TITLE: possible to find $f^*$ satisfying $\; \sum_{i=1}^{N} f(g(x,i)) = f^*\left(\sum_{i=1}^{N}g(x,i)\right)$
QUESTION [0 upvotes]: Let $f:\mathbb{R} \rightarrow \mathbb{R},  g:\mathbb{R}^n \times\mathbb{N} \rightarrow \mathbb{R}$
Then is it always possible to find a function $f^*$ which is satisfying
$ \forall x \; \sum_{i=1}^{N} f(g(x,i)) = f^*\left(\sum_{i=1}^{N}g(x,i)\right)$
for any function $f$? 
And is there any remarkable relationship between $f$ and $f^*$?

REPLY [0 votes]: In general no (if $N\ge 2$). Set $F(x) := \sum_1^Nf(g(x,i))$. Then you want to have $f^*$ such that $f^*(\sum_1^Ng(x,i)) = F(x)$. Now assume that for $x,y$ you have that $\sum_1^Ng(x,i) = \sum_1^Ng(y,i)$ but $F(x)\neq F(y)$. Then there is obviously no chance to find such an $f^*$.
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