This maybe a trivial question, but I don't seem to find the answer. I want to show that for two vectors $x'=(x_1', ...)$ and $x''=(x_1'', ...)$, such that all elements are in $[0,1]$, the following statement is true:
$\Pi_i(\lambda x' + (1-\lambda) x'')\leq \lambda \Pi_i( x') + (1-\lambda)\Pi_i( x'')$,
where $\Pi$ is the product operatorand $\lambda \in [0,1]$.
Inspired by this post, I tried for $x'=(x_1', x_2')$ (product of only two elements) and found, by computing the Hessian matrix that $x'x''$ is indeed convex on $[0,1]$.
My intuition is that it is still true for more general products but I am not sure how to show it.
Also, if I am wrong I would actually be curious to know if the product is actually concave.
No in general it's not true unfortunately (in $\mathbb{R}^{n}$) but it is indeed convex on $\mathbb{R}_{++}^{n}$. Consider $f(x_1,\cdots,x_n)=x_1x_2\cdots x_n>0$. Then $\frac{\partial^2f}{\partial x_i \partial x_j} = x_1x_2\cdots x_{i-1}x_{i+1}\cdots x_{j-1}x_{j+1}\cdots x_n=\frac{f}{x_ix_j}$. Therefore the elements of Hessian are $[H]_{ij}=\frac{f}{x_ix_j}$. For an arbitrary vector $v\in \mathbb{R}_{++}^{n}$ we have
$$v^THv=\sum_{i=1}^{n}\sum_{j=1}^{n} f\frac{v_i}{x_i}\frac{v_j}{x_j}(1-\delta(i-j))>0$$
Where $\delta(i-j)=\begin{cases}1 &i=j\\ 0 &O.W.\end{cases}$ is the Kronecker delta function. Since $v^THv$ is a positive form on $\mathbb{R}_{++}^{n}$, thus the function is convex in $\mathbb{R}_{++}^{n}$.
To generalize a little further, it is still convex in $\mathbb{R}_{+}^{n}$ but not strictly convex, because there is the possibility that one or more of $x_i$'s be zero, therefore the Hessian is P.S.D. .