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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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    @Mathtag, yes it is like Filippo said.2012-10-24

1 Answers 1

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As for each fixed $y$, the map $x\mapsto \langle y,x\rangle$ is continuous (by Cauchy-Schwarz inequality), and the sequence $\{\sum_{j=1}^Na_jf_j\}$ is Cauchy by the assumption $\sum_n|a_n|^2<\infty$, the step is allowed.