Return to Answer

added 2 characters in body

Source Link

edited Dec 30, 2010 at 8:59

60.5k
4
122
269

I'd say that the closure of the integer-valued polynomials (IVP) by uniform convergence on bounded sets is the whole set $C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$ of continuous functions mapping integers to integers.

Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $p^K_n\big(\binom x s \big)$$pf^k _n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.

Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $p^K_n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.

Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $pf^k _n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.

Source Link

created Dec 30, 2010 at 8:34

Pietro Majer

60.5k
4
122
269

Let $r\le s\in\ \mathbb{N}$. Then $\big|\binom x s \big| < 1 $ holds for all $x \in ]0,r[$. Consider the IVP $f_n(x):=\frac 1 2 (x + x^n)$, that maps $]-1,1[$ into itself, and converges to $\frac x 2 $ uniformly on compact sets of $]-1,1[$. So th $k$-fold iterate $f_n^k:=f_n\circ\dots\circ f_n$ converges to $\frac x {2^k}$, and for all $p\in\mathbb{Z}$ the sequence of IVP $p^K_n\big(\binom x s \big)$ converges to $\frac p {2^k} \binom x s $ as $n\to \infty$ on $]0,r[$. By density of the dyadic rationals, and by linearity, we get at least any polynomial multiple of $\binom x r$ as a uniform limit of IVP on compact sets of $]0,r[$, and of course, any polynomial taking integer values on $x=0,1\dots, r$ as well. By density, we also get any function $f\in C^0(]0,r[)$ that takes integer values on the integer points; in particular the restriction to $]-1,r[$ of any $f\in C^0\big((\mathbb{Z},\mathbb{R}),(\mathbb{Z},\mathbb{R})\big)$. By translation invariance, and since $r$ is arbitrary, this is true for any interval in place of $]0,r[$, and we conclude by a diagonal argument.