# Extension of polynomial functions

Let $P$ be an interval in $\mathbf{R}$, $n \in \mathbf{N}$. Assume that a function $f: P \rightarrow \mathbf{R}$ satisfies $\Delta^{n+1}_h f(x)=0$ for every $x \in P$ and every $h>0$ such that $x+(n+1)h \in P$. I want to know whether there exists a function $F: \mathbf{R} \rightarrow \mathbf{R}$ such that $\Delta^{n+1}_h F(x)=0$ for every $x \in \mathbf{R}$ and every $h>0$,and $F|_P=f$.

Thanks

-
1) This looks suspiciously like your homework 2) What exactly is this Delta operator? 3) Can you provide some background, where does this problem occur, why do you want to ḱnow this, what have you tried and failed at ... that would encourage people to even read the question really carefully. –  Konrad Voelkel Sep 22 '11 at 11:59
$\Delta^n_h$ is the difference operator defined by $\Delta_h f(x)=f(x+h)-f(x)$ , $\Delta^{n+1}_h f(x)=\Delta_h (\Delta^n_h f(x))$. My question is related to the following result: It is now that if a function $f : R \rightarrow R$ satisfies $\Delta^{n+1}_h f(x)=0$ for $x\in R$, $h>0$ and is bounded on a subset of positive Lebesgue measure then it is a polynomial of order $\leq n$. –  rts Sep 22 '11 at 13:33

## 1 Answer

Yes, such an extension does exist.

Let $\{ e_i \}_{i \in I}$ be a basis of $\mathbb{R}$ over $\mathbb{Q}$. Pick any finite subset $e_1, ..., e_k$ of the $e_i$s, and look at the restriction of $f$ to $P' = (\mathbb{Q}e_1 \oplus \cdots \oplus \mathbb{Q}e_k) \cap P$. It's easy to see that there is a unique extension of $f$ to a polynomial $\mathbb{Q}e_1 \oplus \cdots \oplus \mathbb{Q}e_k \rightarrow \mathbb{R}$ of degree at most $n$, where by polynomial, I mean a polynomial in the coefficients on $e_1, ..., e_k$ (for uniqueness, you can use that $P'$ is Zariski-dense in $\mathbb{Q}e_1 \oplus \cdots \oplus \mathbb{Q}e_k$).

By the uniqueness of this polynomial, we see that if instead of choosing $e_1, ..., e_k$ we had chosen some subset of them, we would still get the same function on the vector space spanned by the subset. Thus, the following definition for $F$ is well-defined: for any $x \in \mathbb{R}$, write $x$ as a linear combination of finitely many $e_i$s $e_1, ..., e_k$, and evaluate the polynomial extension of $f$ to $\mathbb{Q}e_1 \oplus \cdots \oplus \mathbb{Q}e_k$ at $x$.

To see that this satisfies $\Delta^{n+1}_hF(x) = 0$ for any $x,h$, choose a finite set $e_1, ..., e_k$ with which we can represent both $x$ and $h$, and then note that on the restriction to $\mathbb{Q}e_1 \oplus \cdots \oplus \mathbb{Q}e_k$, $F$ is equal to a polynomial of degree at most $n$, and thus $\Delta^{n+1}_hF(x) = 0$.

-