Determine this set is convex or not.

Question

I have to determine the following set is convex or not.

$$S=\{x \in \mathbb{R}^n_{++} : \prod_{i=1}^n x_i \ge 1\},$$ where $\mathbb{R}^n_{++}= \{(x_1,\ldots,x_n) : x_1,\ldots, x_n > 0\}$.

I think $S$ is not convex set then I have tried to find some counterexamples but they turned out to be false.

Any help would be appreciated.

Answer 1

Lemma Assume $f$ is a concave function defined in some convex subset $C\subset\mathbb{R}^n$. Then for every $c$ the set $$A=\{x\in C| f(x)\geq c\}$$ is convex.

Proof If $x,y\in A$, then for $0\leq\lambda\leq 1$ the point $\lambda x+(1-\lambda y)\in C$ because $C$ is convex. Moreover, since $f$ is concave, $$f(\lambda x+(1-\lambda)y)\geq \lambda f(x)+(1-\lambda)f(y)\geq c$$ Hence the point $\lambda x+(1-\lambda y)\in C$ is also in $A$, so $A$ is convex. Q.E.D

Put $C=\mathbb{R}^n_{++}$, and define $$f(x)=\sum_{i=1}^{n}\log x_i\quad (x\in C)$$ Then the set $S$ in the given question is $$S=\{x\in C| f(x)\geq 0\}$$

For every $1\leq i\leq n$, the function $f_i(x)=\log x_i$ is concave on $C$, hence $f$ is also concave, as a sum of concave functions. It follows from the Lemma that $S$ is convex.

Answer 2

First consider the case when $n=2$. Consider a function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = 1/x$ with $\mathbf{dom} f = \mathbb{R}_{++}$. Then, notice that

\begin{align} S &= \{ (x_{1}, x_{2}) \in \mathbb{R}^{2}_{++}: x_{1}x_{2} \geq 1\} \\ &= \{ (x, t) \in \mathbb{R}^{2}_{++}: xt \geq 1\} \\ &= \{ (x, t) : x \in \mathbf{dom} f, t \geq f(x) \} \\ &= \mathrm{epigraph}(f). \end{align} Since $f$ is convex on $\mathbf{dom} f$, its epigraph is convex and therefore the set $S$ is convex.

Can you generalize the above argument to general $n$?