I am trying to study an optimisation problem under constraints. The point is that all my constraints are linear as well as all terms of my objective function except one.
This guy : $$ \alpha^{\lfloor x \rfloor} (\beta + \alpha (x - \lfloor x \rfloor)) $$ is clearly non-linear and non-convex (here $\alpha > 1$ and $\beta>0$ are just parameters). I would like to prove that my optimisation problem is quasiconvex so I have some questions :
If I prove my function is quasiconvex, is it true that the sum of a linear function and my quasiconvex function is quasiconvex ?
Do you have any idea on how to show that my function is('nt) quasiconvex ?
I think that $\alpha ^{\lfloor x \rfloor}$ is quasiconvex because the function "floor" is quasiconvex and the composition of an increasing function with a quasi convex function is quasi convex. However even if I prove every term is respectively quasiconvex, I can say that the sum is also quasiconvex.
Thank you !
It is not true. $2x+cos(x)$ is quasiconvex as it is strictly increasing on $\mathbb{R}$, but $cos(x)=(2x+cos(x))+(-2x)$ is clearly not quasiconvex.
A quick plot with a=2, b=1, and x=[-5,5] clearly shows that your fonction is not quasiconvex anyways, a more rigorous demonstration would be done by showing that the function increases on $[k,k+1[$ and that $f(k)<f(k-\epsilon)$ for $\epsilon$ small but positive.
For more information on quasiconvex functions, Boyd optimization book is accessible for free online : http://stanford.edu/~boyd/cvxbook/