Does a set of linearly independent vectors exist that can not be orthogonalized using Gram Schmidt I am reading Fourier Series and Orthogonal Functions by Harry Davis  Theorem 10 page 54 states "A (linearly independent) sequence satisfies the condition of finality if and only if it is orthogonal."  Finality (I think) here means the  coefficients of the vectors in the sum must be recalculated every time you try to add another term to the summation.  Does this mean there are linearly independent vectors that can not be orthogonalized using the Gram Schmidt process?   Why wouldn't the process work?
 A: Davis says that a sequence $(u_1,u_2,...)$ is "final" essentially if no vectors $u_j$ influence the norm of a linear combination $\|a_1 u_1 + ... + a_i u_i\|$ for any $i < j$. This is true if and only if the sequence is orthogonal.
This doesn't say anything about orthogonalization, or the Gram-Schmidt process.
Your concerns are justified. You can't generally take a traditional basis (or Hamel basis), in which every element of your space is uniquely a finite linear combination, and hope to orthogonalize it. In fact, infinite-dimensional Hilbert spaces never have orthogonal Hamel bases.
(The reason is that if $\{e_i\}_{i \in I}$ is such a set, and if you take a sequence $e_1,e_2,e_3,...$ from it of pairwise distinct elements and form the series $$v = \sum_{k=1}^{\infty} \frac{1}{k^2} e_k,$$ then $v$ is not orthogonal to any $e_k$. This is a contradiction, because $v$ should be orthogonal to all but the finitely many $e_i$'s from which it is a linear combination.)
On the other hand, the Gram-Schmidt process works fine for orthogonalizing a countable sequence. (Nothing goes wrong.) This is one proof that an infinite-dimensional Hilbert space can never have countable dimension.
