Prove that for any finite set of increasing and onto functions $g_i:[0,1] \to [0,1], i=1,2,..,n$, there is no function $h\neq 0$ with bounded variation such that:
$$\sum_{i=1}^n h\circ g_i = 0$$ .
I could prove this for n=2 as follows:
we can always assume $g_1(x)=x, \forall x$, and we have $h(x)+h(g_2(x))=0 \; \forall x$.
suppose $h(x_0)\neq 0$. then make the sequence $x_{n+1}=g_2(x_n)$, we have $h(x_{n+1})=-h(x_n)$, l.e. $h(x_0)=-h(x_1)=h(x_2)...$. Moreover the sequence $x_n$ is monotone on a compact interval, therefore it converges to a value x, now by looking at the behaviour of $h$ around $x$, we can see that it can't have bounded variation.
Annex result :
if $h(x)$ is convex on $[0,1]$ we can apply Jensen's inequality :
$$\sum_{i=1}^{n}h(g_i(x))=0\geq n\operatorname{ h}\Big(\frac{\sum_{i=1}^{n}g_i(x)}{n}\Big)$$
With equality iff $g_i$ are all equals (trivial case) so we suppose that $g_i$ are not all equals we have :
$$\sum_{i=1}^{n}h(g_i(x))=0> n\operatorname{ h}\Big(\frac{\sum_{i=1}^{n}g_i(x)}{n}\Big)$$
Putting $y=\frac{\sum_{i=1}^{n}g_i(x)}{n}$ we have :
$$0>h(y)\quad \forall\,y\,\operatorname{such that} 0\leq y \leq 1$$
So we get a contradiction since we have :
$$0=\sum_{i=1}^{n}h(g_i(x))<0$$
The same things can be done with $h(x)$ concave
So we conclude that the function $h(x)$ is neither concave or convex .