TITLE: Example of a basis which is not a Riesz basis?
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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.