# functions in reals such that inequality holds

• Feb 11th 2011, 05:42 AM
maximus101
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
• Feb 12th 2011, 05:11 AM
Opalg
Quote:

Originally Posted by maximus101
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 $\displaystyle f'(x)$ ?
• Feb 14th 2011, 07:53 AM
maximus101
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 $\displaystyle f'(x)$?
• Feb 14th 2011, 08:27 AM
Opalg
Quote:

Originally Posted by maximus101
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.
• Feb 14th 2011, 08:47 AM
maximus101
Quote:

Originally Posted by Opalg
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.