1. ## [Solved] Continuity and f(c)=c

Thanks in advance for any help. I think I might just be missing something simple.

Problem:
Let f be a continuous function who's domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1]. Prove that there exists some value c in [0,1] such that f(c)=c .

Hints Provided:
1) consider the function g(x)=x-f(x) . Argue that g is continuous on [0,1] . Draw some inference about g(0) and g(1), specifically, are they positive or negative. Apply Intermediate Value Theorem to the function g.

What I have so far:

F(x) must be equal to x, ie. f(x)=x

0<=c<=1
0<=f(c)<=1
f(0)<=f(c)<=f(1)

Using the above I have proved that both f(c) and c are in the interval [0,1], but not that f(c)=c.

I kind of feel like I am not supposed to assume an actual function for f (not assume that f(x)=x), but I am not sure how to proceed then.

I'm not quite sure where the hint regarding g(x) fits in. If f(x) is in fact x then g(x)=x-x = 0, so neither positive or negative.

2. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
Thanks in advance for any help. I think I might just be missing something simple.
Problem:
Let f be a continuous function who's domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1]. Prove that there exists some value c in [0,1] such that f(c)=c .
Hints Provided:
1) consider the function g(x)=x-f(x) . Argue that g is continuous on [0,1] . Draw some inference about g(0) and g(1), specifically, are they positive or negative. Apply Intermediate Value Theorem to the function g.
Note that $\displaystyle g(0)=-f(0)\le 0~\&~g(1)=1-f(1)\ge 0.$
Thus $\displaystyle \exists~ c\in [0,1]$ so $\displaystyle g(c)=0.$ WHY?

3. ## Re: Continuity and f(c)=c

Thanks for the quick response!

I get that $\displaystyle g(0)=-f(0)\le 0~\&~g(1)=1-f(1)\ge 0.$
but not why $\displaystyle g(c)=0.$

Is this right?:

Originally Posted by Intermediate Value Theorem
If f is a real-valued continuous function on the interval [a, b], and N is a number between f(a) and f(b), then there is a c ∈ [a, b] such that f(c) = N.
Using the theorem, I come up with a value of N that is equal to 0 because:
$\displaystyle g(0)\le N \le g(1)$
and because N is a value greater than $\displaystyle g(0)\le 0$ and less than a value $\displaystyle g(1)\ge 0$.

so $\displaystyle c \in [0,1]$ such that $\displaystyle g(c)=0$.
$\displaystyle g(c)=0=c-f(c)$

So then c can be equal to any value [0,1] and g(c)=0 will still hold true, right? If so... that's the answer to the problem, I think.

4. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
I get that $\displaystyle g(0)=-f(0)\le 0~\&~g(1)=1-f(1)\ge 0.$
but not why $\displaystyle g(c)=0.$
Is this right?:Using the theorem, I come up with a value of N that is equal to 0 because:
$\displaystyle g(0)\le N \le g(1)$
and because N is a value greater than $\displaystyle g(0)\le 0$ and less than a value $\displaystyle g(1)\ge 0$.
so $\displaystyle c \in [0,1]$ such that $\displaystyle g(c)=0$.
$\displaystyle g(c)=0=c-f(c)$
That is correct.

5. ## Re: Continuity and f(c)=c

Awesome! Thank you!

6. ## Re: Continuity and f(c)=c

You need to note that if f(0)= 0 we are done so we can assume that f(0) is NOT 0. And since f maps [0, 1] to [0, 1], we must have f(0)> 0 whence g(0)= 0- f(0)< 0.
If f(1)= 1 we are done so we can assume that f(1) is NOT 1 which means that f(1)< 1 and then g(1)= 1- f(1)> 0.

Since f is continuous, so is g and so there exist c such that g(c)= 0.

7. ## Re: Continuity and f(c)=c

I'm... not sure I get it. We're assuming that $\displaystyle f(0) \neq 0$ even though we later prove that $\displaystyle f(c) = c$ so $\displaystyle f(0) = 0$? Same for $\displaystyle f(1)$?

