For convex functions $f(x)>1,g(x)>1, x\in \mathbb{R}$, is the product $(f\cdot g)(x)$ necessarily convex?

Question

From what I understand, this is how it is:

assume $f\cdot g$ is concave.
then, $$(f\cdot g)(0)>1$$ by this and the assumption, the product must be less than $1$ for some real $x$. thus, at least one of the functions at $x$ must be less than $1$, which brings to a contradiction.

hence, $(f\cdot g)(x)$ is convex $\square$.

is this train of thought correct?

That a function is not concave does not imply that it is convex. — Martin R, May 24 '20 at 07:39

score 2 · Accepted Answer · answered May 24 '20 at 07:58

2

As mentioned in the comments, your arguments are wrong.

For a counterexample, take $$\begin{cases} f(x) &= 2 +(x-2)^2\\ g(x) &= 2 +(x+2)^2 \end{cases}$$

which are both convex and greater than one while $f \cdot g$ is not convex.

answered May 24 '20 at 07:58

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For convex functions $f(x)>1,g(x)>1, x\in \mathbb{R}$, is the product $(f\cdot g)(x)$ necessarily convex?

1 Answers1