Split a function into finitely many functions Consider a function $\phi : \mathbb{R}^+ \rightarrow \mathbb{R}^+$. $\phi$ is convex and increasing. 
Can we come up with functions $\phi_i : \mathbb{R}^+ \rightarrow \mathbb{R}^+, i \in \{1,2,\cdots,n \}$ such that $$ \phi(\sum_{i=1}^{n} \theta_i) = \sum_{i=1}^{n} \phi_i( \theta_i), $$ where $n$ is finite and $\theta_i \in \mathbb{R}^+$ for all $i$?
If yes, what properties must $\phi_i$'s satisfy? 
 A: Suppose we have $\phi(\sum_{i=1}^{n} \theta_i) = \sum_{i=1}^{n} \phi_i( \theta_i)$ for all $\theta_i$. Let $s=\sum_{i=1}^{n} \theta_i$. Replacing $\theta_j$ by $\theta_j + d$ for just one value of j and eliminating the unchanged terms we see
$\phi(s + d) - \phi(s) = \phi_j(\theta_j + d) - \phi_j(\theta_j)$.
This is for any j and any s and $\theta_j$ ($\theta_j < s$), so 
$\phi_j(s+d)-\phi_j(s) = \phi_j(t+d)-\phi_j(t)$
for any positive s, t, d. Similarly for $\phi$.
A function $\mathbb{R} \rightarrow \mathbb{R}$ with this property is not necessarily linear (though I don't know about $\mathbb{R}^+ \rightarrow \mathbb{R}^+$) but an increasing one must be.
Note also that if we put $\theta_i = a/(n-1)$ for some a and let $b_1=\sum_{i=2}^{n} \phi_i( \theta_i)$ then, for all x,
$\phi(x+a) = \phi_1(x) + b_1$.
$\phi(x+a)-\phi(x)$ is independent of x so, letting $c_1 = \phi(x+a) - \phi(x) +b_i$ we get
$\phi_1(x) = \phi(x) + c_1$
Similarly we get c_i such that $\phi_i(x) = \phi(x) + c_i$ for the other values of i.
So $\phi$ and the $\phi_i$'s are all linear functions with the same gradient, differing only by constants.
PS. "Linear" can be ambiguous. I mean functions of the form f(x) = mx + c, whose graph is a straight line, but not necessarily passing through the origin. 
