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 $C$?

  • 2
    $\begingroup$ $a < x < b, |x| < max(abs(a), abs(b))$ $\endgroup$ May 23 '12 at 6:04

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

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

  • 4
    $\begingroup$ You may want to use \max in math-mode instead of max in future. $\endgroup$
    – user17762
    May 23 '12 at 6:15
  • 1
    $\begingroup$ Thanks. I didn't realize I could do that. $\endgroup$
    – Eugene
    May 23 '12 at 6:18

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