# Inner product and infinite sum

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!

• How do you use the condition: $H$ is a Hilbert space? – M. Strochyk Oct 24 '12 at 12:54
• 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. – Tomás Oct 24 '12 at 13:10
• 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. – Hans Oct 24 '12 at 13:30
• $H$ is complete space. – M. Strochyk Oct 24 '12 at 13:37
• @Tomás: Is this true even if we have infinite sum $\sum_{n=1}^{\infty}$? – Mathtag Oct 24 '12 at 14:34

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.