Convexity of a sum of functions

Question

I need to check whether a functions is convex. The function is sum over fractions $ S(c, \sigma, r) = \sum_n \frac{\mu_n}{c(\mu_n^2 + \omega^2)}$ where $\mu_n = \frac{r\lambda_n + \sigma}{c}$ with $\lambda_n, \omega \in \Re^+$. I checked to Hessian with mathematica, but this gets huge and is not realy leading somewhere, i think... and other ideas?

Answer 1

Everything is positive here. We have

$$ \frac{\mu_n}{c(\mu_n^2 + \omega^2)} = \frac{\frac{r\lambda_n + \sigma}{c}}{c((\frac{r\lambda_n + \sigma}{c})^2 + \omega^2)} = \frac{r\lambda_n + \sigma}{(r\lambda_n + \sigma)^2 + c^2\omega^2} $$

Setting $h_n\equiv r\lambda_n + \sigma$ and $g \equiv c^2\omega^2$ we can write

$$S(c, r, \sigma) = \sum_n \frac{h_n}{h_n^2 + g} = \sum_n \left(h_n +\frac {g}{h_n}\right)^{-1}$$

Consider the function $f(c, r, \sigma) = g/h_n$. We have (prime for order of derivative, subscripts for variables)

$$\begin{align} f'_c = g'/h_n & &f''_c = g''/h_n\\ f'_r=-g\lambda_n/h^2_n& & f''r=2g\lambda^2_n/h^3_n\\ f'_\sigma=-g/h^2_n& & f''_{\sigma}=2g/h^3_n\\ f''_{cr}=-g'\lambda_n/h^2_n & &f''_{c\sigma}=-g'/h^2_n & & f''_{r\sigma}=2g\lambda_n/h^3_n \end{align}$$

So the determinant of the Hessian of $f$ is $$\Big |H_f\Big|=\left|\begin{matrix} g''/h_n & &-g'\lambda_n/h^2_n & &-g'/h^2_n \\ -g'\lambda_n/h^2_n & &2g\lambda^2_n/h^3_n && 2g\lambda_n/h^3_n\\ -g'/h^2_n && 2g\lambda_n/h^3_n & & 2g/h^3_n \\ \end{matrix}\right|$$

Noting that $g'=\frac{2}{c}g$ , that $g''=\frac{2}{c^2}g$,and taking out common factors we have

$$\Big |H_f\Big|=h_n^{-7}\lambda_n^2\left|\begin{matrix} \frac{2}{c^2}g & &-\frac{2}{c}g & &-\frac{2}{c}g \\ -\frac{2}{c}g & &2g && 2g\\ -\frac{2}{c}g && 2g & & 2g \\ \end{matrix}\right|$$

$$=h_n^{-7}\lambda_n^22^3g^3c^{-2}\left|\begin{matrix} 1 & &-1 & &-1 \\ -1 & &1 && 1\\ -1 && 1 & & 1 \\ \end{matrix}\right| $$

$$=h_n^{-7}\lambda_n^22^3g^3c^{-2}(-1)^2\left|\begin{matrix} 1 & &1 & &1 \\ 1 & &1 && 1\\ 1 && 1 & & 1 \\ \end{matrix}\right| =0$$

...the zero holding also for principal minors and permutations. So the function $f(c, r, \sigma) = g/h_n$ is convex. $h_n$ is also convex, and the sum of two convex functions is convex. So $h_n +\frac {g}{h_n}$ is a convex function in $(c, r, \sigma)$. Now all you have to figure out is whether the reciprocal of a convex function, i.e.$\left(h_n +\frac {g}{h_n}\right)^{-1}$, is also convex. If it is then, again, the sum of convex functions is convex so the whole of your expression will be convex.