# Applying the Extreme Value Theorem

• Feb 9th 2008, 07:59 AM
Joel24
Continuity - minimum and maximum
Hi. I'd really appreciate any help you can give me on this.

If $f: \Re \to \Re$ is a continuous, periodic function, then prove $f$ has a maximum and minimum.

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A periodic function is where if you have any K>0, then for all $x \in \Re$, $f(x + K) = f(x)$. And we need to show that a set $S = \{ f(x) | x \in \Re \}$ has a maximum and minimum.

I'm thinking that we need to show it is bounded first of all. But I have no idea how I could do that. I'm just jotting down some thoughts, so I don't know whether they are relevant.

$| f(x+K) - f(x) | = 0 < \epsilon$ for all $\epsilon > 0$

This question in the textbook is under the section of the "The extreme value theorem" so I think that must be relavent.
Regards,
Joel.
• Feb 9th 2008, 10:35 AM
Plato
Do you know the theorem that says: If $f:\mathbb{R} \to \mathbb{R}$ is continuous then on any closed interval $[a,b]$ f attains a maximum and a minimum in that interval?
If so, then what about the interval $[0,K]$?
If not, then you do need to prove that theorem.
• Feb 9th 2008, 11:37 AM
Joel24
Quote:

Originally Posted by Plato
Do you know the theorem that says: If $f:\mathbb{R} \to \mathbb{R}$ is continuous then on any closed interval $[a,b]$ f attains a maximum and a minimum in that interval?
If so, then what about the interval $[0,K]$?
If not, then you do need to prove that theorem.

Thank you, Plato.

I think I mentioned that theorem in my post. If I'm not mistaken that is the extreme value theorem (at least that's what is written in a textbook - not sure if that is the universal name for it).

Over the interval [0, K], I'm not sure. Over that interval, does $f(0+K) = f(0)$? and $f(K+K) = f(K)$ ? Don't know where to go with this. *embarrassed*
• Feb 9th 2008, 12:17 PM
Plato
Quote:

Originally Posted by Joel24
Over that interval, does $f(0) = 0$? and $f(K) = K$ ? So the maximum is K and minimum is 0?

No absolutely not. Apply the theorem to f. Then there are points $
p \in [0,K]\,\& \,q \in [0,K]$
such that $\forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p)$.

Now you need to use the property of a periodic function to prove that these are the maximum and minimum for all reals.
• Feb 9th 2008, 12:34 PM
Joel24
Quote:

Originally Posted by Plato
No absolutely not. Apply the theorem to f. Then there are points $
p \in [0,K]\,\& \,q \in [0,K]$
such that $\forall x \in [0,K] \Rightarrow \quad f(q) \leqslant f(x) \leqslant f(p)$.

Now you need to use the property of a periodic function to prove that these are the maximum and minimum for all reals.

Yes, I realised and edited my mistake just before your post. My concern is that these max and minimums don't hold. (I know they do, but I don't understand why)

A function is periodic if there exists a K>0 such that for all real x, $f(x+K) = f(x)$

We need to show that for all real x, $f(x+K) \leq f(p)$

But we only considered the period [0,K]. There has to be a number greater than K in the set of real numbers. What if k=98. Then consider the interval [-7,K+99]

$f(K+99) = f(99)$ which could be greater than $f(p)$ . :S

Apologies if I'm going off track. I'm just trying to understand this.
• Feb 9th 2008, 12:51 PM
Plato
K is the period of f. If n is an integer then $\forall x\left[ {f(x + nK) = f(x)} \right]$.
Lets say that $t > K$ then using floor function, $j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = t - Kj$.
Is it clear that $r \in [0,K]\,\& \,t = jK + r$.
Can you finish?
• Feb 9th 2008, 01:02 PM
Joel24
Quote:

Originally Posted by Plato
K is the period of f. If n is an integer then $\forall x\left[ {f(x + nK) = f(x)} \right]$.
Lets say that $t > K$ then using floor function, $j = \left\lfloor {\frac{t}{K}} \right\rfloor \,\& \,r = \frac{t}{K} - j$.
Is it clear that $r \in [0,K]\,\& \,t = jK + r$.
Can you finish?

Yes, I think so. Thank you very much.

What does $j = \left\lfloor {\frac{t}{K}} \right\rfloor$ mean?
• Feb 9th 2008, 02:15 PM
Plato
Quote:

Originally Posted by Joel24
What does $j = \left\lfloor {\frac{t}{K}} \right\rfloor$ mean?

The floor function is often called the greatest integer function.
$\left\lfloor {\frac{{99.1}}{{11}}} \right\rfloor = 9$.
Please, see my edit of the reply above: $r = t - Kj$.
• Feb 9th 2008, 02:31 PM
ThePerfectHacker
Nowhere in your initial post do you mention the function is periodic on a closed interval. Consider $f(x) = \tan x$ whenever it is defined an $f(x)= 0$ elsewhere then $f$ does not attain a maximum or minimum eventhough the function is periodic.
• Feb 9th 2008, 03:14 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
Nowhere in your initial post do you mention the function is periodic on a closed interval.

But the post made it clear that the function is continuous on the reals.