0

$f:X\rightarrow \mathbb{R}$ where $X\subset \mathbb{R}^n$. All the second order partial derivatives are non-negative i.e $\frac{\partial^2 f}{\partial x_{i}^2}\geq 0$ and $X=\{(x_{1},x_{2},..,x_{n}):\sum_{i=1}^{n}x_{i}=1 \}.$

Can say anothing about the convexity of $f(x_{1},x_{2},...,x_{n})$ ?

Rodrigo de Azevedo
  • 1
Dipankar Mondal
  • 195
  • It is, of course, convex in each variable separately. But as pointed out in the answer by @daw, it is not convex in general. For that, you need the matrix of second partials $\partial^2f/\partial x_i\partial x_j$ to be positive semi-definite. – Harald Hanche-Olsen Apr 20 '17 at 07:03

1 Answers1

1

No. Take $f(x_1,x_2)=x_1x_2$. Then on $X$ it holds $x_2=1-x_1$, and the function reduces to $x_1(1-x_1)$, which is non-convex.

daw
  • 49,113
  • 2
  • 38
  • 76