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Let $f$ be a real valued continuous function on the interval $[0,1]$ such that $$\int_0^1 x^n f(x) dx=0$$

for $n=0, 1, 2 ...$. Show that $f$ is identically zero.

mathsucker
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    Can you use Stone Weierstrass theorem? –  May 18 '15 at 03:05
  • Stone-Weierstrass as @John stated. This one shows up a lot eh? – matt biesecker May 18 '15 at 03:06
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    We've had basically the same question two days ago, have a look here: http://math.stackexchange.com/questions/1284317/if-f0-1-to-mathbfr-is-continuous-and-that-int-01-fxxk-dx-0-f – aexl May 18 '15 at 03:10

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In particular, this means that if $Q(x)$ is any polynomial, $\int_0^1Q(x)f(x)dx=0$

Consider a sequence $P_n$ of polynomials converging uniformly on $[0,1]$ to $f$ due to Stone-Weierstrass theorem. Then $\int_0^1P_n(x)f(x)dx=0$ for all $n$.

Since $P_n\to f$ uniformly, $P_nf\to f^2$ uniformly, hence $\int_0^1 P_n(x)f(x)dx\to \int_0^1f^2(x)dx$ uniformly. But $\int_0^1P_n(x)f(x)dx=0$!! So $\int_0^1 f^2(x)dx=0\implies f=0$ as $f$ is continuous.

Landon Carter
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