# Is every linear independent set which has dense span in Hilbert space a Schauder basis?

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?

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

• 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. But I guess from the answer you provided, I think the answer is negative. Commented Jul 4, 2022 at 16:02
• Right, I thought you might have intended to ask that, so I included an answer for this variant. Do you know how to accept an answer? Commented Jul 5, 2022 at 4:09
• Just want to point out even if $f$ is not real analytic, it could get a sequence of real analytic functions converging to it in $L^2$ sense. The convergence need not be pointwise. Commented Jun 2, 2023 at 2:37
• This post seems to say when it will happen, math.stackexchange.com/questions/4329949/… Commented Jun 2, 2023 at 2:42