Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that if $f$ is convex in $(a,b)$ it is Lipschitz in $[c,d]$ where $a \lt c \lt d \lt b$.

Here's what I have so far:

Let $t_1,t_2 \in \mathbb{R}$ such that $a \lt t_2 \lt c \lt d \lt t_1 \lt b$ and let $x_1,x_2 \in [c,d]$.

Because $f$ is convex I know that $$\dfrac{f(c)-f(t_2)}{c-t_2} \lt \dfrac{f(x_2)-f(x_1)}{x_2-x_1} \lt \dfrac{f(t_1)-f(d)}{t_1-d}$$

I think I'm almost there, but what is the right $C$?

share|improve this question
1  
$a < x < b, |x| < max(abs(a), abs(b))$ –  TenaliRaman May 23 '12 at 6:04

1 Answer 1

up vote 3 down vote accepted

Just let $C$ be the max of the absolute value of the functions on the far left and far right of the inequalities, ie:

$$C = \max \left\{\left|\dfrac{f(c)-f(t_2)}{c-t_2}\right|, \left|\dfrac{f(t_1)-f(d)}{t_1-d}\right|\right\}$$

share|improve this answer
    
You may want to use \max in math-mode instead of max in future. –  user17762 May 23 '12 at 6:15
    
Thanks. I didn't realize I could do that. –  Eugene May 23 '12 at 6:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.