= Solution
First check that the proposed space is closed under the stated operations: $|\alpha a_i+b_i|^2\le2\alpha^2a_i^2+2b_i^2$, and summing gives a finite result. For any finite $N$, the <Cauchy-Schwarz inequality> gives
$$
\sum_{i=1}^N|a_ib_i|
\le\left(\sum_{i=1}^Na_i^2\right)^{1/2}
\left(\sum_{i=1}^Nb_i^2\right)^{1/2}
\le\|A\|_2\|B\|_2.
$$
The left side is an increasing, bounded sequence of real numbers. Its <limit> exists, proving <absolute convergence> of $\sum_i a_ib_i$. This justifies termwise addition and scalar multiplication in the defining sum. Symmetry and <bilinearity> follow, while $\langle A,A\rangle=\sum_i a_i^2\ge0$, with equality only when every $a_i=0$. Thus it is a real <inner product> on the real <l2 sequence space>.
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