1. ## Proof problem

Hi,

Can anybody offer any help/ideas with the following question:

Suppose f is a differentiable function with | f '(x) | ≤ 1 for all x ε R . Show that

| f(x) - f(y) | | x - y | for all x,y ε R .

Will using the mean value theorem help with this proof?

2. Hello,
Originally Posted by jackiemoon
Hi,

Can anybody offer any help/ideas with the following question:

Suppose f is a differentiable function with | f '(x) | ≤ 1 for all x ε R . Show that

| f(x) - f(y) | | x - y | for all x,y ε R .

Will using the mean value theorem help with this proof?

Of course oO

Let x and y in R, $x \neq y$

MVT says that there exists c between x and y such that :
$f'(c)=\frac{f(x)-f(y)}{x-y}$

Take absolute values of each side :
$\frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|$

But we know that $|f'(c)| \leq 1$, for any c in R.

Therefore...