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Math Help - A result of differentiability

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    A result of differentiability

    Let's f be a real function on R .Prove if f be differentiable in t with derivative A then :
    For every e > 0 there be a d >0 such that for every open interval (x1,x2) containing t whose length |x2-x1| is smaller than d ,| (f(x2)- f(x1))/(x2-x1) - A | < e
    Is its inverse true? Could you provide a contradictory example ? What conditions if f had that would be true?
    This is useful to better insight to definition of derivatives of measures in topology.
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  2. #2
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    Quote Originally Posted by mazaheri View Post

    Let f be a real function on \mathbb{R}. Prove that if f is differentiable in t with derivative A then :


    For every \epsilon > 0 there is a \delta > 0 such that for every open interval (x_1,x_2) containing t

    whose length |x_2-x_1| is smaller than \delta: \ \ \left| \frac{f(x_2)- f(x_1)}{x_2-x_1} - A \right| < \epsilon.
    let \epsilon > 0. then by definition there exists \delta > 0, such that if |x-t| < \delta, then:  \left|\frac{f(x) - f(t)}{x-t}- A \right| < \epsilon. \ \ \ (1)

    now let x_1 < t < x _2 and x_2 - x_1 < \delta. then obviously: t - x_1 = |x_1 - t| < \delta, and x_2 - t = |x_2 - t| < \delta.

    thus by (1): \ \ \ \left|\frac{f(t)-f(x_1)}{t-x_1} - A \right| < \epsilon, and \left|\frac{f(x_2)-f(t)}{x_2 - t} - A \right| < \epsilon, which will give us:

    |f(t)-f(x_1)-(t-x_1)A | < \epsilon(t - x_1), and |f(x_2)-f(t) - (x_2 - t)A| < \epsilon ( x_2 - t). thus by the triangle

    inequality: |f(x_2)-f(x_1)-(x_2 - x_1)A| \leq |f(x_2)-f(t) - (x_2 - t)A| \ + \ |f(t) - f(x_1) - (t - x_1)A | <

    \epsilon(t - x_1) + \epsilon(x_2 - t)= \epsilon(x_2 - x_1), which after dividing both sides by |x_2 - x_1|=x_2 - x_1, gives you:

    \left|\frac{f(x_2)-f(x_1)}{x_2 - x_1} - A \right| < \epsilon. \ \ \ \square

    Is its inverse true? Could you provide a contradictory example ?
    the converse is not true. as an counter-example define f(x)=0, \ \forall x \neq 0, and f(0)=1, and choose A=0.

    then for any non-zero x_1 \neq x_2 and \epsilon>0: \ \left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-A \right|=0 < \epsilon, but f is not differentiable at x=0.
    Last edited by NonCommAlg; June 30th 2008 at 11:55 PM.
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