# Inner product of an infinite sum

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

-
When you say "Is it = $\sum_{n=1}^\infty \sum_{n=1}^\infty |a_{n}|^{2}|b_{n}|^{2}$", the two sums should use different indices. – Zev Chonoles Mar 16 '11 at 0:31
Right, I fixed it. – user8315 Mar 16 '11 at 0:33

You could simplify it some this way: $$\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle=\lim_{N \to \infty} \lim_{M\to \infty} \left \langle \sum_{n=1}^N a_{n}b_{n},\sum_{m=1}^M a_{m}b_{m}\right \rangle = \lim_{N \to \infty} \lim_{M\to \infty} \sum_{n=1}^N \sum_{m=1}^M \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle$$ $$= \sum_{n=1}^\infty \sum_{m=1}^\infty \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle=\sum_{n=1}^\infty \sum_{m=1}^\infty a_{n}b_{n} \overline{a_{m}b_{m}}$$
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 $1*1 + 1*(-1) + \cdots = 0$.