## Interpolating between piecewise linear functions, with a family of smooth functions

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.

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.

-
 How about something like this: fix $\phi$ a smooth symmetric nonnegative bump function with total integral one, extend $F,G$ by zero to all of $\mathbf{R}$, and then form $f_t(x)=\frac{(1-t)^2}{t}\int_{\mathbf{R}}F(x+y)\phi(y\frac{1-t}{t})dy+\frac{t^2}{1-t}\int_{\mathbf{R}}G(x+y)\phi(y\frac{t}{1-t})dy.$ This is clearly decreasing as a function of $x$ (differentiating under the integral sign gives nonpositive integrands), and it satisfies $f_0=F,f_1=G$, but I haven't examined what properties of $\phi$ are required for monotonicity in $t$ to be true or plausible... – David Hansen Aug 24 2011 at 3:55 You may find something interesting in the so called Nurbs: en.wikipedia.org/wiki/… – Kirill Shmakov Mar 1 2012 at 17:44

Interesting problem! This is not an actual answer, just a comment. I believe some minor condition may be missing, since it seems to me that the problem as posed does not necessarily have a solution. Set for example $F:[0,2)\to\mathbb R$, $F|_ {[0,1)}\equiv 1$, $F|_ {[1,2)}\equiv 0$ and $G:[0,2)\to\mathbb R$, $G|_ {[0,1)}\equiv 2$, $G|_ {[1,2)}\equiv 1$. These two functions satisfy conditions 1-4 but a smooth family you're looking for would have to contain some functions discontinuous at point $x=1$ because of condition 2 which asks that the family be strictly increasing. Maybe the inequalities for $L$ should be strict? Or maybe you should add to $E$ those points where $F(x)$ equals the left or right limit of $G$ there? Maybe I'm just missing something. In the latter case, I kindly ask the moderators to delete this comment.