Proving the Existence of ##x## for ##f(x)=x##

[Lee33]

Homework Statement

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?

 

[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?

 

[Lee33]

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

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

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

 

[haruspex]

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

 

[Lee33]

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

 

[brmath]

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

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

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

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.

 

[Lee33]

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

 

[brmath]

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

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

 

[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 $f(c) = x$?

 

[haruspex]

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

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)?

 

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

 

[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?

 

[brmath]

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?

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?

 

[haruspex]

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 $f(c) = x$.

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].

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?

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...

 

[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.

 

[brmath]

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.

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.

 

[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?

 

[haruspex]

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.

How exactly? Which form of the IVT are you relying on?

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 $f(0) = 0$

Why? f(x) = (1+x)/2 satisfies those conditions, but not f(0)=0.

 

[Lee33]

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

 

[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?

 

[haruspex]

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?

f is not stated to be onto [0, 1].

 

[brmath]

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?

You are closing in on it. We consider an f which is >x for every x $\in$ [0,1]. You can easily do that for 0 < x < 1. But if f maps [0,1] into [0,1] it must be zero somewhere and it must be 1 somewhere. Also it cannot get below zero or above 1. So to stay above x on (0,1) it must be that f(0) = 0, and also that f(1) = 1. If it assumes the value 0 or 1 anywhere else on [0,1] then it can't stay above x for all x in [0,1], but will be above somewhere and below somewhere, which puts you back in the IVT case. But if f(0) = 0 then f(x) = x at x = 0.

The same exact thing applies if you try to make f(x) < x for every x in [0,1]. You get forced into f(0) = 0 and f(1) = 1. So f(x) = x at those points.

In my original answer I didn't even consider f(1) because I had already stated that f(0) = 0 and that was enough to show there was an x such that f(x) = x.

To see all this graphically, try graphing f(x) = $x^2$ on [0,1]. Where does it cross x? Suppose you try f(x) = $x^2$ + 1. You'll see that it is definitely above x for every x $\in$[0,1]. But you will also see that it doesn't fulfill the hypotheses for f. Why not?

Hope it is clearer now.

 

[Lee33]

It is clearer, thank you very much! I will work more examples on this to get a better feel for it. Thanks again!

 

[haruspex]

Fwiw, here's my method.
If there is no x s.t. f(x)= x then in particular f(0) > 0 and f(1) < 1. Defining g(x) = f(x) - x, it follows that g(0)>0 and g(1)<0. But g is continuous, so there exists an x s.t. g(x) = 0 (IVT). For that x, f(x) = x.

 

[brmath]

Fwiw, here's my method.
If there is no x s.t. f(x)= x then in particular f(0) > 0 and f(1) < 1. Defining g(x) = f(x) - x, it follows that g(0)>0 and g(1)<0. But g is continuous, so there exists an x s.t. g(x) = 0 (IVT). For that x, f(x) = x.

very nicely said

 

[Lee33]

haruspex - Now I see why you introduced g(x) = f(x) - x so that you can use IVT. Sorry for my ignorance. Thank you both for all the help!