Hej,
I have the function $\frac{x}{y}$ on the domain $R_{++}$. The Hessian matrix is - as I have calculated it - positive semidefinite. But I'm not really sure, if the function is really convex at all on the domain.
Thanks for any help.
Hej,
I have the function $\frac{x}{y}$ on the domain $R_{++}$. The Hessian matrix is - as I have calculated it - positive semidefinite. But I'm not really sure, if the function is really convex at all on the domain.
Thanks for any help.
Something went wrong with your calculation, because the Hessian matrix $ \begin{pmatrix} 0 & -1/y^2 \\ -1/y^2 & 2/y^3 \end{pmatrix} $ has negative determinant.
There is a similar function with positive semidefinite Hessian in the positive quadrant, namely $v(x,y)=x^2/y$. The Hessian is $ \begin{pmatrix} 2/y & -2x/y^2 \\ -2x/y^2 & 2x^2/y^3 \end{pmatrix} = \frac{2}{y^3}\langle x,y \rangle \otimes \langle x,y \rangle \ge 0 $ Given its simple form, I wonder if there is any "obvious" reason for the convexity of $v$.