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Math Help - Finding functions

  1. #1
    Super Member Showcase_22's Avatar
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    Finding functions

    So far, I have this:

    Proving |f(x)-f(y)| is continuous:

    Need to show: \forall \ \epsilon>0 \exists \delta>0 s.t \ |y-x| < \delta \ \Rightarrow |f(x)-f(y)|< \epsilon.

    |y-x|< \delta \Rightarrow |y-x|^2< \delta^2

    Therefore: |f(x)-f(y)| \leq (y-x)^2< \delta^2

    Let \delta^2=\epsilon

    Hence |f(x)-f(y)| \leq (y-x)^2< \delta^2=\epsilon so continuity is proved.

    However, i'm sure this helps but I can't see why. I've also tried expanding the right and left hand sides but that doesn't help either.

    Does anyone have any hints?
    Last edited by Showcase_22; February 7th 2009 at 07:02 AM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,

    You're right, the functions f satisfying |f(x)-f(y)|\leqslant (x-y)^2 for all x,y\in\mathbf{R} are continuous on \mathbf{R}. In fact they are even differentiable on \mathbf{R}. Can you show it ?
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  3. #3
    Super Member Showcase_22's Avatar
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    |f(x)-f(y)| \leq (x-y)^2

    \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \leq \frac{h^2}{h}=h \rightarrow 0.

    Setting h=y-x.

    Therefore:

    |f(x)-f(y)| \leq (x-y)^2 \ \forall x,y \in \mathbb{R}

    Hence |f(x)-f(y)| \leq 0

    0 \leq f(x)-f(y) \leq 0 \Rightarrow \ f(x)=f(y)

    Therefore f(x) and f(y) are the same number.

    Therefore function f is a constant polynomial.
    Last edited by Showcase_22; February 7th 2009 at 06:42 AM.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    |f(x)-f(y)| \leq (x-y)^2

    \lim_{h \rightarrow 0} \frac{f(x+h)}{h} \leq \frac{h^2}{h}=h \rightarrow 0.
    You forgot one term: \lim_{h \rightarrow 0} \frac{f(x+h){\color{blue}-f(x)}}{h} \leqslant \frac{h^2}{h}=h \rightarrow 0.
    Anyway this does not show that f is differentiable at x since it is possible for \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty and \lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0 to hold simultaneously.

    Hint: use the same approach to show that |f'(x)|\leqslant 0.
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  5. #5
    Super Member Showcase_22's Avatar
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    Sorry about the first part, it was a typo.

    Quote Originally Posted by flyingsquirrel View Post
    You forgot one term: [tex]
    Anyway this does not show that f is differentiable at x since it is possible for \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty and \lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0 to hold simultaneously.
    Since we're taking the modulus of f(x)-f(y), wouldn't this imply |-\infty| \leq 0 giving +\infty \leq 0 which is a contradiction?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Since we're taking the modulus of f(x)-f(y), wouldn't this imply |-\infty| \leq 0 giving +\infty \leq 0 which is a contradiction?
    If you are thinking about saying that \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\leqslant 0 implies \left| \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\leqslant 0, no, it would not lead to a contradiction because a\leqslant  b does not imply |a|\leqslant |b|. (the reason is that t\mapsto |t| is not an increasing function)
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