# Hilbert basis - Alternative definition with infinite linear combinations

Can I define a Hilbert basis as follows?

A Hilbert basis $B$ of a Hilbert space $\mathcal{H}$ is a orthonormal set in $\mathcal{H}$ with $$\mathcal{H} = \left\{ \sum_{i=1}^\infty \alpha_i b_i \,\middle|\, \alpha_i \in \mathbb{C}, b_i \in B \right\}$$

# Background

The definition I know is the following: A Hilbert basis $B$ is defined as a maximal orthonormal set in $\mathcal{H}$.

I know that the (topological) closure of the span of $B$ generates $\mathcal{H}$:

$$\mathcal{H} = \text{cl}(\text{span}(B))$$

I also know that the span only generates finite linear combinations of $B$, i.e.,

$$\text{span}(B)= \left\{ \sum_{i=1}^n \alpha_i b_i \,\middle|\, \alpha_i \in \mathbb{C}, b_i \in B, n \in \mathbb{N} \right\}$$

Finally, I know from this question that I can generate $\mathcal{H}$ by a "more liberal span", that also produces infinite linear combinations.

$$\mathcal{H} = \left\{ \sum_{i=1}^\infty \alpha_i b_i \,\middle|\, \alpha_i \in \mathbb{C}, b_i \in B \right\}$$

However, I do not know if the converse holds, i.e., if my definition is sufficient to characterize all Hilbert bases.

• I really don't think the term "Hilbert basis" is standard here - better just to say "orthonormal basis". (For example, the "Hilbert basis theorem" is not about orthonormal bases in a Hilbert space...) Jun 15 '18 at 13:37

There is a problem with what you have written since $\sum \alpha_i b_i$ is not defined for all choices of $\alpha_i$'s and $b_i$'s. Actually, the series converges in the norm iff $\sum |\alpha_i|^{2} <\infty$ (assuming that $b_i$'s are distinct). If you put this condition on the coefficients the assertion is correct. In other words, $\mathcal H =\{\sum_{i=1}^{\infty} \alpha_i b_i : \{b_i\} \subset B,\sum_{i=1}^{\infty} |\alpha_i|^{2} <\infty\}$
any element $v$ of the Hilbert space can be expressed by a series $$v=\sum_{i=0}^\infty a_ib_i$$ where $b_i$ are the elements of a Schauder basis and $\{a_i\}$ is a sequence of scalars.