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Let $X$ be an (infinite-dimensional) inner product space with an (countable) orthogonal system $\lbrace e_i \rbrace_{i \in \mathbb{N}}$. The field is $\mathbb{R}$ or $\mathbb{C}$. Here X may not be a Banach space. Suppose further that $Span(\lbrace e_i \rbrace_{i \in \mathbb{N}})$ is dense in $X$, do we still have infinite series representation of each element? i.e., I'm wondering if $\forall x \in X$, $\exists \lbrace a_n \rbrace \subseteq \mathbb{R}$ (or $\mathbb{C}$) such that $$x = \sum_{n=1}^{\infty} a_ne_n, \forall x \in X$$

If $X$ is a Hilbert space, then above is just a trivial application of Hilbert basis. I'm quite interested in the situation where $X$ is not complete. Any help or idea is appreciated.

Stack_Underflow
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    Complete $X$ to a Hilbert space. Then ${e_i}$, properly normalized, is an orthonormal basis, so any $x$, even in the completion (and therefore certainly when $x \in X$), can be written in the desired form. – David Gao Mar 05 '24 at 02:32
  • @DavidGao That's simple and elegant. Thank you! – Stack_Underflow Mar 05 '24 at 02:36
  • Related, but requires proof ... if $X$ is a separable inner product space, then there exists an orthonormal system in $X$. – GEdgar Mar 05 '24 at 02:47
  • @GEdgar Yes, I think this could be proved by countably invoking the Gram-Schmidt Process on the dense subset? – Stack_Underflow Mar 05 '24 at 04:07
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    @GEdgar Personally, I think the more interesting fact is that you can’t do that without separability assumptions. The separable case is just Gram-Schmidt. – David Gao Mar 05 '24 at 04:10
  • @GEdgar Thanks, then I think with non-separability, one might use Zorn's Lemma.... – Stack_Underflow Mar 05 '24 at 04:23
  • @Stack_Underflow: No, one might not use Zorn's Lemma. In an inner product space that is not complete, you can find a maximal orthonormal set - but it doesn't have to have the property that its linear span is dense. The theorem that the linear span of a maximal orthonormal set is dense uses completeness of the space. However, if you start with a countable dense subset and do Gram-Schmidt on it you're guaranteed that the result will have the density property. – Chad K Mar 05 '24 at 14:40
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    @ChadK ... here is a counterexample for the non-separable case https://math.stackexchange.com/a/201149/442 – GEdgar Mar 05 '24 at 16:06

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If $\ x\in X\ $ let $$ x_n\stackrel{\text{def}}{=}\sum_{i=1}^n\frac{\langle x,e_i\rangle}{\|e_i\|^2}e_i\ . $$ for $\ n\in\mathbb{N}\ .$ Then $\ \big\langle x-x_n, e_i\big\rangle=0\ $ for all $\ i\le n\ .$ Since $\ Span\big(\big\{e_i\big\}_{i\in\mathbb{N}}\big)\ $ is dense in $\ X\ ,$ then for any $\ \epsilon>0\ $ there exists $\ y=\sum_\limits{i=1}^m\xi_ie_i\in Span\big(\big\{e_i\big\}_{i\in\mathbb{N}}\big)\ $ such that $\ \|x-y\|<\epsilon\ .$ Then for any $\ r\ge m\ $ \begin{align} \epsilon^2&>\|x-y\|^2\\ &=\big\|x-x_r+x_r-y\big\|^2\\ &=\big\|x-x_r\|^2+2\mathfrak{Re}\big\langle x-x_r, x_r-y\big\rangle+\big\|x_r-y\big\|^2\\ &=\big\|x-x_r\|^2+\big\|x_r-y\big\|^2\ , \end{align} because $\ x_r-y\ $ is a linear combination of $\ \big\{e_i\big\}_{i=1}^r\ ,$ all of which are orthogonal to $\ x-x_r\ .$ It follows from this that, for all $\ r\ge m\ ,$$\ \big\|x-x_r\big\|<\epsilon\ ,$ and hence, from the arbitrariness of $\ \epsilon\ ,$ that $$ x=\lim_{n\rightarrow\infty}x_n=\sum_{i=1}^\infty\frac{\langle x,e_i\rangle}{\|e_i\|^2}e_i\ . $$

lonza leggiera
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