Let $E$ be a (right) Hilbert $A$-module where $A$ is unital $C^*$-algebra. I call a subset $S$ of $E$ to orthonormal if $\langle x,x\rangle=1$ ($1$ denotes the unit in $A$) for all $x\in S$ and $\langle x,y\rangle=0$ for all $x\ne y$. Then Zorn's lemma guarantees the existence of maximal orthonormal set in $E$. Let $\{e_i\}$ be a maximal orthonormal set. My question is- Does the following identity hold for $x\in E$? $$x=\sum\limits_{i\in I}e_i\langle e_i,x\rangle$$
If I know that the series $\sum\limits_{i\in I}e_i\langle e_i,x\rangle$ converges in $E$, then the above equality holds, because then $x-\sum\limits_{i\in I}e_i\langle e_i,x\rangle$ will be orthogonal to $e_j$ for all $j$, hence $x=\sum\limits_{i\in I}e_i\langle e_i,x\rangle$.
I can prove that $\left\langle x-\sum\limits_{i\in I'}e_i\langle e_i,x\rangle,x-\sum\limits_{i\in I'}e_i\langle e_i,x\rangle\right\rangle\ge0\implies |x|^2\ge\sum\limits_{i\in I'}|\langle e_i,x\rangle|^2$ for any finite $I'\subseteq I$.
Does that imply the series $\sum\limits_{i\in I}e_i\langle e_i,x\rangle$ converge in $E$? Can anyone help me in this regard? Thanks for your help in advance.
Here is a counter-example for real C*-algebras. There are similar examples for complex ones too, but the phenomenon is easier to understand in the real case.
Consider the real C*-algebra $C(S^2)$ formed by all real-valued functions on the 2-sphere $S^2$. Also consider the vector space $E$ formed by all continuous vector fields on $S^2$ (i.e. continuous sections of the tangent vector bundle $TS^2$).
Pointwise multiplying vector fields by continuous functions yields a natural $C(S^2)$ module structure on $E$. Moreover, providing each tangent space $T_pS^2$ with its standard inner-product $\langle \cdot, \cdot\rangle _p$ (i.e., considering $S^2$ with its standard Riemannian structure), we can define a $C(S^2)$ valued inner product $\langle \cdot, \cdot\rangle _{C(S^2)}$ on $E$, namely $$ \langle \xi , \eta \rangle _{C(S^2)}(p) = \langle \xi (p), \eta (p)\rangle _p, \quad (\xi ,\eta \in E, \ p\in S^2). $$
Now, it is not hard to see that if $\{e_i\}_i$ is an orthonormal basis for $E$, then, for each $p$ in $S^2$, one has that $\{e_i(x)\}_i$ will be an orthonormal basis for $T_pS^2$, but this contradicts the fact that the tangent bundle of $S^2$ is not trivial!
In fact it is even impossible to find a single non-vanishing vector field (see the Hairy Ball theorem of algebraic topology). In the context of the above comments, this says that indeed there is no "unit vector" in this Hilbert module.