span of maximal orthonormal sets in a Hilbert space

Question

I am working through Walter Rudin's Real and Complex Analysis and I'm having trouble understanding something about Theorem 4.18:

I understand the proof of the theorem, but I'm confused about the relationship between conditions (i) and (ii). If $ \{u_\alpha\}$ is a maximal orthonormal set in H, that means that there are no extra vectors in H that we can add to $ \{u_\alpha\}$ which is orthogonoal to every vector in $ \{u_\alpha\}$. For example, $R^3$ equipped with the dot product is a Hilbert space, and $\{u_\alpha\}=\{(1,0,0),(0,1,0),(0,0,1)\}$ would be a maximal orthonormal set in $R^3$. But in this case, the set of all finite linear combinations of members of $\{u_\alpha\}$ would not just be dense in H, but is equal to H. My question is, wouldn't this be the case for every Hilbert space, and if not, could someone provide a good example to help build intuition as to how that works?

Answer 1

If $H$ is finite dimensional, then clearly (i) and (ii) are equivalent, since $\{u_\alpha\}$ is finite and spans $H$.

However if $H$ is of infinite dimension, the space of all finite linear combinations of elements of $\{u_\alpha\}$ will not be equal to $H$. Indeed, since $\{u_\alpha\} $ must be countable, we can rewrite w.l.o.g. $ \{u_\alpha\}\equiv (u_n)_{n\in\mathbb N}$ and consider the element $$u:= \sum_{n=1}^\infty 2^{-n}\frac{u_n}{\|u_n\|} $$ By considering partial sums, one can show that $u$ is a well-defined element of $H$, however by definition it can not be written as a linear combination of finitely many elements of $(u_n)_n $.

(To make the above more concrete, you may take $H\equiv$ $\ell_2$ and consider the maximal orthonormal set given by $\{u_n = (\delta_{n,k})_{k\in\mathbb N}\}$, in which case $u=(2^{-k})_{k\in\mathbb N}$, which is indeed in $\ell_2$ but not a finite linear combination of the $u_n$'s)