Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


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.

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

1 Answer 1

up vote 1 down vote accepted

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 $v$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.