0
$\begingroup$

A function $f(x)$ in some domain $a\leq x \leq b$ is convex if and only if for any $x_1 < x_2 < x_3$ from domain $[a,b]$,

$$\frac{(f(x_2)-f(x_1))}{(x_2-x_1)} \leq \frac{(f(x_3)-f(x_1))}{(x_3-x_1)} \leq \frac{(f(x_3)-f(x_2))}{(x_3-x_2)}.$$

I am aware that for any $(x_1,x_2)$ for a convex function $f(x_1) \geq f(x_1)+f'(x_1)(x_2-x_1)$. How do I proceed further to prove the above inequality?

$\endgroup$
  • $\begingroup$ You can use your definition of convexity to show that a line from $(a,f(a))$ to $(b,f(b))$ with $a<b$ is a (non-strict) lower bound for $f$ for all $x<a$ or $b<x$ - i.e. $x$ not in $[a,b]$. So secant lines are lower bounds. You have an inequality for tangent lines - which are limits of secant lines. The rest is just analysis. $\endgroup$ – Milo Brandt Feb 9 '15 at 14:47
  • $\begingroup$ Draw a figure and you shall see that the stated inequalities are an immediate consequence of the definition of convexity. If necessary write $x_2$ as $x_2=(1-t)x_1+t x_3$ with $0<t<1$. $\endgroup$ – Christian Blatter Feb 9 '15 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.