4.7

Let $f$ be a continuous function on $I = [-1, 1]$ with the property that $\int_{I} x^n f(x) \ dx = 0$ for $n = 0, 1, ...$. Show that $f$ is identically 0.

PROOF

So, since $|f|$ is integrable, we can use Weierstrauss' Approximation theorem as follows. Since $\int_{I} x^n f(x) \ dx = 0$ for any non-negative n, then for any polynomial P, $\int_{I} P(x) f(x) \ dx = 0$. By Weierstrauss' Approximation theorem, we know that we may approach f with polynomials, hence taking a sequence of polynomials $P_n$ approaching f, then $\int_{I} \lim_{n} P_n(x) f(x) dx \leq \lim\inf \int_{I} P_n(x) f(x) dx$ by Fatou's Lemma, hence $\int_{I} (f(x))^2 dx \leq 0$, hence $f^2$ is 0 almost everywhere, hence so too is f.

Written by Nicholas Stull

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett