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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.

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  • $\begingroup$ @gerw it is dense, see Müntz–Szász theorem. $\endgroup$
    – Yimin
    Feb 18, 2016 at 0:30
  • $\begingroup$ My fault. But then, you can use this density result: It implies that $\int_0^1 f \, v \, \mathrm{d}x = 0$ for all $v \in L^2(0,1)$. Hence, $f \equiv 0$. $\endgroup$
    – gerw
    Feb 18, 2016 at 7:44

1 Answer 1

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Suppose $\int_{0}^{1}f(t)t^{2l}dt=0$ for all $l=0,1,2,3\cdots$. Using a power series, $$ \int_{0}^{1}f(t)\cos(n\pi t)dt=0,\;\;\; n=0,1,2,3,\cdots. $$ Therefore, the even extension $f_e$ of $f$ to $[-1,1]$ has all $0$ Fourier coefficients. So $f_e=0$ a.e.. Because $f$ is smooth on $[0,1]$, then $f\equiv 0$ on $[0,1]$.

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  • $\begingroup$ I thought about this before. My another question is if we are provide $f$ is orthogonal to all $x^{2l+1}$, $l\ge 0$, can I get the same result? according to Muntz Szasz, it seems $x^0$ must be included. $\endgroup$
    – Yimin
    Feb 18, 2016 at 3:34
  • $\begingroup$ Anyway, this proof works. $\endgroup$
    – Yimin
    Feb 18, 2016 at 3:34
  • $\begingroup$ @Yimin : In this other case, look at the odd extension $f_o$ of $f$ and the corresponding sin series. You'll get $f_o=0$ a.e., but that won't give $f(0)=0$. However, by continuity, you get that point, too. $\endgroup$
    – Disintegrating By Parts
    Feb 18, 2016 at 13:58

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