1
$\begingroup$

Let $f(x_1,...,x_n)$ be $C^0$ continuous function $R^n\to R$ defined on a compact domain $A\subset R^n$. Let $f$ be strictly monotonously increasing w.r.t. every argument in the domain of definition. I'm looking for a method to continue $f$ to the whole $R^n$ as $C^0$ continuous function strictly monotone w.r.t. every argument.

Particularly, I have a computational problem with the values of a function $f(x,y)$ sampled on 2D grid. The function is given in a compact domain of a grid and it is strictly monotonous w.r.t. $x$ and $y$ in the domain. It should be continued to the whole grid, with the only requirement of being continuous and strictly monotone w.r.t. every argument.

In a special case, when the domain is a square, I can construct an explicit formula for such continuation. So what I miss is a formula or an algorithm to continue the function to the bounding box of the domain, supporting the monotonic property above.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Can you give the precise definition of "strictly monotonously increasing w.r.t. every argument"? The obvious definition would be "for each $k$ and each fixed $x_1, \dots, x_{k-1}, x_{k+1}, \dots, x_n$, the map $x \mapsto f(x_1, \dots, x_{k-1}, x, x_{k+1}, \dots, x_n)$ is strictly increasing on its domain." But then this can't possibly work; if your domain $A$ is, say, the diagonal of the unit square in $\mathbb{R}^2$, then every function has this property, but most of them will not extend. $\endgroup$
    – Nate Eldredge
    Commented Sep 8, 2016 at 13:03
  • $\begingroup$ I agree with your definition, all variables are fixed except of one, then the one is changed in its domain, i.e., until $(x_1,...,x_n)$ is still in $A$. The diagonal from your example is a degenerate case, since the intersection of $A$ and a line $y=Const$ is a single point. Strict monotony in a point has no practical sense. The domains encountered in my 2D problem are full-dimensional compact regions without holes (simply connected). Some of them are convex, some not. I would like to have an algorithm completing this function in the bounding box of the domain, preserving the monotonic (cont.) $\endgroup$
    – Igor
    Commented Sep 8, 2016 at 16:03
  • $\begingroup$ property and working both for convex and non-convex domains. May be, for some functions defined on some non-convex domains such continuation is not possible, in this case I would expect from the algorithm a no-go report. As a brute force algorithm I can imagine is to represent monotony conditions for all points on the grid (outside A , in the bounding box) as linear inequalities, supply some linear objective function and pass everything to a linear programming (LP) solver. The only doubt is a large number of points on the grid and a worst case complexity of LP solver. (cont.) $\endgroup$
    – Igor
    Commented Sep 8, 2016 at 16:03
  • $\begingroup$ So I am still looking for an algorithm of lower complexity or an explicit formula for such continuation. $\endgroup$
    – Igor
    Commented Sep 8, 2016 at 16:03

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .