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Math Help - functions in reals such that inequality holds

  1. #1
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    functions in reals such that inequality holds

    for which functions
    f : R \rightarrow R

    \forall x,y \in R

    does

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

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

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

    does

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

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

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

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

    so next I'm not sure how to connect this with 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:

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

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

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

    if \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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