# Smooth function with $f^{(n)}=0$ is polynomial

by philmcole   Last Updated January 16, 2018 22:20 PM

Let $I \subseteq \mathbb R$ be an interval and $f:I \to \mathbb R$ a function. Show that $f$ is a polynomial if and only if $f$ is smooth and there exists an $n \in \mathbb N$ so that $f^{(n)} = 0$.

I think I've got the forward direction:

Let $f(x) = \sum_{k=0}^n a_n x^n$ be a polynomial. We know that monomials are differentiable and the derivative is a monomial with one degree less. Since a polynomial is only a linear combination of monomials and the derivative is linear, we have that polynomials are differentiable and the derivative is also a polynomial with one degree less. Therefore we get that $f^{(n)}$ exists and is a constant. Since the derivative of a constant function exists and is the zero function we get that $f^{(n+1)} = 0$. The zero function is differentiable with derivative the zero function, so we get that $f$ is smooth.

But I struggle with the reverse direction, where I have to show that $f$ is indeed no other function than a polynomial. I should make use of the following corollary: $f$ is constant iff $f$ is differentiable and $f'=0$.

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Let is prove that for all $n\geq1$ the following statement holds:

$(P_n)$ if a smooth function $f$ is such that $f^{(n)}=0$, then $f$ is a polynomial.

The statement $(P_1)$ holds: if $f^{(1)}=0$, that is, if $f'=0$, then we know that $f$ is constant, so that it is a polynomial.

Suppose now that $(P_n)$ holds and let $f$ be a smooth function such that $f^{(n+1)}=0$. Then $g=f'$ is a smooth function such that $g^{(n)}=0$ and the fact that we are assuming that $(P_n)$ holds implies that $g$ is a polynomial. But then $f$, which is a primitive of $g$, is also a polynomial. We thus see that $(P_{n+1})$ holds if $(P_n)$ holds.

Mariano Suárez-Álvarez
January 16, 2018 21:55 PM