Originally Posted by
HallsofIvy Assuming that you are, in fact, talking about an infinite dimensional vector space and that $\displaystyle \{e_i\}$ is an orthonormal vector space, then those are just extensions of the corresponding notions for finite dimensional vector spaces.
In particular, if $\displaystyle u= \sum_{i= 0}^\infty \alpha_ie_i$, we can determine each of the coefficients, $\displaystyle \alpha_i$, by taking the inner product of u with $\displaystyle e_i$: $\displaystyle <u,e_i>= <\sum_{j=1}^\infty \alpha_j e_j, e_i>= \sum_{j=1}^\infty \alpha_j <e_j, e_i>$. But $\displaystyle <e_j, e_i>= 1$ if i= j, 0 otherwise so that sum collapses down to the single ualue, $\displaystyle <u, e_i>= \alpha_i$. That's where line (2) comes from: Since $\displaystyle \alpha_i= <u,e_i>$, $\displaystyle u= \sum_{i=0}^\infty \alpha_ie_i= \sum_{i=0}^\infty <u,e_i>e_i$.
The norm, in any innerproduct space, is defined by $\displaystyle ||u||= \sqrt{<u,u>}$ so if $\displaystyle u= \sum_{i=1}^\infty <u, e_i>e_i$, then $\displaystyle ||u||= \sqrt{<\sum_{i=1}^\infty \alpha_iu_i, \sum{j=1}^\infty \alpha_j u_j>}$$\displaystyle = \sqrt {\sum_{i=1}^\infty\sum_{j=1}^\infty \alpha_i\alpha_j <e_i,e_j>}$. But since, again, since $\displaystyle e_i$
are orthonormal, $\displaystyle <e_i, e_j>$ = 1 if i= j, 0 otherwise, so that sum reduces to $\displaystyle \sqrt{\sum_{i=1}^\infty \alpha_i^2}$ which, as above, is exactly the same as $\displaystyle \sqrt{\sum_{i=1}^\infty |<v,e_i>|^2}$.