The relation between the convexity and the monotonicity of a function Assume the function $ F:\; ] 0,+\infty)^{n} \rightarrow] 0,+\infty) $
is  convex and $ F(x_{1},x_{2},\cdots, x_{n}) = 0 $ if and only if $ x_{i} = 0 \text{ for } i=1, 2 , \ldots , n. $ I would like to know if it is true that the function $\mu$ defined by:
$$ \mu ( x_ { 1 },x_ { 2 },\ldots, x_ { n} ) = F \left( \mu _ { 1 } \left(x_ { 1 } \right) , \mu _ { 2 } \left(x_ { 2}\right) , \ldots , \mu _ { n } \left( x_ { n} \right) \right)\quad \forall ( x_ { 1 },x_ { 2 },\ldots, x_ { n})\in X = ] 0,+\infty)^{n}  ,$$
is a nondecreasing function, where each $\mu _ { i}:\; ] 0,+\infty)\rightarrow] 0,+\infty)\text{ for } i=1, 2 , \ldots , n $ is nondecreasing and convex.
If $n\ge2$ then $F$ has to be zero, if I understand the assumptions right.

Suppose $F(x)\ge0$ for all $x\ge0$, and $F(x)=0$ if $x_i=0$ for one $i\in \{1\dots n\}$. Then $F(x)=0$ for all $x$:

Take two points $\hat x$ and $\tilde x$, both have one zero entry, so $F(\hat x)=F(\tilde x)=0$. Take a point $x$ on the line connecting $\hat x$ and $\tilde x$. Then by convexity, $F(x)\le0$. Since $F(x)\ge0$, $F(x)=0$ follows. Since every point $x\ne 0$ is on at least one such line,
$F\equiv0$ follows.
Professor A.Aghajani answered me as follow:

The reason is that the function "$F$" satisfies the assumptions of Thm 1.3 should be increasing with respect to each component. To see this, for simplicity assume "$F$" is one variable, and remember that "$F$" is convex, nonnegative and $F(0)=0.$ Now assume $x&lt;y$, then there is a positive $"t&lt;1"$ such that $x=ty.$ Now we can write, using the convexity:
$$F(x)= F(ty)=F( (1-t).0 + ty )&lt; (1-t)F(0)+t F(y)=0+tF(y)&lt;F(y).$$
So we proved "$F$" is increasing, now if "$\mu$" is a measure then it should be increasing (as a set function) thus $F(\mu)$ is a composition of two increasing function which is also increasing...