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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 ${\bf x}$?

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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 $x_3$.

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  • $\begingroup$ 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

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