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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.

Andrea Camisa
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2 Answers2

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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.

Red shoes
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  • Thank you for the clarification. What additional assumption should I put on the functions? Or, maybe I should say: are there situations in which this statement is true? – Andrea Camisa Apr 25 '17 at 20:38
  • @AndreaCamisa I don't know. The hessian of $f$ must be Positive definite , maybe you could get something from it. – Red shoes May 30 '17 at 07:50
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In general, the statement that product of convex functions are convex is false.

Consider the following case for $x\in \mathbb{R}$:

$f_{a}(x) = (x-2)^{2}$, $f_{b}(x) = (x+2)^{2}$

Then, $f(a,b) = f_{a}f_{b}(x) = (x-2)^{2}(x+2)^{2}$ is clearly non-convex. [I can't post images yet but I wanted to show a plot in Wolfram Alpha][1]

It is, however, possible to impose constraints on your convex functions to make their product convex (e.g. both non-decreasing/increasing in addition to non-negativity). Please refer to this discussion for more details.

EDIT

As comment correctly pointed out, $f(a,b) = f_{a}(x)f_{b}(x) \neq f_{a}f_{b}(x)$. Therefore the product should be $f(a,b) = f_{a}(x)f_{b}(y) = (x-2)^2 (y+2)^2$ which is still non-convex.

Tingkai Liu
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    This is not a product of two convex functions !! variables are separate! It's just a two_variable function. But still f(a,b) is not necessarily convex, see my counter example. – Red shoes Apr 25 '17 at 17:13
  • @Redshoes good catch! Apologies for the silly mistake. – Tingkai Liu Jun 09 '18 at 20:33