Suppose $f(x)-f(y)=h(g(x)-g(y))$, then $h$ is linear? Background $f,g$ are real functions, $h$ is increasing, satisfying:
$$f(x)-f(y)=h(g(x)-g(y)),$$
What function is $h$?
Motivation: At a first glance, $h$ can be any function, since $f,g$ are arbitrary functions. However, things are not that simple.
Hypothesis: $h$ must be linear.
My try: Let $y=c$, where $g(c)=0$ (Suppose this zero exist)
$$f(x)-f(c)=h(g(x)-g(c))$$
So,
$$f(x)=h(g(x)-g(c))+f(c)$$
Then, $h(g(x)-g(y))=h(g(x)-g(c))+f(c)-(h(g(y)-g(c))+f(c))$
$$h(g(x)-g(y))=h(g(x)-g(c))-h(g(y)-g(c))$$
Since $g(c)=0$,
$$h(g(x)-g(y))=h(g(x))-h(g(y))$$
By definition, $h$ is linear functional. That is, $h(x)=ax+b$.
However, I have no idea how to prove this without assuming that $g(c)=0$ for some constant $c$.
 A: Fix $c$. If
$$f(x)-f(y)=h(g(x)-g(y))$$
then
$$f(x)-f(y)=h(g_{2}(x)-g_{2}(y))$$
where
$$g_{2}(x)=g(x)-g(c)$$.
I think that completes your argument.
Neat question.
A: You do not give precise informations about $f, g, h$, so let us assume that all are functions $\mathbb R \to \mathbb R$. They satisfy
$$f(x)-f(y)=h(g(x)-g(y)) .\tag{1}$$
Let us first observe that on both sides of $(1)$ only the differences $f(x) - f(y)$ and $g(x) - g(y)$ occur. Thus for any two numbers $c,d$ the functions $f_c(x) = f(x) + c$ and $g_d(x) = g(x) + d$ also satisfy $(1)$. This means that w.l.o.g. we may assume that $f(0) = g(0) = 0$. With $y = 0$ we get
$$f(x) = h(g(x))  \tag{2}$$
and therefore
$$h(g(x)) - h(g(y))  = h(g(x) - g(y)) \tag{3}$$
Clearly, $(3)$ is equivalent to $(1)$  - if $(3)$ is satisfied, we may define $f$ via $(2)$. The crucial condition $(3)$ only depends on the image $G  =g(\mathbb R)$. In fact, it says that that for all $a, b \in G$ we have
$$h(a) - h(b) = h(a - b) . \tag{4}$$
If we do not have restrictions on $g$, then $G$ can be any nonempty subset of $\mathbb R$ containing  $0 = g(0)$. Condition $(4)$ gives a restriction for $h$ only on the set $G^* = \{ a - b \mid a, b \in G \}$. This is a symmetric set ($G^*
 = -G^*$) containing $0$. Outside of $G^*$ we cannot say anything about $h$. But what we can say is that $h$ is an odd function on $G^*$ (take $a  = 0$). In particular, $h(0) = 0$.
Now assume that $g$ and $h$ are continuous. Then either $G = \{0\}$ (and $G^* = \{0\}$) or $G$ is an interval. In the first case $h$ can be any function with $h(0) = 0$. In the second case $G^*$ is a symmetric interval. Pick $r  \in G^*$, $r > 0$. We get $h(r) - h(r/2) = h(r - r/2) = h(r/2)$, thus $h(r/2) = h(r)/2$. Inductively we get $h(r/2^n) = h(r)/2^n$. By another induction we get
$$h(k\cdot r/2^n) = k/2^n \cdot h(r) \tag{5}$$
for $k=1,\ldots,2^n$. Note that $h((k+1)\cdot r/2^n) = h(k\cdot r/2^n  -(-r/2^n)) =  h(k\cdot r/2^n) - h(-r/2^n) = h(k\cdot r/2^n) + h(r/2^n)$. The set of all $k\cdot r/2^n$ is dense in $[0,r]$, thus $h(x) = x/r \cdot h(r)$ for $x \in [0,r]$. This means that $h$ is $\mathbb R$-linear on $[-r,r]$, $h(x) = \lambda x$. But formula $(5)$ is true for all $r' \in G^*$ with $r' > 0$. Since $r'/2^m \in [0,r]$ for some $m$, we see that $h$ is linear on $G^*$.
If $h$ is not required to be continuous, then it may have exotic behavior. Here is an example. Assume that $g$ is surjective, i.e. $G = \mathbb  R = G^*$. Let $\{b_\alpha\}_{\alpha \in A}$ be a basis of $\mathbb R$ as a vector space over $\mathbb Q$. We may assume that all $b_\alpha > 0$ (otherwise replace $b_\alpha$ by $-b_\alpha$). Pick $\alpha_0$ such that the sets $A^- =\{\alpha \in A \mid b_\alpha \le b_{\alpha_0}\}$ and $A^+ =\{\alpha \in A \mid b_\alpha > b_{\alpha_0}\}$ are nonempty. Define
$$h(b_\alpha) = \begin{cases} b_\alpha  & \alpha \le \alpha_0 \\ 2b_\alpha  & \alpha >\alpha_0 \end{cases}$$
This determines a unique $\mathbb Q$-linear map $h : \mathbb R \to \mathbb R$, thus it satisfies $(4)$. It is not $\mathbb R$-linear and not continuous. But I  must admit that I am not sure whether it is increasing.
