I have a question about orthogonality. Suppose we have a smooth function $f(x)\in C^{\infty}([0,1])$, and we are given that
$$\int_0^1 f(x) x^{2l} dx = 0$$
for all $l = 0,1,\dots$.
Can we conclude that $f(x)= 0$?
Thanks in advance.
Update: It is easy to see that $f$ cannot have finite zeros, thus not analytic. Otherwise, suppose $f$ changes sign at $x_1, x_2, \dots x_n$, we can construct a polynomial
$$P(x) = \Pi_{i=1}^n (x^2 - x_i^2)$$
has the same changing-sign behaviour as $f(x)$. And
$$\int_0^1 P(x) f(x) = 0$$
it is a contradiction. since $P(x)f(x) > 0$ or $P(x)f(x) < 0$ a.e.