Because the assignment was to prove that $\displaystyle f(c)=c$, why are we assuming the opposite?

8. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
I'm... not sure I get it. We're assuming that $\displaystyle f(0) \neq 0$ even though we later prove that $\displaystyle f(c) = c$ so $\displaystyle f(0) = 0$? Same for $\displaystyle f(1)$?
Because the assignment was to prove that $\displaystyle f(c)=c$, why are we assuming the opposite?.
No you did not show. You showed that $\displaystyle f(c)=0$ for at least one $\displaystyle c\in[0,1]$.
You cannot know that $\displaystyle c=0\text{ or }c=1$.

9. ## Re: Continuity and f(c)=c

Am I right in thinking we can assume what HallsofIvy stated because if $\displaystyle f(0)=0$, then
$\displaystyle g(0)= 0 - f(0) = 0 - 0 = 0$
&
$\displaystyle g(1)= 1 - f(1) = 1 - 1 = 0$
^
And that does not allow us to continue on to find $\displaystyle g(c)=0$

Ok... so, taking that into account, the function $\displaystyle f(c)=c$ is only true along the domain (0,1), not [0,1], which means that $\displaystyle f$ becomes a piecewise defined function? Or that there's no way to know that (at the moment). But I did (like you said) successfully prove that $\displaystyle f(c)=c$ for at least one $\displaystyle c \in [0,1]$, which was the assignment.

$\displaystyle g(c) = 0 = c - f(c)$
$\displaystyle 0 = c - f(c)$
$\displaystyle f(c)=c$

10. ## Re: Continuity and f(c)=c

(Not a double post, separated intentionally as I am introducing a new question)
I'm not sure why I'm having such a hard time getting this

If I could trouble you for some further help:
This is part 2 of the problem, I tried to use what I got in part 1, but...

Part 2: Give an example to show that the result of Part 1 is not true if $\displaystyle f$ is not continuous, that is, give a example of a function $\displaystyle f$ whos domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1], but $\displaystyle f$ is not continuous and $\displaystyle f(x) \neq x$ for any $\displaystyle x$ in [0,1].

For part of my answer of part 1, I argued that $\displaystyle g(x) = x - f(x)$ is continuous because $\displaystyle g(x)$ becomes a polynomial (specifically a binomial), and polynomials are continuous at all points in their domain. I understand that if $\displaystyle f(x)$ was not continuous, $\displaystyle g(x)$ would not be, and therefore it would invalidate the rest of my work.

I'm having a hard time coming up with another $\displaystyle f(x)$ with the specified domain (that is not continuous). If a function has a domain and range of [0,1], is there any function thats not continuous? I know that $\displaystyle f(x)=x$ is not the only possible function (for example, $\displaystyle f(x)=x^2$ would satisfy the requirements, but it is also continuous).

11. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
If a function has a domain and range of [0,1], is there any function thats not continuous?
Piecewise defined function are often not continuous.

12. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
Am I right in thinking we can assume what HallsofIvy stated because if $\displaystyle f(0)=0$, then
$\displaystyle g(0)= 0 - f(0) = 0 - 0 = 0$
&
$\displaystyle g(1)= 1 - f(1) = 1 - 1 = 0$
^
That just goes back to my original confusion...

Ok... so, taking that into account, the function $\displaystyle f(c)=c$ is only true along the domain (0,1), not [0,1], which means that $\displaystyle f$ becomes a piecewise defined function? Or that there's no way to know that (at the moment). But I did (like you said) successfully prove that $\displaystyle f(c)=c$ for at least one $\displaystyle c \in [0,1]$, which was the assignment.

$\displaystyle g(c) = 0 = c - f(c)$
$\displaystyle 0 = c - f(c)$
$\displaystyle f(c)=c$
I don't think you really understand any of this.
Hall's suggestion is this: IF either $\displaystyle f(0)=0\text{ or }f(1)=1$ then stop. You have found your $\displaystyle c$.

So suppose that neither of those is true. We have $\displaystyle g(x)=x-f(x)$, so that $\displaystyle g$ is clearly a continuous function.
Moreover, $\displaystyle g(0)=0-f(0)<0\text{ and }g(1)=1-f(1)>0$.
Thus $\displaystyle g(0)<0<g(1)$ so by the intermediate value theorem $\displaystyle \exists c\in [0,1]$ such that $\displaystyle g(c)=0.$
But $\displaystyle 0=g(c)=c-f(c)$, therefore $\displaystyle f(c)=c$.

13. ## Re: Continuity and f(c)=c

Oh! I get it. I'm not sure why I was hung up the way I was. Its not that $\displaystyle f(c) \neq c$ for 0 and 1. You have to make that assumption to be able to prove that $\displaystyle f(c)=c$
Ok. Cool. Thanks!

emakerov, I read your post and went "oh, duh!" XD haha, thank you.
So I came up with the following:
$\displaystyle f(x) = \begin{cases} 1 & \text{if } x = 0 \\ \frac{1}{\left | x-1 \right |} & \text{if }0 < x < 1\\ 0 & \text{if }x = 1 \end{cases}$

so then:

$\displaystyle g(x)=x-f(x)$
$\displaystyle g(0)=0-1 \text{ and } g(0) \geq -1$
$\displaystyle g(1)=1-0 \text{ and } g(1) \leq 1$

Intermediate Value Theorem cannot be applied because $\displaystyle g(x)$ is no longer a continuous function (even though all i did was mess with the one value).... and... therefore the result of part 1 ($\displaystyle f(c)=c$) is no longer true.

14. ## Re: Continuity and f(c)=c

Originally Posted by FrozenDragoon
Part 2: Give an example to show that the result of Part 1 is not true if $\displaystyle f$ is not continuous, that is, give a example of a function $\displaystyle f$ whos domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1], but $\displaystyle f$ is not continuous and $\displaystyle f(x) \neq x$ for any $\displaystyle x$ in [0,1].
Originally Posted by FrozenDragoon
So I came up with the following:
$\displaystyle f(x) = \begin{cases} 1 & \text{if } x = 0 \\ \frac{1}{\left | x-1 \right |} & \text{if }0 < x < 1\\ 0 & \text{if }x = 1 \end{cases}$
This does not satisfy the problem's requirements because the codomain of f is not [0, 1] (it's $\displaystyle [1,\infty)\cup\{0\}$).

Think about it in terms of graphs. You need to draw one or several lines inside the square [0, 1] x [0, 1] so that they don't intersect the diagonal y = x. You can shift around portions of these lines or even individual points, as long as the result is a graph of a function (i.e., there is one and only one point above every $\displaystyle x\in[0, 1]$). For example, can you make minimal changes to the graph of f(x) = 1/2 so that it does not intersect the diagonal?

15. ## Re: Continuity and f(c)=c

Well, the graph of f(x)=1/2 is just a straight line on y=.5, but if you make f(x)=(1/2)x, then they don't intersect, because the slope is less. Except where f(0)=0. So just define it piecewise for that x, right?

But if we use the function f(x)=(1/2)x, then is the co-domain still [0,1]? With a domain of [0,1], f(x) will never be a value above .5 (though I suppose that would be the range, instead).
(f(x)=(1/2)x+1/2 also works, but with the same problem, just on 1).
^
But both of those functions are continuous, so I'd need something like
$\displaystyle f(x) = \begin{cases} .5 & \text{if } 0\leq x < .5 \\ .3 & \text{if }.5\leq x < \leq 1 \end{cases}$

Plots on Wolfram Alpha

So then, most of the work transfers, that there is no way to get a $\displaystyle f(c)=c$ using $\displaystyle g(x)$ now that $\displaystyle g(x)$ is not continuous.

Page 1 of 2 12 Last