What is the condition for this equality to hold? Let $H$ be a Hilbert space and $\{e_n\}$ an orthonormal sequence.
If $x\in H$ satisfies $\sum_{n=1}^\infty |\langle x,e_n\rangle|^2=1$, what can I say about the vector $x$? 
 A: \begin{align}
\sum_{n=1}^N |\langle x,e_n\rangle|^2 & = \sum_{n=1}^N \| \langle x,e_n\rangle e_n \|^2 \text{ because } \{e_n\}_{n=1}^N \text{ are unit vectors} \\[10pt]
& = \left\| \sum_{n=1}^N \langle x, e_n\rangle e_n \right\|^2 \text{ because the terms are orthogonal} \\[10pt]
& = \|x\|^2 - \left\| x-\sum_{n=1}^N \langle x,e_n\rangle e_n \right\|^2 \text{ Why? See below.}
\end{align}
In inner product spaces one has $\|a+b\|^2 = \|a\|^2 + \langle a,b\rangle + \langle b,a\rangle + \|b\|^2$ (where the two middle terms are each other's complex conjugates), so if $a,b$ are orthogonal then $\|a+b\|^2 = \|a\|^2 + \|b\|^2.$ Therefore the justification of the last equality displayed above will be that
$$
\sum_{n=1}^N \langle x,e_n\rangle e_n \text{ is orthogonal to } x - \sum_{n=1}^N \langle x,e_n\rangle e_n.
$$
To see that they are orthogonal, compute their inner product.
We must therefore conclude that
$$
\sum_{n=1}^N |\langle x,e_n\rangle|^2 \le \|x\|.
$$
Since this is true of every value of $N,$ and the terms are nonnegative, we must have
$$
\sum_{n=1}^\infty |\langle x,e_n\rangle|^2 \le \|x\|.
$$
The conclusion about $x$ is therefore that its norm is $\ge 1.$
If $\{e_n : n=1,2,3,\ldots\}$ were assumed to be an orthonormal basis rather than only an orthonormal set, then we could draw a stronger conclusion.
