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 $\mathbb{R}$).
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3$\begingroup$ In what context did you encounter this problem? What are your thoughts on it? $\endgroup$– Ayman HouriehCommented Mar 11, 2013 at 21:12
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1$\begingroup$ The important result is that two polynomials which agree at infinitely many values are equal. $\endgroup$– Thomas AndrewsCommented Mar 11, 2013 at 21:18
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$\begingroup$ the context in which it occurs, is to begin studying analytic functions in the complex variable course. $\endgroup$– Alex MedCommented Mar 11, 2013 at 21:28
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3 Answers
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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 $I$.
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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 $\mathbb{R}$.
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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.
answered Mar 11, 2013 at 21:29