Take the 2-minute tour ×
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.

Am I correct to say that the following function is convex?

$$\begin{align} & f(x,y)=-\sqrt{xy} \\ & x>0,y>0 \\ \end{align}$$

After computing the Hessian:

$$ Hf =\left[ \begin {array}{cc} 1/4\,{\frac {{y}^{2}}{ \left( xy \right) ^{ 3/2}}}&1/4\,{\frac {xy}{ \left( xy \right) ^{3/2}}}-1/2\,{\frac {1}{ \sqrt {xy}}}\\ 1/4\,{\frac {xy}{ \left( xy \right) ^ {3/2}}}-1/2\,{\frac {1}{\sqrt {xy}}}&1/4\,{\frac {{x}^{2}}{ \left( xy \right) ^{3/2}}}\end {array} \right]$$

Which simplifies to:

$$Hf=\left[ \begin {array}{cc} 1/4\,{\frac {y}{x\sqrt {xy}}}&-1/4\,{\frac {1}{\sqrt {xy}}}\\ -1/4\,{\frac {1}{\sqrt {xy}}}&1/4 \,{\frac {x}{y\sqrt {xy}}}\end {array} \right]$$

And taking the determinant:

$$det(Hf)=0 \ \ \ \ \forall \ x,y \in \Re^+$$

Which is inconclusive.

Will need another method, namely $ z^T (H f) z \ $


See solution in the answer below for continuation:

Aside: And, extending this to an n-dimensional problem: $$f(x_1,x_2,...,x_n)=-\sqrt[n]{x_1x_2...x_n}$$ $$x_i \gt 0 \ \ \ i=1,2,3,...,n$$

Will also yield a convex function.

Thanks.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

The function is convex, but note that zero determinant is not a sufficient condition to give positive semi-definite (or negative semi-definite). With $z^T=(a,b)$, I get $$z^T H\!f z=\frac{1}{4(xy)^{3/2}}(ay-bx)^2,$$ and so $H\!f$ is positive semi-definite, giving a convex function.

share|improve this answer
    
I just checked your result and it looks to be correct. I was, for some reason, thinking of affine functions and a zero matrix. Thank you. –  user42538 Nov 18 '12 at 0:18

The $n$-dimensional function $f(x_1,x_2,...,x_n)=-\sqrt[n]{x_1x_2...x_n}$ where $x_i \gt 0 \ \ \ i=1,2,3,...,n$ is called arithmetic average (except for a negative sign).

Yes, it is convex. The Hessian matrix $\mathsf{H}f$ is

$$\mathsf{H}f = \left ( \frac{f(x_1,...,x_n)}{n^2x_ix_j}(1-\delta_{ij}) \right)_{ij} \quad i,j = 1,...,n$$

where $\delta_{ij} = 1$ if $i=j$, $0$ otherwise. It is easy to show that for any $d \in \mathbb{R}^n$, $d^T\mathsf{H}fd \ge 0$.

This function is actually Example 4.3.3 of the text book "Fundamentals of Convex Analysis" by Jean-Baptiste Hiriart-Urruty & Claude Lemarechal.

share|improve this answer

Your Answer

 
discard

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.