# Continuous map

1. Nov 9, 2013

### Lee33

1. The problem statement, all variables and given/known data

Let $f$ be a continuous map from $[0,1]$ to $[0,1].$ Show that there exists $x$ with $f(x)=x.$

2. The attempt at a solution

I have $f$ being a continuous map from $[0,1]$ to $[0,1]$ thus $f: [0,1]\to [0,1]$. Then I know from the intermediate value theorem there exists an $x$ with $f(x)=x$ but I don't know how to formally prove it?

2. Nov 9, 2013

### brmath

We know that if g(x) = x then g(0) = 0 and g(1) = 1. Suppose f(x) > x on (0,1]. In order to map [0,1] into [0,1] f(0) must exist and you have no choice for it but f(0) = 0 (why not?). So you have f(0) = g(0).

Suppose f(x) < x on (0,1]. what could you say?

Then suppose f(x) > x on some portion of the interval and < x on the rest of the interval. What does that tell you?

3. Nov 9, 2013

### Lee33

What can I say about $f(x)<x$ on $(0,1]$? That $f(1) = g(1)$?

Then by the IVT there exists a $c$ such that $f(c) = x$?

4. Nov 9, 2013

### haruspex

Consider g(x) = f(x) - x. Continuous? IVT? Any solutions to g(x) = 0?

5. Nov 9, 2013

### Lee33

haruspex - I've seen $g(x) = f(x) - x$ before when using IVT. Why is it so useful to consider this first?

6. Nov 9, 2013

### brmath

No you can't say that. Consider the case where f(x) = x/2. It is < x on (0,1].

That's right. So you got to use your theorem. But there are those 2 odd cases where you have to think about what is going on instead of clobbering them with a theorem.

People look at f(x) -x because they want to apply theorems about the zeros of functions. In this case it isn't necessary.

7. Nov 10, 2013

### Lee33

brmath - Thanks for clarifying. One last question, what can I say about $f(x)<x$ on $(0,1]$?

8. Nov 10, 2013

### brmath

Does it intersect x on (0,1]? And if not look at my answer for f(x) > x on (0,1].

9. Nov 10, 2013

### Lee33

Brmath - Can I say, since the interval $[0,1]$ in connected and the function $f$ is continuous then the image of $f$ is connected thus by the IVT there exists a c such that $f(c) = x$?

Last edited: Nov 10, 2013
10. Nov 10, 2013

### haruspex

Not essential, of course, but it does make it quite easy.
Suppose there is no x s.t. f(x)= x. So f(0) > 0, f(1)<1. What does that tell us about g(0) and g(1)?

11. Nov 10, 2013

### Lee33

Suppose there is no x s.t. f(x)= x. So f(0) > 0, f(1)<1. What does that tell us about g(0) and g(1)?

That $g(0) \ne f(0)$?

12. Nov 10, 2013

### Lee33

If you guys don't mind. Can you give a detailed explanation of this problem? I really am trying to understand this.

What I get so far is that since $f$ is continuous and the interval $[0,1]$ is connected then I know from IVT there exists a $c$ such that $f(c) = x$.

What I don't understand is why introduce a new function $g(x)$ with $f(x)$ and then imply $f(0)>0$, $f(1)<1$ and so on. I don't see the connection?

13. Nov 10, 2013

### brmath

You have 3 potential cases:(1) f(x) > x for every x in the interval [0,1];(2) f(x)< x for every x in the interval [0,1];(3) f(x) is sometimes greater than x and sometimes less than x in the interval [0,1]. You correctly stated that the intermediate value theorem proves that f(x) must equal x somewhere if we have the third case.

That brings us to the other two cases. What if f(x) > x for every x on [0,1]. Then your theorem would not be true. So you have to say, well either the theorem isn't true and my prof messed up or it isn't possible for f(x) >x on [0,1]. I went with the idea that it isn't possible, and f(x) and x have to cross at least at 0. Same for f(x) < x. Please make sure you understand this argument. The endpoints are different from the middle points of the interval.

Next, the g(x). I said g(x) = x not because we really need another function running around there, but because it was the easiest way for me to denote the value of x at 0. Probably there was some way of saying it which wouldn't have caused confusion for you. Perhaps you can reword things without the g(x) -- it might clarify things for you.

Finally I want to say there is a reason why the problem took place on the closed interval [0,1]. The reason is that the theorem is not true on the open interval (0,1). Can you see why?

14. Nov 10, 2013

### haruspex

I don't see how that follows. f(x) = x/2 satisfies that, but there's no solution to f(c)=1 for c in [0,1].
There might be some confusion here because brmath and I each introduced a g(x), but they're different. Your question here seems to be in response to my post.
You agree that if there is no x s.t. f(x)= x then f(0) > 0 and f(1) < 1, yes? It follows (defining g(x) = f(x) - x) that g(0)>0 and g(1)<0. But g is continuous, so...

15. Nov 10, 2013

### Lee33

Brmath - Thank you clarifying!

Haruspex - I am not to fond with using $g(x) = f(x) - x$ here because if this question popped on a test I would never have guessed to use it.

16. Nov 10, 2013

### brmath

Lee33,

The question isn't whether you use f(x) -x = 0 (Haruspex) or f(x) = x (brmath). They are exactly equivalent; it's really about our individual ways of thinking about things.

What's important is to think through the pieces of the question. It's much easier to do that if you have spent lots of time building mathematical proofs than if you are just starting out. So don't go down the road of "I never would have thought of that". You've done this problem, and in the future you will think of it. I promise you the best mathematicians in the world have "never would have thought of that" experiences.

In this case, you could have drawn a bunch of different f's and looked at how they relate to x. That might have led you to the issue of f > x throughout the interval.

What I would add is that they are trying to teach you the intermediate value theorem, but by doing it for [0,1] rather than (0,1) they complicated matters a bit, because the closed interval drags in issues connected to the endpoints. Personally I prefer to annoy students with one issue at a time.

17. Nov 10, 2013

### Lee33

Thanks again brmath!

So to make sure I understand, this is what I know thus far:

If $f(x)> x$ at some points of $x\in [0,1]$ and $f(x)<x$ at other points $x\in [0,1]$ then we can use the IVT and we are done.

Now, like you said, consider when $f(x) >x$ for all $x\in [0,1]$. Now, to map $[0,1] \to [0,1]$ we must have that $f(0)$ must exist and since $f$ is continuous function such that $f: [0,1]\to [0,1]$ we have that $f(0) = 0$ which contradicts $f(x) < x$ for all $x\in [0,1]$.

Now consider when $f(x) < x$ then in order to map $f: [0,1]\to [0,1]$ we must have the function $f(1)$ exists but this is not that case since $f(x)< x$ which contradicts the choice of $[0,1]$. Hence contradiction.

Is this correct?

18. Nov 10, 2013

### haruspex

How exactly? Which form of the IVT are you relying on?
Why? f(x) = (1+x)/2 satisfies those conditions, but not f(0)=0.

19. Nov 10, 2013

### Lee33

Haruspex - This question is a lot harder than I expected. I am going to revise my proof now.

20. Nov 10, 2013

### Lee33

Haruspex -

f(x) = (1+x)/2 does satisfy those conditions but it does not satisfy mapping $0$ since no $x\in [0,1]$ will make $f(x) = (1+x)/2 = 0$ which is what I said in my proof, that it is a contradiction?