# A result of differentiability

• Jun 30th 2008, 07:25 PM
mazaheri
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.
• Jun 30th 2008, 11:44 PM
NonCommAlg
Quote:

Originally Posted by mazaheri

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

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

whose length $\displaystyle |x_2-x_1|$ is smaller than $\displaystyle \delta: \ \ \left| \frac{f(x_2)- f(x_1)}{x_2-x_1} - A \right| < \epsilon.$

let $\displaystyle \epsilon > 0.$ then by definition there exists $\displaystyle \delta > 0,$ such that if $\displaystyle |x-t| < \delta,$ then: $\displaystyle \left|\frac{f(x) - f(t)}{x-t}- A \right| < \epsilon. \ \ \ (1)$

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

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

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

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

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

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

Quote:

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

then for any non-zero $\displaystyle x_1 \neq x_2$ and $\displaystyle \epsilon>0: \ \left|\frac{f(x_2)-f(x_1)}{x_2-x_1}-A \right|=0 < \epsilon,$ but $\displaystyle f$ is not differentiable at $\displaystyle x=0.$