# Proving fixed point(s) exist w/ intermediate value theorem

1. Oct 4, 2016

### cmkluza

1. The problem statement, all variables and given/known data
Use the Intermediate Value Theorem to prove that any continuous function with domain [0,1] and range in [0,1] must have a fixed point.
2. Relevant equations
Intermediate Value Theorem (IVT) states that if a function $f(x)$ of domain [$a,b$] takes values $f(a)$ and $f(b)$, then it also takes any value between $f(a)$ and $f(b)$.

Fixed points are any points on the line $y=x$.
3. The attempt at a solution
I'm awful at proofs. I can understand this statement at a very abstract level, and it appears to be true. But obviously that's not proof. By the IVT, any continuous function, $f(x)$, with domain [0,1] takes all values between $f(0)$ and $f(1)$, right? I don't know where to go from here.

Are there any tips anyone can offer on how best to approach this proof?

2. Oct 4, 2016

### andrewkirk

Consider the function $g(x)=f(x)-x$.
Is it continuous?
What can we say about $f(c)$ if $g(c)=0$?
Can you use IVT to prove that $g$ must be zero somewhere on [0,1]?

3. Oct 4, 2016

### Staff: Mentor

The statement to prove is equivalent to saying that the graph has to intersect the line y = x somewhere.

4. Oct 4, 2016

### cmkluza

You'll have to forgive me, I'm pretty slow.

$g(x)$ should be continuous. If $g(c)=0$ then $f(c)=c$. Therefore, there is a fixed point so long as $g$ is zero at some point.

By IVT, I can say that there exist all values, $a$, where $g(0) \leq a \leq g(1)$.

Am I heading in the right direction? I don't see where I can use IVT to prove that $g$ must be zero somewhere on [0,1]. Perhaps I'm looking at IVT in the wrong way?

In the simplest terms, I understand IVT to state that there exist all values between the two $y$ values that exist the beginning and end of the domain of a continuous function. How can I leverage this to say that $g$ must be zero at some point? Sorry if I'm missing something extremely obvious.

5. Oct 4, 2016

### andrewkirk

What are the possible ranges of $g(0)$ and $g(1)$? Can you use that to work out whether 0 must lie between $g(0)$ and $g(1)$?

6. Oct 4, 2016

### cmkluza

The range of the function should be [0,1], from the problem statement, right? Does that have any implications as far as what $g(0)$ and $g(1)$ could be?

7. Oct 5, 2016

### Staff: Mentor

Sure. Both g(0) and g(1) have to lie somewhere in that interval.

8. Oct 5, 2016

### cmkluza

In order for $g(0)$ or $g(1)$ to encompass zero, and given that $0 \leq g(x) \leq 1$, wouldn't that imply that either $g(0)$ or $g(1)$ would have to equal zero? I don't see how I can prove that.

9. Oct 5, 2016

### Staff: Mentor

What do you mean by "encompass zero"?
No. Consider $y = g(x) = -\frac {1}{2}x^2 + x + \frac 1 2$ on the interval [0, 1]. Neither g(0) nor g(1) equals 0.

10. Oct 5, 2016

### andrewkirk

Try to reason through it. Start with $g(0)$. Since $g(x)=f(x)-x$, that is equal to $f(0)-0$. What is the range of possible values for that?

Then do the same thing for $g(1)$. Hint: it will not be the same range as for $g(0)$. Once you have written down the range of possible values for $g(0)$ and the range of possible values for $g(1)$ the next step may become apparent.

11. Oct 5, 2016

### cmkluza

I'm sorry, I'm not sure on this. We have $g(1)=f(1)-1$. Since $f(x)$ lies on [0,1] that means that $g(1)$ must either be zero or negative. Therefore, it's not bounded by that interval ([0,1]), but $f(x)$ is. Is that what you were saying? I think I got mixed up with the bounds and the newly defined function $g$.
I think I finally have it. $g(0)=f(0)$, and since $f(x)$ is in [0,1]. then $g(0)=f(0)$ is either zero or positive. By my logic above, $g(1)$ is either zero or negative. Therefore, by IVT, there exists some value, $c$, between 0 and 1 where $g(c) = f(c) - c = 0$, which satisfies the proof, right?

12. Oct 5, 2016

### andrewkirk

Almost. You need to state, before you apply the IVT, that 0 lies between g(0) and g(1). That follows from the middle sentence of your post, but it needs to be stated before you apply IVT.

Depending on how relaxed your lecturer is, they may want you to also prove that $g$ is continuous rather than assuming it to be obvious. That takes two steps, given that $f$ is continuous. The first is to prove that the function $x\mapsto x$ is continuous which, if you haven't already been given it, is easy to show with an $\epsilon,\delta$ argument. For the second step there is a basic theorem about continuous functions that can be used.

13. Oct 5, 2016

### Staff: Mentor

You're mixing up what I said with a reply that came later.

I was replying to what you said in post #6:

14. Oct 6, 2016