convex function of n convex function Assume $f$ is convex function that takes $n$ arguments, and we know $f$ is non-decreasing in every component.
Assume we have another $n$ convex functions: $h_1,...,h_n$
Does that composition $f(h_1(x_1),\ldots,h_n(x_n))$ is convex?
I assume it does, but I also know that the composition of convex function isn't necessarily convex so I'm having hard time to prove it.
 A: Let $\alpha\in]0,1[$ and let $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ be in $\mathbb{R}^N$. To show convexity of your function, it suffices to show
\begin{equation}
f\left(h_1(\alpha x_1 + (1-\alpha)y_1), \ldots, h_n(\alpha x_n + (1-\alpha) y_n)\right)\leq \alpha f(h_1(x_1),\ldots,h_n(x_n)) + (1-\alpha)f(h_1(y_1),\ldots,h_n(y_n)).
\end{equation}
Proof idea: For every $i\in \{1,\ldots,n\}$, since $h_i$ is convex, we know
\begin{equation}
h_i(\alpha x_i + (1-\alpha) y_i)\leq \alpha h_i(x_i) + (1-\alpha)h_i(y_i).
\end{equation} Since we know $f$ is increasing in each component, we use this inequality in every component to find
\begin{equation}
f\left(h_1(\alpha x_1 + (1-\alpha)y_1), \ldots, h_n(\alpha x_n + (1-\alpha) y_n)\right)\leq
f(\alpha h_1(x_1)+ (1-\alpha)h_1(y_1), \ldots, \alpha h_n(x_n) + (1-\alpha)h_n(y_n)).
\end{equation}
Notice that the righthand side above is just $f$ evaluated at $\alpha (h_i(x_i))_{1\leq i\leq n} + (1-\alpha) (h_i(y_i))_{1\leq i\leq n}.$ Using convexity of $f$ in $\mathbb{R}^n$ will yield the desired inequality :)
