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How can I prove that a function $f(x,y)= \frac{x^2}{y}$ is convex for $ y \gt 0$?

I take the Hessian matrix of $\displaystyle \frac{x^2}{y}$, and I got:

$$H = \displaystyle\pmatrix{\frac{2}{y} & -\frac{2x}{y^2} \\-\frac{2x}{y^2} & \frac{2x^2}{y^3}}$$ further more, we have: $$H = \frac{2}{y^3}\displaystyle\pmatrix{y^2 & -xy \\-xy & x^2}$$ I need to prove that H is semi-define positive matrix for y>0. I think I am close to the answer. But lack some knowledge to prove it since I try to calculate the $det(H)$ and it ends up equal to 0. Anyone can help? Thanks.

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2 Answers 2

up vote 2 down vote accepted

You're basically done:

A symmetric matrix $(a_{ij})_{i,j\le n}$ is positive semidefinite iff the subdeterminants $\det(a_{ij})_{i,j\in H}$ are $\ge 0$ for all $H\subseteq\{1,2,..,n\}$.

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Omg.. You're right... I update my response.. –  Berci Feb 19 '13 at 10:13

It's a little easier to work with $$\frac{1}{2}y^3H=\begin{pmatrix} y^2 & -xy \\ -xy & x^2\end{pmatrix} $$ which you may recognize as the tensor product of $(y,-x)$ with itself. This is the first way to see that it is positive semidefinite.

Second way: $$\begin{pmatrix} s & t \end{pmatrix} \begin{pmatrix} y^2 & -xy \\ -xy & x^2\end{pmatrix}\begin{pmatrix} s \\ t \end{pmatrix} =(sy-tx)^2 \ge 0 $$ for all $s,t\in\mathbb R^2$.

Third way: the trace is $x^2+y^2>0$ and the determinant is zero. Make your conclusion about the eigenvalues.

Fourth approach involves Sylvester's criterion which must be modified for semi-definite matrices to include all principal minors, not only the leading ones. For size 2 this is not too bad: two corners and the determinant. For larger matrices this quickly becomes a hassle.

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+1 for the clear and multiple approaches. –  Amzoti Feb 19 '13 at 1:43

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