Non-unique way of expressing vector in Hilbert space

Question

Let $H$ be a Hilbert space, $(b_{i})_{i\in\mathbb{N}}$ are linear independent set with dense span. If $b_{i}$ fail to be a Schauder basis, does that mean there exists $x\in H$ such that $x$ can be expressed non-uniquely by $b_{i}$? In other words, $x=\sum_{n=1}^{\infty}a_{n}b_{n}=\sum_{n=1}^{\infty}a'_{n}b_{n}$, where $a_{n}, a'_{n}$ are different.

By linear independence, I think it is quite impossible to happen, but things are weird in infinite dimension case, so I am not sure.

Edit Assume $x$ also takes a form $\sum_{n}\sum_{k}^{n}c_{nk}b_{k}$, can we conclude $\sum_{n}c_{nk}=a_{k}$?

There is no guarantee that every vector $x$ is of the form $\sum a_nbn$. — Kavi Rama Murthy, Jul 05 '22 at 11:34
@geetha290krm If $x$ has the representation $=\sum_{n}a_{n}b_{n}$, can we conclude that if $x=\sum_{n}\sum_{k}^{n}c_{nk}b_{k}$ also, then $\sum_{n}c_{nk}=a_{k}$? — Ken.Wong, Jul 05 '22 at 12:01
The comments seem to be ignoring that the question is about uniqueness! — Ruy, Jul 05 '22 at 13:56

score 1 · Accepted Answer · answered Jul 05 '22 at 15:47

Uniqueness may indeed fail. For an example, let $H$ be a separable Hilbert space with orthonormal basis $\{e_n\}_{n\geq 1}$, and define, for every $n\geq 0$, $$ f_n=\left\{\matrix{ \hfill\displaystyle-\frac{e_1}2, & \text{if } n=0,\cr e_n- \displaystyle\frac{e_{n+1}}2, & \text{if } n\geq 1,\cr }\right. $$ Then it is clear that $\{f_n\}_{n\geq 0}$ is a linearly independent set. Moreover $$ \sum_{n=0}^\infty 2^{-n}f_n= -\frac{e_1}2 + \sum_{n=1}^\infty 2^{-n}(e_n- \frac{e_{n+1}}2) = $$$$ = -\frac{e_1}2 + \sum_{n=1}^\infty 2^{-n}e_n- \sum_{n=1}^\infty 2^{-(n+1)}e_{n+1} = $$$$ = -\frac{e_1}2 + \sum_{n=1}^\infty 2^{-n}e_n- \sum_{n=2}^\infty 2^{-n}e_{n} = 0. $$ This provides an alternative way to represent zero besides $\sum_{n=0}^\infty 0\ f_n$.

Non-unique way of expressing vector in Hilbert space

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