We know that for general separable Banach space, not every linear independent set which is dense is a Schauder basis. For separable Hilbert space, is every linear independent set(not necessary orthogonal) that is dense a Schauder basis of the Hilbert space?
My guess is yes, because every linear independent set can be made orthogonal by the Gram-Schmidt procedure, but I am not sure about the situation before the procedure.
Edit
I made a terrible mistake, what I intend to ask is if the span of linear independent set is dense, is the set a Schauder basis?
Best Answer
The answer is negative. A dense set $D$ in a Hilbert or Banach space $X$ can never be a Schauder basis, because the representation is not unique. For each $F \subset D$ finite and $x \in x$, we can write $x$ as a sum of series of elements from $D_0=D \setminus F$. Choose $d_1 \in D_0$ such that $\|x-d_1\|<1/2$. If $d_1,\ldots,d_{k-1} \in D_0$ have been chosen, then select $d_k \in D_0$ so that $\|x-\sum_{j=1}^k d_j\|<2^{-k}$. Then $$d=\sum_{k=1}^\infty d_k \,.$$
A variant of the question (which is often asked) is:
``For separable Hilbert space $H$, does every linearly independent set(not necessary orthogonal) that has a dense span, necessarily contain a subsequence that is a Schauder basis of $H$?''
The answer is still negative, but for a different reason. The sequence of powers $1,x,x^2,\ldots$ has a dense span in $L^2[-1,1]$. But no subsequence of the powers can be a Schauder basis, because if $f$ in $L^2[-1,1]$ is not real analytic, it is not the sum in $L^2$ of a power series. Indeed, power series converge uniformly inside their radius of convergence, and outside it the summands do not tend to zero in L^2.
https://en.wikipedia.org/wiki/Schauder_basis