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Let $f$ be a continuous function with $f: [0,1] \to \mathbb{R}$ such that $$\int_{0}^{1}f(x)x^{n}dx = 0$$ for all non negative $n$. Prove that $f = 0$.

I tried to think this problem like this: if $f=0$ then it is obvious. If $f$ is positive also I can prove that $f = 0$. In general I started thinking by Weirstrass approximation theorem (every continuous function can be approximated by a polynomial). I don't know if I'm going in the right direction or not. Can someone please tell me the proper proof of this.

Arun Sharma
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    Here's a way to do it with Weierstrass' approximation theorem. – mrf May 16 '16 at 13:32
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    If $f(x)\ne0$ at some point then there exist a small region near this point that it is nonzero. Then you use Weierstrass to construction a triangle inside this region and zero outside. Then the integral will be non-zero, contradicting your given condition. – velut luna May 16 '16 at 13:33
  • https://en.wikipedia.org/wiki/Moment_problem – Rodrigo de Azevedo May 18 '16 at 21:58

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According to Weierstrass Approximation Theorem in mathematical analysis every continuous function can be approximated by a sequence of polynomials, so $f$ itself is the limit of a sequence of polynomials. Take ${P_{n}}$ as the desired sequence. So for any positive number $\epsilon$, there is a big enough number $N$ so that for each $n>N$ we have \begin{equation} \lvert f-P_{n}\rvert < \epsilon \end{equation} for every $x$ in $(0,1)$, which means that for $n>N$ \begin{equation} P_{n}-\epsilon<f<P_{n}+\epsilon \end{equation} where $P_{n}$ is a polynomial of finite degree. Take $A$ and $B$ be the set of points where $f$ has positive and negative values respectively. Without loss of generality (why?) we can assume that $A\cup B= (0,1)$. So \begin{equation} \int_{A}f(x)P_{n}dx - \epsilon \int_{A}f(x)dx<\int_{A}f^{2}(x)dx < \int_{A}f(x)P_{n} + \epsilon\int_{A}f(x)dxdx \end{equation} and \begin{equation} \int_{B}f(x)P_{n}dx + \epsilon \int_{B}f(x)dx<\int_{B}f^{2}(x)dx < \int_{B}f(x)P_{n} - \epsilon\int_{B}f(x)dx \end{equation} Combining the two equations, we have: \begin{equation} \int_{0}^{1}f(x)P_{n}dx - \epsilon \int_{A}f(x)dx + \epsilon \int_{B}f(x)dx<\int_{0}^{1}f^{2}(x)dx < \int_{0}^{1}f(x)P_{n} + \epsilon\int_{A}f(x)dxdx - \epsilon \int_{B}f(x)dx \end{equation} but according to the assumption:( $\int_{0}^{1}f(x)x^{n}dx = 0$ for any non-negative $n$ ), we have $\int_{0}^{1}f(x)P_{n}dx = 0$ (why?). So \begin{equation} -C\epsilon <\int_{0}^{1}f^{2}(x)dx < C\epsilon \end{equation} Where
$$ C = \int_{A}f(x)dx - \int_{B}f(x)dx$$ Therefore $$ \lvert \int_{0}^{1}f^{2}(x)dx \rvert < C\epsilon $$.

Since $C$ is a constant number and $\epsilon$ is an arbitrary number we conclude that $$ \lvert \int_{0}^{1}f^{2}(x)dx \rvert = 0$$ but $f^2$ is a positive function on $(0,1)$ so the above equation implies that $f(x)=0$ on $(0,1).$