Example of a basis which is not a Riesz basis? I'm looking for an example of a countable "basis" $B=(\phi_i)_{i\in I}$ in a real Hilbert space $\mathcal{X}$ which is not a Riesz basis. So, we only require that the closure of the span of $B$ is $\mathcal{X}$ and that, for every $N\in\mathbb{N}$ and every cardinality-$N$ subset $\{i_1,\ldots,i_N\}$ of $I$,
\begin{equation}
(\forall (\alpha_{i_1},\ldots,\alpha_{i_N})\in\mathbb{R}^N)\quad 
\alpha_{i_1}\phi_{i_1} + \alpha_{i_2}\phi_{i_2}+\cdots+\alpha_{i_N}\phi_{i_N}=0\Rightarrow (\alpha_{i_j})_{1\leq j\leq N}\equiv 0.
\end{equation}
(note that, by "span" I mean finite linear combinations.)
A few observations / questions

*

*This must occur in infinite-dimensional $\mathcal{X}$.


*I think this would mean that the basis decomposition operator $L\colon\mathcal{X}\to\ell_2\colon x\mapsto(\langle x\,|\,\phi_i\rangle)_{i\in I}$ is unbounded?
 A: Start with an orthonormal basis $ \{e_n\}_n$ and put
$\phi_n=e_n/n$, or $\phi_n=ne_n$.  A less trivial example would be
$$
\phi_n =e_n/n + \sum_{i=1}^{n-1}e_i.
$$

EDIT. Here is a concrete criteria for identifying, or debunking,  Riesz bases.
Recall that a sequence $\{\phi_n\}_n$  in a Hilbert space $H$ is called a Riesz basis if it spans a dense subspace
of $H$, and there are positive  constants
$c$ and $C$  such that
$$
  c\left(\sum_n|x_n|^{2}\right)\leq
  \left\Vert \sum_nx_n\phi_n\right\Vert ^{2}\leq
  C \left(\sum _n|x_n|^{2}\right),
  \tag 1
  $$
for every finitely supported sequence $\{x_n\}_n$ of scalars.
This is obviously equivalent to the fact that the correspondence
$$
  T : (x_n)_n \in  \ell^2 \mapsto \sum_nx_n\phi_n\in  H
  $$
defines a (not necessarily isometric) isomorphism from  $\ell^2$  onto $H$.
Theorem.  Let $\{\phi_n\}_n$ be a sequence  in  $H$ spanning  a dense subspace.  Then $\{\phi_n\}_n$ is a Riesz
basis iff the matrix
$$
  A=\{\langle \phi_j,\phi_i\rangle \}_{i, j}
  $$
represents an invertible operator on $\ell^2$.
Proof.  Assuming that $\{\phi_n\}_n$ is a Riesz basis, let $T$ be the operator defined above.
Denoting by
$\{e_k\}_k$ the standard orthonormal basis of $\ell^2$, notice that the matrix $A=\{a_{i, j}\}_{i, j}$ representing the
operator $T^*T$ is given by
$$
  a_{i,j}=  \langle T^*T(e_j),e_i\rangle  = \langle T(e_j),T(e_i)\rangle  = \langle \phi_j,\phi_i\rangle .
  $$
Since $T$ is invertible, if follows that $T^*T$ is also invertible, so this concludes the proof of the "only if" part.
Conversely, suppose that $A=\{\langle \phi_j,\phi_i\rangle \}_{i, j}$
represents an invertible operator on $\ell^2$.  Then, for every finitely supported sequence
$x=\{x_n\}_n$  of scalars we have that
$$
  \langle Ax,x\rangle  =
  \sum_{i,j} \langle \phi_j,\phi_i\rangle x_j\overline{x_i} =
  \sum_{i,j} \langle x_j\phi_j,x_i\phi_i\rangle  =
  \left\|\sum_ix_i\phi_i\right\|^2.
  $$
This shows that $A$ is a positive operator and then $B:=A^{1/2}$ is an invertible  self-adjoint operator  satisfying
$$
  \|Bx\|^2 =
  \langle Bx,Bx\rangle  =
  \langle B^2x,x\rangle  =
  \langle Ax,x\rangle  =
  \left\|\sum_ix_i\phi_i\right\|^2.
  $$
Combining this with the fact that
$$
  \|B^{-1}\|^{-1}\|x\| \leq      \|Bx\|\leq     \|B\|\|x\|,
  $$
we deduce that
$\{\phi_n\}_n$ satisfies (1) and hence is  a Riesz basis.  QED

This said, it is very easy to build examples of linearly independent sets spanning a dense subspace which are not  Riesz
bases.  A typical obstruction for this would   be when the matrix  $A$ above has rows or columns which are not square summable.
A: Let $\{v_1,v_2,\dots\}$ be a complete orthonormal basis for $\mathcal{X}$. Let $w_1=v_1$, $w_2=\frac{1}{\sqrt 2}(v_1+v_2)$ and more generally for $k=3,4,\dots$ let $w_k=\frac{1}{\sqrt k}(v_1+v_2+\cdots v_k)$. Note that $\|w_k\|=1$. Since $v_k=\sqrt k w_k-\sqrt{k-1} w_{k-1}$, the closure of the span of $\{w_1,w_2,\dots\}$ is $\mathcal{X}$. Moreover, since $L(v_k)= (0,0,\dots,-\sqrt{k-1},\sqrt{k},0,\dots)$, $\|L(v_k)\|= \sqrt{k^2+(k-1)^2}$ so that the mapping into $\ell_2$ is unbounded.
