Show the set of element in $\Bbb R^n$ which the product of each element greater than $1$ is convex

Question

I know it might be easy for you, but I have tried to show that the following set is convex for more than three days.

$$A := \left\lbrace (a_1,\dots,a_n) \in \mathbb{R}^n : x_i \geq 0, i = 1,\dots,n, \prod_{i=1}^{n} x_i \geq 1 \right\rbrace$$

I can prove only the first condition ($x_i \geq 0$) that is $(\lambda x_i +(1-\lambda) y_i)\geq 0$ for all $i$ but I cannot address how come of

$$\prod_{i=1}^{n}(\lambda x_i +(1-\lambda) y_i) \geq 1$$

Could you hint or tell me some to let me figure out of this? Thank you in advance.

Answer 1

You can take the logarithm of both sides for the second condition:

$$ \prod_{i=1}^n x_i \geq 1 \Leftrightarrow \sum_{i=1}^n \log (x_i) \geq 0 \Leftrightarrow -\sum_{i=1}^n \log (x_i) \leq 0. $$

This is the sublevel set of a convex function, which is itself a convex set.

Answer 2

You want to show that the boundary of the set is convex, and for that you need to show that the Hessian of $\prod x_i$ is convex. The $ij$-th element of the hessian is $\prod_{k\neq i, j} x_k.$ (the diagonal elements are zero). Divide through by $\prod x_k,$ to get the $ij$-the element equal to $\frac1{x_i}{x_j}.$ Let $a_k = \frac1{x_k}.$ Your problem now reduces to showing that the matrix $A$ whose $ij$-th element is $a_i a_j,$ where the $a_k$ are positive, is positive definite. But that matrix is $D(J-I)D,$ where $D$ is the diagonal matrix of $a_k$s, so the question reduces to checking that $J-I$ is positive definite. Its eigenvalues are $n-1$ and $0,$ so it is, indeed, positive (semi) definite whenever $n\geq 1.$