# Example of a basis which is not a Riesz basis?

*Solution:*

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.

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.