Is the multiplication of convex function by a linear function convex?

Question

I have a set of convex functions $f_{ij}:\mathbb{R}_+^n\mapsto\mathbb{R}$ for all $i,j\in \{1,\ldots,n\}$.

If I defined the following functions $g_{ij}:\mathbb{R}_+^{n\times n}\times\mathbb{R}_+^n\mapsto\mathbb{R}$ by

$$ g_{ij}(\mathbf{x},\mathbf{y})=x_{ij}f_{ij}(\mathbf{y}), $$

can this function be convex?

On $\mathbb{R}_+$, $y=(x-1)^2$ is convex, $y=(x-1)$ is convex, whereas $y=(x-1)^3$ is not... — Jean Marie, Nov 03 '16 at 22:50
@det You are right, I was abused by your very general title. — Jean Marie, Nov 03 '16 at 22:57

user251257 · Accepted Answer · 2016-11-03T23:12:42.527

3

Consider $f(y)=y$ on $\mathbb R_+$, which is obviously convex and $$ g(x,y) = xf(y) = xy $$ on $\mathbb R^2_+$. Then, $g$ is not convex, as its hessian is not positive semidefinite.

Or a direct proof: $$ g(1\pm x, 1\mp x) = 1 - x^2 < 1 = g(1,1). $$

edited Nov 03 '16 at 23:12

answered Nov 03 '16 at 23:01

user251257

9,229

The Hessian is it $\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$? If so, then $\left(x, y\right)\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix}=2xy$. $g$ is convex iff $x^\top H x \geq 0$. – Ribz Nov 04 '16 at 01:09
@det yes. And the rhs is not non-negative in general. – user251257 Nov 04 '16 at 01:11
But $(x,y)\in\mathbb{R}_+^2$. – Ribz Nov 04 '16 at 01:12
@det no. Please check the definition of positive semidefinite again. – user251257 Nov 04 '16 at 01:13
Ah ok, I mix the variable of the function and the vector in the definition of positive semidefinite. Thanks. – Ribz Nov 04 '16 at 01:16
Does it change anything if the domain was $[0, 1]^{n\times n}\times [0,1]^n$ ? – Ribz Nov 04 '16 at 16:10
@det affine scaling won't affect convexity. – user251257 Nov 04 '16 at 17:39

Is the multiplication of convex function by a linear function convex?

1 Answers1