If an Hilbert space is separable does it mean $\text{span}(\{e_k\})=\mathcal{H}$ or $\text{cl}\,\text{span}(\{e_k\})=\mathcal{H}$?

139 Views Asked by At

If an Hilbert space is separable it exists a complete orthonormal sequence $\{e_k\}\subset\mathcal{H}$ and that means $$ x = \sum\limits_k \langle x, e_k\rangle e_k \,\forall x\in\mathcal{H} $$ so that the set of all $x$, meaning $\mathcal{H}$, should be a subset of $\text{span}(\{e_k\})$, defined as $$ \text{span}(\{e_k\}) \doteq \left\{ \sum\limits_k \alpha_k e_k :\alpha_k\in\text{C} \right\} $$ At the same time $\alpha_k e_k$ is an element of $\mathcal{H}$ for all $\alpha_k\in\text{C}$ and so it is every combination of these elements, meaning that $\text{span}(\{e_k\})\subseteq\mathcal{H}$.

By these inclusion relations I should conclude $\text{span}(\{e_k\})=\mathcal{H}$, but instead I see written $\text{cl}\,\text{span}(\{e_k\})=\mathcal{H}$ (see for example "Hilbert spaces with applications - Debnath, Mikusinki" third edition pag. 114, where span is defined with a finite summation, or the ending part of the proof of the theorem 4.8.15 pag. 185).

What is my incomprehension?

1

There are 1 best solutions below

1
On BEST ANSWER

Often the linear algebraic span is defined to consist of all finite linear combinations of the form $\sum_{k=1}^n \alpha_k e_k$. In an infinite-dimensional Hilbert space, you may need an infinite series to express a vector; such a vector belongs to the closure of the span (i.e. it is a limit point). In a nutshell, working with the closure allows you to bring in concepts of convergence to deal with the possibility of being infinite-dimensional.