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Math Help - [SOLVED] A mean value theorem problem

  1. #1
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    [SOLVED] A mean value theorem problem

    This one has been stumping me for quite some time know...
    Given f:[a,b]\rightarrow[a,b] continuous on [a,b] and differentiable on (a,b) such that there exists an \alpha\in(0,1) so that |f'(x)|<\alpha for all x\in(a,b), show that there exists a unique point c\in[a,b] such that f(c)=c.

    I can show that if there is a point, then it is unique, but showing the point exists is challenging.
    Hint: Pick x_0\in[a,b] and let x_{k+1}=f(x_k).
    I have used repeated applications of the mean value theorem to achieve the following inequality
    |x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|
    Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by nqramjets View Post
    This one has been stumping me for quite some time know...
    Given f:[a,b]\rightarrow[a,b] continuous on [a,b] and differentiable on (a,b) such that there exists an \alpha\in(0,1) so that |f'(x)|<\alpha for all x\in(a,b), show that there exists a unique point c\in[a,b] such that f(c)=c.

    I can show that if there is a point, then it is unique, but showing the point exists is challenging.
    Hint: Pick x_0\in[a,b] and let x_{k+1}=f(x_k).
    I have used repeated applications of the mean value theorem to achieve the following inequality
    |x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|
    Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!

    ahh but it is

    \alpha \in (0,1) so as k increases

    \alpha ^k \to 0, \mbox{ an } k \to \infty


    Just pick k large enough such that \alpha^k =\frac{\epsilon}{x_1-x_0}
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  3. #3
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    Quote Originally Posted by nqramjets View Post
    This one has been stumping me for quite some time know...
    Given f:[a,b]\rightarrow[a,b] continuous on [a,b] and differentiable on (a,b) such that there exists an \alpha\in(0,1) so that |f'(x)|<\alpha for all x\in(a,b), show that there exists a unique point c\in[a,b] such that f(c)=c.

    I can show that if there is a point, then it is unique, but showing the point exists is challenging.
    Hint: Pick x_0\in[a,b] and let x_{k+1}=f(x_k).
    I have used repeated applications of the mean value theorem to achieve the following inequality
    |x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|
    Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!
    Am I missing something here? You want to prove f(c)= c so that is an "intermediate value" theorem, not a "mean value" theorem.

    You need neither that the function is differentiable nor the bound on the derivative to prove "existence", perhaps to prove "uniqueness".

    This is a well known proof:
    Suppose f:[a,b]->[a,b] is continuous on [a, b]. If f(a)= a, then we are done. So we can assume f(a)> a. If f(b)= b, we are done. So we can assume f(b)< b. Let h(x)= f(x)- x. Then h(a)= f(a)-a> 0 and h(b)= f(b)- b< 0. But h(x) is also continuous so there must exist c such that h(c)= f(c)- c= 0. Therefore f(c)= c.
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  4. #4
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    It turns out this really is Cauchy, but only because if you write out |x_m-x_n| you get a partial geometric sum. The problem is that |x_{k+1}-x_k|<\epsilon doesn't necessarily make x_k Cauchy. In this though, it does. This is what was making me struggle.
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  5. #5
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    Hey this is slick, thanks. And yes, to show uniqueness you do need MVT.
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