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Let $\{f_{n}\}_{n=1}^{\infty}$ be an orthogonal sequence of nonzero functions in a Hilbert space $H$ with inner product $\langle f,g\rangle_{H}=\int_{-\infty}^{\infty}f(x)g(x)dx$. Show that for any sequence of numbers $\{a_{n}\}$, with $\sum_{n}|a_{n}|^{2}<\infty$ and $\sum_{n}a_{n}f_{n}=0$ then $a_{n}=0$ for all $n$.

I tried the following: Let $\{a_{n}\}$ be a sequence, with $\sum_{n}|a_{n}|^{2}<\infty$ and $\sum_{n}a_{n}f_{n}=0$. Then pick any $f_{m}$, and take inner product with the sum: $$0=\langle f_{m}, \sum_{n}a_{n}f_{n} \rangle= \sum_{n}a_{n}\langle f_{m}, f_{n} \rangle=a_{m}\langle f_{m}, f_{m} \rangle $$ wich implies that $a_{m}=0$ for all $m$. But I'm a little worry about taking the sum out the inner product, it is like interchanging order of sum and integral, since

$$\langle f_{m}, \sum_{n}a_{n}f_{n} \rangle= \int_{-\infty}^{\infty} \sum_{n}a_{n}f_{m}(x)f_{n}(x)dx=\sum_{n}a_{n} \int_{-\infty}^{\infty} f_{m}(x)f_{n}(x)dx=\sum_{n}a_{n} \langle f_{m}, f_{n} \rangle $$

Did I miss anything? Do I need to worry about this? I think I should use that $\sum_{n}|a_{n}|^{2}<\infty$ somewhere!

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    How do you use the condition: $H$ is a Hilbert space?2012-10-24
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    You dont have to worry about the definition of the inner product. If it is an inner product, you just have to use the properties of an inner product.2012-10-24
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    The series converges in the Hilbert norm, right? Then I agree with Tomás, the question does not depend on wether $H$ is a function space or not. Your argument seems correct to me (you use continuity of the scalar product) and the assumption that $(a_n)$ is square summable is redundant I think.2012-10-24
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    $H$ is *complete* space.2012-10-24
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    @Tomás: Is this true even if we have infinite sum $\sum_{n=1}^{\infty}$?2012-10-24
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    @Mathtag, yes it is like Filippo said.2012-10-24

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