I have a function that is of the form

$C({\bf x}) = c_1\left(a_1x_1 + b_1x_1^2\right) + c_2\left(a_2(x_1-x_2) + b_2(x_1-x_2)^2\right) + c_3\left(a_3(x_2-x_3) + b_3(x_2-x_3)^2\right)$

where each $c_i:\mathbb{R}\to\mathbb{R}$ is strictly convex and strictly increasing and all $a_i,b_i>0$. I want to determine if $C({\bf x})$ is strictly convex in ${\bf x} = (x_1,x_2,x_3)$. Is $c_1\left(a_1x_1 + b_1x_1^2\right)$ strictly convex in ${\bf x}$?


No, $c_1(a_1 x_1+ b_1 x_1^2)$ is not strictly convex in $\bf x$, because it is constant in $x_2$ and $x_3$.

But (and this is a hint) $c_2\left(a_2(x_1 - x_2) + b_2 (x_1 - x_2)^2\right)$ is strictly convex in $x_2$ and $c_3\left(a_3 (x_2 - x_3) + b_3 (x_2 - x_3)^2\right)$ is strictly convex in $x_3$.

  • $\begingroup$ Thanks. The fact that $c_1(\cdot)$ is not strictly convex in ${\bf x}$ makes sense, but it is strictly convex in $x_1$, right? So if each $c_i$ is strictly convex in each $x_i$ and we know that the sum of strictly convex functions is strictly convex, can we conclude that $C({\bf x})$ is strictly convex in ${\bf x}$? $\endgroup$ – jonem Mar 19 '14 at 6:41
  • $\begingroup$ The three functions are each convex, so their sum is convex. In any given direction, at least one of the three functions is strictly convex. From that it's easy... $\endgroup$ – Robert Israel Mar 19 '14 at 6:56
  • $\begingroup$ I see it now, thanks for your direction. $\endgroup$ – jonem Mar 19 '14 at 7:14
  • $\begingroup$ Sorry to be a pain, but I have another question. I understand why $c_2$ is strictly convex is $x_2$, but why is it not also strictly convex in $x_1$? $\endgroup$ – jonem Mar 19 '14 at 16:24
  • $\begingroup$ It is strictly convex in $x_1$. I didn't say it wasn't. $\endgroup$ – Robert Israel Mar 20 '14 at 1:19

Your Answer

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

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