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Suppose that $f \in L^{1}[0,1]$ and $\int_{0}^{1}x^nf(x)\, dx=0$ for $n=0,1,2,\dots$ Does that imply that $f=0$ a.e.?

I think that there will be a counterexample but it is hard to find out.

JohnD
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Guillermo
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    Take $f(0)=42$ and $f(x)=0$ for $x\in(0,1]$. Then for all $n\in\mathbb{N}$, $\int_0^1x^nf(x),\mathrm{d}x=0$, yet $f$ is not nil (only nil almost everywhere). – gniourf_gniourf Dec 20 '14 at 18:46
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    What is your definition of "$=0$"? Is it "$=0$" in the sense of $L^1$ (that is, $=0$ a.e.) or "$=0$" in the sense of functions (that is, $=0$ everywhere)? In the latter case the answer is no, through simple examples whose support is one point. In the former case I think the answer is yes. The easiest argument is probably approximation by continuous functions. – Ian Dec 20 '14 at 18:47
  • http://mathoverflow.net/questions/32369/functions-orthogonal-to-xn – Matthew Towers Dec 20 '14 at 18:49
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    @mt_ Good reference. But $f \in L^{1}$ doesn't have to be in $L^2$. – Guillermo Dec 20 '14 at 19:00
  • $$ \sum_{n\ =\ 0}^{\infty}{1 \over n!}\int_{0}^{1}x^{n},{\rm f}\left(, x,\right),{\rm d}x = 0 $$ – Felix Marin Dec 20 '14 at 19:15

1 Answers1

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Your assumptions imply that $\int p(x) f(x) \, dx = 0$ for all polynomials $p$.

Now, by the (Stone)-Weierstraß Theorem, every continuous function $g \in C([0,1])$ is a uniform limit of polynomials. This yields (how exactly?) $\int g(x) f(x) \, dx = 0$ for all continuous functions $g \in C([0,1])$.

Now, every $g \in L^\infty ([0,1]) \subset L^1([0,1])$ can be approximated (in the $L^1$-norm) by a sequence of continuous functions $(g_n)_n$ (see e.g. Compact support functions dense in $L_1$). For a suitable subsequence (again denoted by $(g_n)_n$), this implies convergence almost everywhere (see e.g. $L^1$ convergence gives a pointwise convergent subsequence).

By considering $h_n := \min \{\Vert g \Vert_\infty, \max\{ - \Vert g \Vert_\infty, g_n\} \}$, you can assume that the sequence $(h_n)_n)$ is uniformily bounded and converges to $g$ almost everywhere.

By dominated convergence (what is the dominating function?), this yields

$$ \int f(x) g(x) \, dx = \lim_n \int f(x) h_n(x) \, dx = 0. $$

But if we now take $g = \chi_{\{x \mid f(x) \geq 0 \}}$, this implies (how?) $f \leq 0$ almost everywhere. An analogous argument yields $f \geq 0$ almost everywhere and hence $f \equiv 0 $ almost everywhere.

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PhoemueX
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