So I have two nonnegative convex functions $f_a:\mathcal{A}\rightarrow\mathbb{R}$ and $f_b:\mathcal{B}\rightarrow\mathbb{R}$ defined on the convex sets $\mathcal{A}\subseteq\mathbb{R}^a$ and $\mathcal{B}\subseteq\mathbb{R}^b$.
If I consider their product, i.e. $f:\mathcal{A}\times\mathcal{B} \rightarrow \mathbb{R}$ defined as $f(a, b) = f_a(a) f_b(b)$, is this function convex? First of all, I should show that $\mathcal{A}\times\mathcal{B}$ is convex, but this is strightforward. Then, I have to show the convexity of $f$.
This is what I tried so far: Consider $a_1, a_2 \in \mathcal{A}$ and $b_1, b_2 \in \mathcal{B}$ and $\lambda \in [0,1]$. I have to show that
$$f(\lambda a_1 + (1-\lambda) a_2, \lambda b_1 + (1-\lambda) b_2) \leq \lambda f(a_1, b_1) + (1-\lambda) f(a_2, b_2)$$
Expanding the LHS and applying the convexity of $f_a$ and $f_b$, I get
$$f(\lambda a_1 + (1-\lambda) a_2, \lambda b_1 + (1-\lambda) b_2) \leq \lambda^2 f(a_1, b_1) + (1-\lambda)^2 f(a_2, b_2) + \lambda(1-\lambda) \left[ f_a(a_1)f_b(b_2) + f_a(a_2)f_b(b_1) \right]$$
Now $\lambda^2 \leq \lambda$ and $(1-\lambda)^2 \leq (1-\lambda)$, but I don't know how to get rid of the mixed terms.
Any help would be appreciated.
Best Answer
One simple counter example is, take $f_a , f_b$ identity functions on $R$.
then $f(x,y)= xy $, which is clearly not a convex function.