I know that generally, a linear-fractional function is quasiconvex, however, I wonder if it is possible to prove this linear-fractional function to be convex, under the following additional conditions?
The function f(x) is formulated as
f(x) = $\frac{a_1x_1+...a_mx_m}{a_1x_1+...a_mx_m+a_{ m+1}x_{m+1}...+a_nx_n}$
the additional conditions are
- $a_1,...a_n$ are constant positive numbers
- m < n
- $0 < x_i <1$, for i = 1...n
- all the variables and constants are in R
If not, under what conditions can the linear-fractional function f(x) be convex?
Thank you
Lets take for simplicity function $f(x,y)=\frac{ax}{ax+by}$.
The second derivatives: $\frac{\partial^2}{\partial x^2}f(x,y) = -\frac{2a^2by}{(ax+by)^3}, \quad\frac{\partial^2}{\partial y^2}f(x,y) = \frac{2ab^2x}{(ax+by)^3}, \quad \frac{\partial^2}{\partial x \partial y}f(x,y) = \frac{ab(ax-by)}{(ax+by)^3}$
The Hessian at point $(x,y)=\left(\frac{b}{a+b}, \frac{a}{a+b}\right)$ is $\begin{pmatrix} (-) & 0 \\ 0 & (+)\end{pmatrix}$ - not a possitive definite matrix. This should proove that function can not be convex.
In my research I am interested in functions of the form $g(x,y) = \frac{1}{1+ax+by}$.
The derivatives are $\frac{\partial^2}{\partial x^2}f(x,y) = \frac{2a^2}{(1+ax+by)^3}, \quad\frac{\partial^2}{\partial y^2}f(x,y) = \frac{2b^2}{(1+ax+by)^3}, \quad \frac{\partial^2}{\partial x \partial y}f(x,y) = \frac{2ab}{(1+ax+by)^3}$.
The Hessian at point $(x,y)$ is $\frac{2}{(1+ax+by)^3}\begin{pmatrix} a^2 & ab \\ ab & b^2 \end{pmatrix} = \frac{2}{(1+ax+by)^3} \begin{pmatrix}a \\ b\end{pmatrix} \begin{pmatrix}a&b\end{pmatrix}$ - a possitive semidefinite matrix in the unit cube. This can be generalized to more than 2 dimensions.