_σ)╝]

Really you just need that f is continuous and lim f(x)->infinity as |x|->infinity. That means there is an M such that for all x>M f(x)>f(0). And there is an m such that for all x<m f(x)>f(0). There's nothing special about f(0), I just picked 0. Now you need the extreme value theorem on [m,M]. I don't think that's overkill. It's what you need.

That is pretty much what I did, in a sence, but I forgot about the extreme value theorem.

Your proof seems a lot easier to formulate; cool!

So from what I see you basically get rid of the big intevals where f(x) is big; form a closed interval and apply EVT. That's cool. You also use the fact that f(x) is bounded on [m,M]. Am i right?

I guess my proof is not too much stuff then (apart from the two sets ). I guess this question calls for a lot of things. I just thought there might be a really clever proof out there.

 

[Dickfore]

What is the behavior of the polynomial as $x \rightarrow \pm \infty$?

 

[╔(σ_σ)╝]

It's infinity. Why do you ask ?

 

[Dickfore]

What is the sign of the infinity on each side?

 

[╔(σ_σ)╝]

Positive.
I already know this.
I don't see the point of these questions. Can you be more direct if you have an alternative solution?
I already solved the problem afterall XD.

Edit

positive; assuming on is the leading coefficient is positive.

 

[Dick]

That is pretty much what I did, in a sence, but I forgot about the extreme value theorem.

Your proof seems a lot easier to formulate; cool!

So from what I see you basically get rid of the big intevals where f(x) is big; form a closed interval and apply EVT. That's cool. You are still using the fact that f(x) is bounded on [m,M]. Am i right?

I'm really just doing the same thing you did with more convenient names. EVT is a form of Bolzano-Weierstrass. I don't need to show f(x) is bounded on [m,M]. Polynomials are continuous. Continuous functions are bounded on closed intervals. That BW or EVT, whatever you want to call it.

 

[Dickfore]

Let $S = \left\{x : f(x) \leq 0\right\}$

S is bounded below since f(x) does not go to $\mathbf{-\infty}$.

Why?

EDIT:

example:

$$f(x) = x^{2} + 1$$

What is your S in this case?

 

[╔(σ_σ)╝]

I'm really just doing the same thing you did with more convenient names. EVT is a form of Bolzano-Weierstrass. I don't need to show f(x) is bounded on [m,M]. Polynomials are continuous. Continuous functions are bounded on closed intervals. That BW or EVT, whatever you want to call it.

Well I called it EVT because you talked about the extreme value theorem. It is a lemma in my book.

Hmm.. BW is the same as EVT? I don't see it! BW talks about sequences while EVT is about well functions. I understand that we can look at f(x_n) as a sequence but I don't see how this relates to a bounded sequence having a convergent subsequence. In fact, BW and completeness is used to prove EVT so I really don't see the direct relationship.

 

[╔(σ_σ)╝]

Why?

EDIT:

example:

$$f(x) = x^{2} + 1$$

What is your S in this case?

The first line of my proof states the assumption that lim(x-> infinity) f(x) = infinity. Which implied that the leading coefficient is positive.

Did you read the proof ? S is some set of x's...

Since the behaviour of f(x) is to approach positive infinity as x gets very large or small I know that the x's such that f(x) <= 0 cannot be unbounded below.

Edit
Keep reading the proof.

I said that if f(x) > 0 for all x then I will replace f(0) with zero in my initial S.

 

[Dickfore]

The first line of my proof states the assumption that lim(x-> infinity) f(x) = infinity. Which implied that the leading coefficient is positive.

Did you read the proof ? S is some set of x's...

Yes, I read the proof. My example is an even degree (second degree) polynomial. Your set S is defined as the values for x for which the function is not positive. What is this set S in my particular example? Is your claim about the boundedness of S justified?

Since the behaviour of f(x) is to approach positive infinity as x gets very large or small I know that the x's such that f(x) <= 0 cannot be unbounded below.

But, I don't. Convince us.

 

[╔(σ_σ)╝]

Yes, I read the proof. My example is an even degree (second degree) polynomial. Your set S is defined as the values for x for which the function is not positive. What is this set S in my particular example? Is your claim about the boundedness of S justified?



