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For complex numbers $a_{n}, b_{n}$, what would be the next step to simplify this expression:

$\left \|\sum_{n=1}^\infty a_{n} b_{n}\right \|^{2}=\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle$

Is it $=\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ !?

where $|a_{n}|^{2}=a_{n}\overline{a_{n}}$

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    Right, I fixed it.2011-03-16

2 Answers 2

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You could simplify it some this way: $\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle=\lim_{N \to \infty} \lim_{M\to \infty} \left \langle \sum_{n=1}^N a_{n}b_{n},\sum_{m=1}^M a_{m}b_{m}\right \rangle = \lim_{N \to \infty} \lim_{M\to \infty} \sum_{n=1}^N \sum_{m=1}^M \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle$ $ = \sum_{n=1}^\infty \sum_{m=1}^\infty \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle=\sum_{n=1}^\infty \sum_{m=1}^\infty a_{n}b_{n} \overline{a_{m}b_{m}}$

There are some details in between that I didn't want to type, and I'm a little iffy about the convergence, but I think it works. You can pass the limit outside the inner product because inner products are continuous. Your expression is missing the cross terms.

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I don't think you can simplify it any further. Your simplification does not work because $\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ is actually an upper bound for your expression. To see this, test the example $1*1 + 1*(-1) + \cdots = 0$.

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    @Approximist, use Cauchy Schwarz.2011-03-16