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.

Let $f(x_1,x_2)=x_1x_2$, with $x_1,x_2 \ge 0$. Is $f$ a convex function? Why?

$\ \\$

Edit (in view of the comments below)

The Hessian matrix is $H=[0\, 1; \,1 \,0]$, which is indefinite (in general). In fact, $x H x'= 2x_1x_2$. However, $2x_1x_2\ge 0$ when $x_1,x_2\ge 0$, so that the Hessian is indeed positive semidefinite when $x_1,x_2\ge 0$. Therefore, is this sufficient to conclude that my $f$ is convex?

share|cite|improve this question
Just apply the definition. –  A.D Jan 25 '13 at 14:47
Or, maybe you know a criterion where the Hessian matrix is negative definite? –  GEdgar Jan 25 '13 at 15:05

3 Answers 3

up vote 1 down vote accepted

The Hessian is indeed indefinite. A $2\times 2$ matrix $A$ is positive semidefinite if and only if $x'Ax\geq0$ for all $x\in\mathbb{R}\times\mathbb{R}$ and not only those in $\mathbb{R}_+\times\mathbb{R}_+$. Hence $f(x,y)=xy$ is not convex.

Indeed, you could try to see $f(1,0)=f(0,1)=0$ while $f(1/2,1/2)=1/4 > 0$.

share|cite|improve this answer
Thank you. This is exactly what I wanted to know. –  user52227 Feb 16 '13 at 14:11

Note that $$\begin{align}f(a(x_1,x_2)+(1-a)(y_1,y_2)) &=f(ax_1+(1-a)y_1 , ax_2+(1-a)y_2) \\ &=(ax_1+(1-a)y_1)(ax_2+(1-a)y_2) \\ &= a^2x_1x_2 +(1-a)^2y_1y_2 +a(1-a)(x_1y_2+x_2y_1) \\ & \leq ax_1x_2 +(1-a)y_1y_2 \\&=af(x_1 ,x_2)+ (1-a)f(y_1,y_2) \end{align}$$

Note that $0<a<1$

share|cite|improve this answer
Thanks. This would prove that $f$ is convex. But I am sorry, can you please explain the $\le$ in your derivation? It's not clear to me –  user52227 Jan 25 '13 at 15:40

Consider two points with $x = a > 0$ and $y > 0$. They lie on a straight line which is both convex and concave. Now consider two points on $x=y,\quad x>0$, here $f=x^2$ which is convex. Now, two points on $x+y=5$ (say) gives you a concave function, so $f$ is clearly neither convex nor concave in your region (or indeed, anywhere).

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.