1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 7, 2012 #1
    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[itex]\in[/itex]ℝ.

    Prove that there exists a unique a[itex]\in[/itex]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 [itex]\exists[/itex]a s.t. lim[itex]_{n→\infty}[/itex]xn = a

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

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

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

    We know that lim[itex]_{n→\infty}[/itex]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. jcsd
  3. Feb 7, 2012 #2


    User Avatar
    Science Advisor

    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, [itex]a_1[/itex] and [itex]a_2[/itex]. Let those be the x and y in "[itex]|f(x)- f(y)|\le c|x- y|[/itex]" and see what you get.
  4. Feb 7, 2012 #3
    [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 [itex]\exists[/itex]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?
  5. Feb 7, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    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.
  6. Feb 8, 2012 #5

    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 + [itex]\sqrt{xn - 2}[/itex]

    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 + [itex]\sqrt{L - 2}[/itex] ? (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?
  7. Feb 8, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper

    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)?
  8. Feb 8, 2012 #7
    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.
  9. Feb 9, 2012 #8


    User Avatar
    Science Advisor

    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.
  10. Feb 9, 2012 #9


    User Avatar
    Science Advisor

    and perhaps the number "c" might give you a clue as to how to choose delta, given epsilon.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook