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

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

2. 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?

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.

• That equivalence with $A$ and the rows or columns not being square summable really helped pull the example together. Thanks for the thorough explanation!
– Zim
Dec 1 '20 at 18:20
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.
• Could you please show why the mapping to $\ell_2$ is unbounded?
• $L(v_k)= (0,0,\dots,-\sqrt{k-1},\sqrt{k},0,\dots)$, so $\|L(v_k)\|= \sqrt{k^2+(k-1)^2}$. Dec 1 '20 at 23:08