Let $X\subseteq\mathbb{R}^n$. I have the following function $f:X\rightarrow\mathbb{R}$: $$ f({\bf x})= \frac{c_1 + \sum_{i=1}^n a_ix_i +\sum_{i=1}^n b_ix_i^2}{c_2+\sum_{i=1}^n d_i x_i}\enspace.$$

All the $a_i$, $b_i$, $c_i$, and $d_i$ are strictly greater than 0, and X is such that $$ c_2+\sum_{i=1}^n d_i x_i>0, \forall {\bf x}\in X\enspace.$$

Is $f$ convex or at least pseudo- or quasi-convex?

Note that $$f({\bf x})= \frac{c_1 + \sum_{i=1}^n a_ix_i }{c_2+\sum_{i=1}^n d_i x_i}+\frac{\sum_{i=1}^n b_ix_i^2}{c_2+\sum_{i=1}^n d_i x_i}$$ and the first term is on the right side is pseudo-linear (hence pseudo-convex, hence quasi-convex) and third term is convex (hence pseudo-convex, hence quasi-convex). I know that the sum of quasi-convex functions is not in general quasi-convex, but I wonder whether something else was known that can help me show that $f$ is.


Never mind: my original function was actually

$$ \frac{\sum_{i=1}^\ell \left(b_i + \sum_{j=1}^\ell a^{(i)}_j x_j\right)^2}{c_1+\sum_{i=1}^n d_ix_i}$$

(for different values of the constants from the original question)

so I can see it as a sum of $g_i^2/h$, where $g_i$ is affine (and actually non-negative in my domain) and $h$ is positive affine. Then each of the $g_i^2/h$ is convex, and so is their sum.


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.