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.

  • $\begingroup$ That equivalence with $A$ and the rows or columns not being square summable really helped pull the example together. Thanks for the thorough explanation! $\endgroup$
    – Zim
    Dec 1 '20 at 18:20
  • $\begingroup$ I'm glad it helped! $\endgroup$
    – Ruy
    Dec 1 '20 at 21:00

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.

  • 1
    $\begingroup$ Could you please show why the mapping to $\ell_2$ is unbounded? $\endgroup$
    – Zim
    Dec 1 '20 at 18:20
  • 2
    $\begingroup$ $L(v_k)= (0,0,\dots,-\sqrt{k-1},\sqrt{k},0,\dots)$, so $\|L(v_k)\|= \sqrt{k^2+(k-1)^2}$. $\endgroup$ Dec 1 '20 at 23:08
  • $\begingroup$ @Hanno I have incorporated the comment. Thanks $\endgroup$ Dec 18 '20 at 9:11

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