# (quasi) convexity $\frac{x}{y}$

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.

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The function is quasilinear, but I'm not sure how to show it correct. –  Masala May 30 '12 at 12:44

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 $v$.