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Thread: Finding functions

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

    So far, I have this:

    Proving $\displaystyle |f(x)-f(y)|$ is continuous:

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

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

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

    Let $\displaystyle \delta^2=\epsilon$

    Hence $\displaystyle |f(x)-f(y)| \leq (y-x)^2< \delta^2=\epsilon$ 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; Feb 7th 2009 at 06:02 AM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,

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

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

    Setting $\displaystyle h=y-x$.

    Therefore:

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

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

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

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

    Therefore function f is a constant polynomial.
    Last edited by Showcase_22; Feb 7th 2009 at 05:42 AM.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    $\displaystyle |f(x)-f(y)| \leq (x-y)^2$

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

    Hint: use the same approach to show that $\displaystyle |f'(x)|\leqslant 0$.
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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 $\displaystyle f$ is differentiable at $\displaystyle x$ since it is possible for $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} =-\infty$ and $\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h){-f(x)}}{h} \leqslant 0$ to hold simultaneously.
    Since we're taking the modulus of $\displaystyle f(x)-f(y)$, wouldn't this imply $\displaystyle |-\infty| \leq 0$ giving $\displaystyle +\infty \leq 0$ 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 $\displaystyle f(x)-f(y)$, wouldn't this imply $\displaystyle |-\infty| \leq 0$ giving $\displaystyle +\infty \leq 0$ which is a contradiction?
    If you are thinking about saying that $\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\leqslant 0$ implies $\displaystyle \left| \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\leqslant 0$, no, it would not lead to a contradiction because $\displaystyle a\leqslant b $ does not imply $\displaystyle |a|\leqslant |b|$. (the reason is that $\displaystyle t\mapsto |t|$ is not an increasing function)
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