# polynomial function locally to globally polynomial.

Let $f:\mathbb{R} \to \mathbb{R}$. Suppose that f is locally polynomial (each $x \in \mathbb{R}$ has a neighborhood $(x-\delta; x +\delta)$ wherein $f$ coincides with some polynomial). Show that f is polynomial (coincides with some polynomial in all $\mathbb{R}$).

• In what context did you encounter this problem? What are your thoughts on it? Commented Mar 11, 2013 at 21:12
• The important result is that two polynomials which agree at infinitely many values are equal. Commented Mar 11, 2013 at 21:18
• the context in which it occurs, is to begin studying analytic functions in the complex variable course. Commented Mar 11, 2013 at 21:28

## 3 Answers

Locally at the origin, $f$ coincides with some polynomial $p$; let $I$ be the largest interval such that $f = p$ on $I$. Notice that $I$ is necessarily open. Suppose $I \neq \mathbb{R}$, for example $s=\sup(I) \notin I$. Locally at $s$, $f=p'$ for some polynomial $p'$. But $p=p'$ on some interval hence $p=p'$ on $\mathbb{R}$: contradiction with the maximality of $I$.

Hint: Show that the set of points, in whose neighborhood the function is given by a fixed polynomial $p$, is both open and closed and use connectedness of $\mathbb{R}$.

Alternatively, for any $x_0$ find $p(x)$ to be a polynomial such that $f(x)=p(x)$ on $(-\delta,+\delta)$. Then take the union of all open intervals that contain $0$ on which $f(x)=p(x)$. This union is open and path-connected. Show that it has no infimum or supremum.