Сlosed, complete and total subsets Could you help me to understand the difference between complete, closed and total subsets in Banach, general normed, Hilbert and general inner-product spaces?
Some definitions that I have found (and they seem confusing):
from Gatech lectures:
A subset $N=\{x_n\}$ of a Banach (or Hilbert) space $X$ is complete (or total, or fundamental) if $\overline{span}(N) = X$.
from Morrison:
A subset $\{x_n\}$ of a Banach space is total if $x^*(x_n)=0; \forall n\in N$ we have $x^*=0$
from Sen (for orthonormal sequences in Hilbert spaces):
An orthonormal system $\{x_n\}$ in $X$ is closed if the subspace spanned by the system coincides with $X$.
An orthonormal system $\{x_n\}$ in $X$ is complete orthonormal system if there is no non-zero $x\in X$ such that $x$ is orthogonal to every element in $\{e_n\}$.
 A: Let $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ be an orthonormal subset of a Hilbert space $H$, and let $x\in H$ be given. Then the following are equivalent:


*

*$\sum_{\alpha} \langle x,e_\alpha \rangle e_{\alpha} =x$.

*$\sum_{\alpha} |\langle x,e_{\alpha}\rangle|^2 = \|x\|^2$.

*$x$ is in the closure of the linear span of $\{ e_{\alpha} \}_{\alpha\in\Lambda}$.


The orthonormal subset $\{ e_{\alpha} \}$ is complete if every $x\in H$ satisfies the above. If you let $M$ denote the subspace of all finite linear combinations of elements of $M$, then the following are equivalent:


*

*$\{ e_{\alpha} \}$ is complete.

*$\sum_{\alpha}\langle x,e_{\alpha}\rangle e_{\alpha}=x$ for all $x\in H$.

*$\sum_{\alpha}\langle x,e_{\alpha}\rangle|^2 =\|x\|^2$ for all $x\in H$.

*The subspace $M$ spanned by the $\{e_{\alpha}\}$ is dense in $X$.

*$\langle x,e_\alpha\rangle =0$ for all $\alpha$ iff $x=0$.
You can abstract to looking only at a subspace, which leads you to consider a subspace $M$ in a Banach space $X$ that is generated by taking finite linear combinations of the set of elements $\{ x_{\alpha} \}_{\alpha\in\Lambda}$. The closure $\overline{M}$ of $M$ consists of every element $x\in X$ such that, for every $\epsilon > 0$, there is a finite linear combination $\sum_{n=1}^{k}\mu_n x_{\alpha_n}$ such that $\|\sum_{n=1}^{k}\mu_n x_{\alpha_n} -x\|< \epsilon$. In the case of orthonormal sets $\{ e_{\alpha} \}_{\alpha\in\Lambda}$, any such $x$ must be equal to $\sum_{\alpha\in\Lambda}\langle x,e_{\alpha}\rangle e_{\alpha}$. But in a Banach space, there may be no way to write the element as such a sum. If the set $\{ x_{\alpha} \}_{\alpha\in\Lambda}$ is total in a Banach space, there is a sequence of finite sums of the $x_{\alpha}$ that converges to $x$, but that does not necessarily give $x$ as an infinite sum.
Having a sum representation in a Banach space $X$ is usually posed in terms of a countable Schauder basis $\{ x_n \}$ where, for every $x\in X$, one assumes the existence of unique constants $\alpha_n$ such that $x = \sum_{n} \alpha_n x_n$. Every complete orthonormal basis of a Hilbert space is a Schauder basis for that Hilbert space. $\ell^1$ has a Schauder basis consisting of the sequences $\{ 1,0,0,0,\cdots \}$, $\{ 0,1,0,0,0,\cdots \}$, $\{ 0,0,1,0,0,\cdots \}$, etc..
Not every Banach space has a Schauder basis. For example, $L^1[-\pi,\pi]$ does not have a Schauder basis. Even though you might guess the Fourier basis $\{ e^{inx} \}_{n=-\infty}^{\infty}$ to be a Schauder basis for $L^1$, it isn't because the Fourier series does not converge in $L^1$ to an arbitrary $f\in L^1$. However, $\{ e^{inx} \}_{n=-\infty}^{\infty}$ is total in $L^1$ because the finite span $M$ of these functions is dense in $L^1$. Of course $\{ e^{inx} \}$ is a Schauder basis for $L^2$ because it's an orthonormal basis of the Hilbert space $L^2$.
For a Banach space $X$, a set of vectors $\{ x_\alpha \}_{\alpha\in\Lambda}$ is total if the subspace $M$ spanned by these vectors is dense in $X$. Equivalently, every $x$ can be approximated arbitrarily closely by a finite linear combination of the $x_\alpha$. Another equivalent is that the only $x^* \in X^*$ for which $x^*(x_\alpha)=0$ for all $\alpha$ is $x^* = 0$. This condition extends (5) for the Hilbert space case because every $x^*$ on a Hilbert space has the representation $x^*(y)=\langle y,x\rangle$ for a unique $x$.
