[![enter image description here][1]][1]

Let $[a,b)\subset\mathbb R$, and $F,G:[a,b)\to\mathbb R$ two decreasing piecewise linear functions so that $F(x)\leq G(x)$ for any $x\in[a,b)$. We assume that:

 1. there is a number $k\in\mathbb N-\{0\}$ and a set of $k+1$ numbers $a=x_0\lt\ldots\lt x_k=b$ partitioning $[a,b)$ in $k$ intervals $[x_{j-1},x_j)$ on which the restrictions of $F$ and $G$ are linear.
 2. the equalizer set $E=Eq(F,G):=\{x\in[a,b)|F(x)=G(x)\}$ is included in the set $\{x_0,x_1,\ldots,x_k\}$.
 3. the restriction of $F$ to the set $E$ is strictly decreasing
 4. for any $j$, $0\lt j\lt k$, there is an open neighborhood $(x_j-\epsilon,x_j+\epsilon)$, and a linear function $L:(x_j-\epsilon,x_j+\epsilon)\to\mathbb R$, so that $G\geq L\geq F$ on $(x_j-\epsilon,x_j+\epsilon)$.

(conditions 2-4 forbid some cases when the problem has no solution)

**Problem:** Find a continuous family of functions $f_t:[a,b)\to\mathbb R$, $t\in[0,1]$ satisfying the conditions:

 1. $f_t$ is smooth and strictly decreasing for any $t\in(0,1)$.
 2. For any fixed $x\in[a,b)-E$, the application $\vartheta_x:[0,1]\to [F(x),G(x)]$, $\vartheta_x(t)=f_t(x)$ is a strictly increasing and bijective smooth function.

It is easy to see that from the condition 2 it follows that:

 - if $0\leq s\lt t\leq 1$, then $f(s)\lt f(t)$ on $[a,b)-E$ (and of course $f(s)=f(t)=F=G$ on $E$).
 - $f_0=F$ and $f_1=G$.

If possible, please also provide some references which can help solving this problem.

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**Update 1:**

So far I tried to construct the functions from known analytical functions, splines and sigmoids, and to use Scwharz-Christoffel to map the region between $F$ and $G$ to a rectangle in the complex plane. While these methods appeared to have some advantages, it seems difficult to show that they really satisfy the required conditions. Anyway, I don't want to reinvent the wheel.


  [1]: https://i.sstatic.net/qsUpP.jpg