Suppose that we are given a smooth function $h:\mathbb{R}^n \to \mathbb{R}$ which satisfies $h \circ F= h \circ G$ for two polynomial functions $F,G:\mathbb{R}^m \to \mathbb{R}^n$ (i.e. each component of $F$ and $G$ is a real polynomial of $m$-variables). Can we approximate $h$ arbitrarily well by polynomial functions $P_h$ which also satisfy $P_h \circ F= P_h \circ G$ (or at least such that these functions are "close")? How about when $F$ and $G$ are not polynomial but smooth?
$\begingroup$
$\endgroup$
9
-
1$\begingroup$ In what norm are you approximating? $\endgroup$– Yemon ChoiCommented Dec 8, 2011 at 18:52
-
$\begingroup$ @Yemon: I want to leave that vague for now. If this can be done in some norm I am interested. $\endgroup$– David CarchediCommented Dec 8, 2011 at 18:56
-
$\begingroup$ @David: since your domain is non-compact, is $h$ supposed to be bounded? If not, then you will need to have a weight in the norm, which makes life tricky. For instance, what are you hoping for in the case $m=n=1$? $\endgroup$– Yemon ChoiCommented Dec 8, 2011 at 19:19
-
$\begingroup$ @Yemon: Sadly, I have to let $h$ be an arbitrary smooth function, so I cannot assume it's bounded. $\endgroup$– David CarchediCommented Dec 8, 2011 at 19:25
-
$\begingroup$ @Yemon: For me though, it's ok not to even have a norm. I just need a reasonably functorial topology on the space of smooth functions for which I can find a convergent sequence of polynomials satisfying the properties I listed. $\endgroup$– David CarchediCommented Dec 8, 2011 at 19:30
|
Show 4 more comments
2 Answers
$\begingroup$
$\endgroup$
In general not. For $m=n=1$, $F$ and $G$ must be of equal degree (otherwise $P_h\circ F$ anf $P_g\circ F$ are of different degree). But even equal degrees don't guarantee anything: if $F(x)=x$, $G(x) = x+1$, then no polynomial $P$ can fulfill $P\circ F=P\circ G$ (no polynomial is periodic).
$\begingroup$
$\endgroup$
1
You could use the topology of uniform convergence on compact sets. By Stone-Weierstrass, any continuous function $h$ can be approximated by a sequence of polynomials $P_n$ in this topology, and then $P_n \circ F - P_n \circ G$ approximates $h \circ F - h \circ G = 0$ in this topology (because if $K \subset {\mathbb R}^m$ is compact, $F(K) \cup G(K)$ is compact).
answered Dec 9, 2011 at 6:34
-
$\begingroup$ The statement "if $P_n$ approximates $h$, then $P_n \circ F-P_n \circ G$ approaches $0$" is true whenever composition and subtraction are continuous maps in whatever topology on smooth functions you're considering. Since most (all?) reasonable topologies will satisfy that property, in general this will not give an extra constraint. $\endgroup$ Commented Dec 9, 2011 at 6:42