# Homework Help: For f(x), prove that there exists a unique a such that f(a) = a

1. Feb 7, 2012

### looserlama

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

Let f: ℝ→ℝ be a function such that there exists a constant 0<c<1 for which

|f(x) - f(y)| ≤ c|x - y|​

for every x,y$\in$ℝ.

Prove that there exists a unique a$\in$ℝ such that f(a) = a.

2. Relevant equations

There's a hint that says: Consider a sequence {xn} defined recurrently by xn+1 = f(xn). Prove that it converges and its limit a satisfies f(a) = a.

3. The attempt at a solution

So I did what the hint said, I defined {xn} and proved that it was Cauchy and therefore converges. The problem is after that, I'm not sure if what I did was right, here's what I said:

{xn} converges

Therefore $\exists$a s.t. lim$_{n→\infty}$xn = a

Therefore if xm = a, then xm+1 = a

Therefore xm+1 = a = f(xm) = f(a)

Therefore $\exists$a s.t. lim$_{n→\infty}$xn = a and f(a) = a

We know that lim$_{n→\infty}$xn is unique

Therefore there exists a unique a such that f(a) = a.

I'm not sure if the limit being unique actually implies that a will be the only number such that f(a) = a ?

Also, I'm not sure if, after I state that there exists a limit, I'm allowed to just say that if xm = a, then xm+1 = a ? Or is there something I should add in there that I'm not explaining?

2. Feb 7, 2012

### HallsofIvy

First, your sequence is not well defined because you have not specified the initial value. You proof shows that the sequence converges to a specific number, but if you start with a different initial value, might it not converge to a different a?

Instead, suppose there are two such numbers, $a_1$ and $a_2$. Let those be the x and y in "$|f(x)- f(y)|\le c|x- y|$" and see what you get.

3. Feb 7, 2012

### looserlama

[STRIKE][/STRIKE]What I was thinking for the sequence is that we could just set x1 = x1, so just for some value of x1 depending on what the function f actually is.

Anyway, I tried just doing it by contradiction like you said, so I have:

Suppose $\exists$a1,a2 such that f(a1) = a1 and f(a2) = a2.

Therefore |f(a1) - f(a2)| = |a1 - a2| ≤ c|a1 - a2| < |a1 - a2|

Which is a contradiction.

Therefore there is a unique a such that f(a) = a.

But is that it? Does that proof hold?

4. Feb 7, 2012

### Dick

If you showed xn is Cauchy then you know it has a limit, call it a. You didn't show that f(a)=a yet. It has nothing to do with xn=a for some n. Take f(x)=x/2 and x0=1. Then the limit is 0, but xn is never 0.

5. Feb 8, 2012

### looserlama

Can't you say that for a recurrently defined sequence that converges, if some term was equal to the limit then the next one would have to be too?

For example, consider the sequence defined as x1 = 5/2 and xn+1 = 2 + $\sqrt{xn - 2}$

Assuming it converges to some L, then can't we say that if some xm = L, then xm+1 will also equal L?

ie, L = 2 + $\sqrt{L - 2}$ ? (Iff lim xn = L)

Also, is the proof by contradiction I did above right?
Cause it seems way simpler than using the hint, but then why would he give us the hint?

6. Feb 8, 2012

### Dick

The proof by contradiction could be a little clearer if you'd say |a2-a1|<=c|a2-a1| implies that |a2-a1|=0 so a2=a1. But no, you can't rely on having any elements of your sequence actually equal to L. Probably won't happen. You have lim xn=a. Your function f is continuous. They didn't say that but you can prove it. So lim f(xn)=f(a). What's the relation between the sequence xn and the sequence f(xn)?

7. Feb 8, 2012

### looserlama

I guess xn+1 = f(xn)
So lim xn+1 = lim f(xn)

So a = f(a)?

What I was getting at before, I guess I didn't explain it very well, was very similar I think.

What I should have said is

Lim xn+1 = lim xn

Therefore a = f(a).

I'm pretty sure that's right.

Anyway, I'm using the proof by contradiction in my assignment.

Thanks a lot to both of you for your help.

8. Feb 9, 2012

### HallsofIvy

If f is continuous then lim f(xn)= f(lim xn)= f(a) so that is correct. However, you still need to prove that any function satisfying the given condition is continuous.

9. Feb 9, 2012

### Deveno

and perhaps the number "c" might give you a clue as to how to choose delta, given epsilon.