But, I don't. Convince us.

Review my edit.


Well if S was unbounded below it means that for, x in S, if x is in any neighbourhood of -infinity then f(x) <0. Which is contrary to my assumption that f(x) gets large as x gets more and more negative.

In my original proof I should have replaced zero with f(0) in the beginning. I did it at the end which was not so logically sound. The proof can to me in peices so as I was typing the proof out I started to notices gaps and I filled them up at the end. I have to admit, I should have organised the proof a lot better.

 

[Dick]

Well I called it EVT because you talked about the extreme value theorem. It is a lemma in my book.

Hmm.. BW is the same as EVT? I don't see it! BW talks about sequences while EVT is about well functions. I understand that we can look at f(x_n) as a sequence but I don't see how this relates to a bounded sequence having a convergent subsequence. In fact, BW and completeness is used to prove EVT so I really don't see the direct relationship.

EVT is BW applied to a compact interval in the real numbers. EVT is corollary of BW. I don't know why I'm saying this. You already said it.

 

[Dickfore]

Ok. So where do you use the set S in your proof?

 

[╔(σ_σ)╝]

I used S to define A. Did you actually read the proof? lol

 

[Dickfore]

I used S to define A. Did you actually read the proof? lol

ok, great job.

 

[╔(σ_σ)╝]

EVT is BW applied to a compact interval in the real numbers. EVT is corollary of BW. I don't know why I'm saying this. You already said it.

I think I see it now.:-)

With BW i could take the range of f on [a,b] and form a convergent subsequence to the max and min of f on [a,b]. Is my thinking correct ?

I am a bit slow today and it's 12:03 am lol.

 

[╔(σ_σ)╝]

ok, great job.

Thanks for been critical with what I wrote. Next time I would try to make the proof more coherent.

 

[Dick]

I think I see it now.:-)

With BW i could take the range of f on [a,b] and form a convergent subsequence to the max and min of f on [a,b]. Is my thinking correct ?

I am a bit slow today and it's 12:03 am lol.

Basically. You'll probably see it really easily tomorrow. Suppose f doesn't have a max. Can it really then be continuous?

 

[╔(σ_σ)╝]

Basically. You'll probably see it really easily tomorrow. Suppose f doesn't have a max. Can it really then be continuous?

No. Since it won't be bounded on the closed interval.

Is something wrong in my previous answer. You said basically XD. So was I wrong?

 

[Dick]

I think I see it now.:-)

With BW i could take the range of f on [a,b] and form a convergent subsequence to the max and min of f on [a,b]. Is my thinking correct ?

I am a bit slow today and it's 12:03 am lol.

You mean is something wrong with this reasoning? You certainly haven't finished the thought. And you are trying to prove there IS a max and min. You can't really start the proof by assuming there is a sequence converging to the max or min.

 

[╔(σ_σ)╝]

You mean is something wrong with this reasoning? You certainly haven't finished the thought. And you are trying to prove there IS a max and min. You can't really start the proof by assuming there is a sequence converging to the max or min.

Well I guess I can use monotone subsequence property to find a monotone sequence converging to the max or min. As long as f is continuous, then it is bounded; so I can actually talk about locating a max and min.

 

[Dick]

Well I guess I can use monotone subsequence property to find a monotone sequence converging to the max or min. As long as f is continuous, then it is bounded; so I can actually talk about locating a max and min.

I thought you were trying to prove that a continuous function on a closed interval is bounded (EVT). Perhaps not?

 

[╔(σ_σ)╝]

No, actually. Well in my book the proof that a continuous function is bounded is separate from EVT. i.e they are two different theorems.

Basically, all I am trying to do is see the relationship between BW and EVT which you talked about. You said they are equivalent but I do see it directly.

 

[Dick]

No, actually. Well in my book the proof that a continuous function is bounded is separate from EVT. i.e they are two different theorems.

Basically, all I am trying to do is see the relationship between BW and EVT which you talked about. You said they are equivalent but I do see it directly.

I just meant that the guts of the proof of the EVT is BW added to the definition of continuity. That's all. Not that they are somehow formally equivalent. And yes, your proof was on the right track. I was thinking EVT was the continuous function bounded theorem. Just confused.