Show that $f(x)\equiv 0$ if $ \int_0^1x^nf(x)\,dx=0$ Let $f \in C[0,1]$. If for each integer $n \ge 0$ we have 
$$ \int_0^1x^nf(x)\,dx=0$$ show that $f(x) \equiv 0$
 A: Another approach. From the assumption,
$$
\int_0^1 p(x)f(x)\,dx = 0
$$
for every polynomial $p$.
By Weierstrass' approximation theorem, every continuous function can be approximated uniformly by polynomials. In particular, let $p_n$ be a sequence of polynomials approximating $f$ uniformly on $[0,1]$. Then
$$
\int_0^1 (f(x))^2\,dx = \lim_{n\to\infty} \int_0^1 p_n(x)f(x)\,dx = 0.
$$
Since $f(x)^2 \ge 0$ this implies $f(x) = 0$ for all $x$. This is standard. If you haven't seen it, one sleek way to do it is to put
$$
I(t) = \int_0^t (f(x))^2\,dx.
$$
Then $I$ is increasing on $[0,1]$ and $I(0) = I(1) = 0$ so $I(t) = 0$ for all $t$. In particular $0 = I'(t) = (f(t))^2$ for $t \in [0,1]$, so $f(t) = $ for all $t\in [0,1]$.
A: The equality implies that $f$ is orthogonal to every polynomial in $L_2[0,1]$, so in particular it is orthogonal to the Legendre polynomials (which form a basis in $L_2[0,1]$). This implies that $f$ must be $0$ a.e., and Since $f$ is continuous, we must conclude that $f \equiv0$.
