If $S=\{v_i\}_{i\in\mathbb N}$ is a (not necessarily orthogonal) Schauder basis for a Hilbert space $H$, must $S$ be an unconditional Schauder basis? I define these terms below because not every source I have found agrees perfectly on the definitions.
On general Banach spaces (where orthogonality is undefined), there exist conditional Schauder bases. But if the Schauder basis is on a Hilbert space and is orthogonal, then it is indeed unconditional. So my question is therefore whether this unconditional property remains if we stay in Hilbert space but orthogonality is removed.
Definitions:
An ordered countable subset $\{v_i\}_{i\in\mathbb N}$ of a Banach space $V$ is a Schauder basis if every $v\in V$ can be written uniquely as a series $v=\sum_{i=1}^\infty a_i v_i$, where convergence is with respect to the norm-induced topology.
A Schauder basis is unconditional if the terms of any convergent series can be rearranged without affecting the sum.
A Hilbert basis is a maximal orthonormal subset of a Hilbert space $H$, possibly uncountable. It is known that even for non-separable Hilbert spaces, there exists a Hilbert basis $B$, and that every $v\in H$ can be expressed uniquely as a sum of a countable subset of $B$, which is always independent of summation order.
Best Answer
Here's a "cheating" way to get a conditional Schauder basis on a Hilbert space. Take an orthogonal Schauder basis $v_i$ and consider a convergent series $\sum\alpha_iv_i$. Now let $\beta_i$ be some conditionally convergent series in $\mathbb R$ and consider the new Schauder basis $w_i = v_i + \frac{\beta_i}{\alpha_i}v_1$. $\sum\alpha_iw_i$ must converge conditionally.