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Thread: functions in reals such that inequality holds

  1. #1
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    functions in reals such that inequality holds

    for which functions
    $\displaystyle f$ $\displaystyle :$ $\displaystyle R$ $\displaystyle \rightarrow$ $\displaystyle R$

    $\displaystyle \forall$ $\displaystyle x,y$ $\displaystyle \in$ $\displaystyle R$

    does

    | $\displaystyle f(y)$ - $\displaystyle f(x)$ | $\displaystyle \mid$ $\displaystyle \leq$ $\displaystyle (y-x)^2$

    hold
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  2. #2
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    Quote Originally Posted by maximus101 View Post
    for which functions
    $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$

    $\displaystyle \forall x,y\in\mathbb{R}$

    does

    $\displaystyle |f(y) - f(x)|\leq (y-x)^2$

    hold
    Hint: Write that condition as $\displaystyle \dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$. What does that tell you about $\displaystyle f'(x)$ ?
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  3. #3
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    Hi, so I did the following:

    $\displaystyle \dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$

    was obtained by $\displaystyle (y-x)^2$=$\displaystyle |y-x||y-x|$ and dividing both sides by $\displaystyle |y-x|$.

    so next I'm not sure how to connect this with $\displaystyle f'(x)$?
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  4. #4
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    Quote Originally Posted by maximus101 View Post
    Hi, so I did the following:

    $\displaystyle \dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|$

    was obtained by $\displaystyle (y-x)^2$=$\displaystyle |y-x||y-x|$ and dividing both sides by $\displaystyle |y-x|$.

    so next I'm not sure how to connect this with $\displaystyle f'(x)$?
    The definition of $\displaystyle f'(x)$ is that it is equal to $\displaystyle \displaystyle\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$. If $\displaystyle \left|\frac{f(y)-f(x)}{y-x}\right| \leqslant |y-x|$ then that limit must be 0.
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  5. #5
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    Quote Originally Posted by Opalg View Post
    The definition of $\displaystyle f'(x)$ is that it is equal to $\displaystyle \displaystyle\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$. If $\displaystyle \left|\frac{f(y)-f(x)}{y-x}\right| \leqslant |y-x|$ then that limit must be 0.
    ok thank you,

    if $\displaystyle \left|\frac{f(y)-f(x)}{y-x}\right| \leqslant |y-x|$ how did we deduce the limit must be 0? And how do we know it is continuous for all x and y.

    does this mean that f is the set of functions with derivative 0 at all x,y? so f is simply a constant.
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