TITLE: multiplication of finite sum (inner product space)
QUESTION [6 upvotes]: I am having difficulty to understand the first line of the proof of theorem 3.22 below. (taken from a linear analysis book)

Why need to be different index, i.e. $m,n$ when multiplying the two sums? This is very basic, but I really need help for explanations.

REPLY [6 votes]: The indices are different because the sums are independent from one another.  If one intends to multiply the sums, then this distinction is a critical one.  
Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction.   Then, we would have incorrectly 
$$\begin{align}
\left(\sum_{n=1}^2a_n\right)\left(\sum_{n=1}^2b_n\right)&=\sum_{n=1}^2\sum_{n=1}^2a_nb_n\\\\
&=2(a_1b_1+a_2b_2)\tag 1
\end{align}$$
However, the correct product that preserves the independence of the summation indices is given by
$$\begin{align}
\left(\sum_{m=1}^2a_n\right)\left(\sum_{n=1}^2b_n\right)&=\sum_{m=1}^2\sum_{n=1}^2 a_mb_n=\sum_{m=1}^2a_m(b_1+b_2)\\\\
&=a_1b_1+a_1b_2+a_2b_1+a_2b_2 \tag 2
\end{align}$$
Evidently, $(1)$ fails to give the correct result.  It has inappropriately mixed the "identities" of the terms to be multiplied and replaced those "cross terms" with an extra "inner-product" term.