# functions in reals such that inequality holds

• Feb 11th 2011, 06:42 AM
maximus101
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
• Feb 12th 2011, 06:11 AM
Opalg
Quote:

Originally Posted by maximus101
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 $f'(x)$ ?
• Feb 14th 2011, 08:53 AM
maximus101
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 $f'(x)$?
• Feb 14th 2011, 09:27 AM
Opalg
Quote:

Originally Posted by maximus101
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.
• Feb 14th 2011, 09:47 AM
maximus101
Quote:

Originally Posted by Opalg
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.