Linear function inverse I have two monotonically increasing piecewise-linear functions:
\begin{align}
F_{L}(s) &= s_L\\
F_{H}(s) &= s_H
\end{align}
and their inverses:
\begin{align}
F_{L}^{-1}(s_L) &= s\\
F_{H}^{-1}(s_H) &= s
\end{align}
How can I find the function :
\begin{align}
F_{M}( (1-h)s_L +hs_H ) &= s \\
\end{align}
where $h \in [0,1]$
Can I calculate $F_M$ in terms of the other functions?
The ranges of $F_L$, $F_H$ and $s$ are in $[0,1]$, if that makes any difference.
 A: I take it that $h$ is given and you're forming the convex combination $F_C=(1-h)F_L+hF_H$ of the two piecewise linear functions, which is again piecewise linear. You know the inverses of $F_L$ and $F_H$, and you want to find the inverse of $F_C$, which you denote by $F_M=F_C^{-1}$.
As Marc has pointed out, $F_C$ is not necessarily invertible. However, assuming that $F_L$ and $F_H$ are (strictly) monotonic in the same direction, $F_C$ is also (strictly) monotonic, an hence invertible.
I don't think you can express its inverse in terms of the other functions; you can just apply the same process that gave you $F_L^{-1}$ from $F_L$ and $F_H^{-1}$ from $F_H$ to $F_C$ to give you $F_C^{-1}$, since $F_C$ is also piecewise linear. If for some reason you can't do that, I don't see how you could use $F_L^{-1}$ and $F_H^{-1}$ to construct $F_C^{-1}$, since you wouldn't know which arguments to apply them to, since a value of $F_C$ could have been assembled from different pairs of values of $F_L$ and $F_H$.
A: With no more information than in the question, there is no guarantee that $(1-h)S_L+hS_H$ is invertible in the first place. For instance whenver $F_L=-F_H$, one gets for $h=\frac12$ that $(1-h)S_L+hS_H=0$. This circumstance should destroy any hope of finding a pleasant formula for $F_M^{-1}$ in this general setting.
