# Thread: Applying the Extreme Value Theorem

1. ## 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.

2. 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.

3. 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*

4. 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.

5. 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.

6. 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?

7. 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?

8. 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: $r = t - Kj$.

9. 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.

10. 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.