# Convexity of cubic vector function?

First question in these forums so go easy on me. I have a function $f_i(x):\mathbb{R}^N\to\mathbb{R}$ which is defined by $$f_i(x) = \frac{(x^TAe_i)x^TAx}{(x^TA+b^T){\bf 1}}$$ where $x\in\mathbb{R}^N$, $b\in\mathbb{R}^N$, $A\in\mathbb{R}^{N\times N}$, $e_i$ is the indicator vector with a $1$ in the $i^{th}$ position, and ${\bf 1}$ is an $N$-dimensional column vector of ones.

I want to determine if $f_i(x)$ is convex in $x$. We are allowed to use the fact that $A$ is positive-semidefinite. Thus I know that $x^TAx$ is convex in $x$, but I'm not sure if this is useful. What is throwing me off the is the cubic dependence on $x$ in the numerator.

Any tips on how to proceed? Thanks!

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Have you considered the one-dimensional case? – user53153 Dec 31 '12 at 3:06
That is useful, it seems that as long as we stay away from the point of discontinuity the function is convex. I suppose a similar statement could be made about $f_i(x)$, that is, if we consider $x$ sufficiently far enough away from $\bar{x}$, with $\bar{x}^TA = -b^T$, could we conclude $f_i(x)$ is convex? – Erik M Dec 31 '12 at 20:02
"Sufficiently far" will have to depend on $A$, since in one dimension $\dfrac{a^2x^3}{ax+1}$ will be non-convex on a large set when $a$ is close to zero. // Furthermore, you have a singularity not just at the point $x^TA+b^T=0$, but on the $n-1$-dimensional hyperplane $(x^TA+b^T)\mathbf{1}=0$. // To get at least some positive result, try to find the 1st and 2nd derivatives of the factor $\dfrac{x^TAe_i}{(x^TA+b^T)1}$; when they are both very small, they will not overcome strict convexity of $x^TAx$ (assuming $A$ is positive definite). I see $\dfrac{x^TAe_i}{(x^TA+b^T)1}$ as the enemy here. – user53153 Jan 1 '13 at 8:02
Thanks Pavel, I'll post here if I come up with a reasonable result. – Erik M Jan 1 '13 at 22:50