Riesz basis

Remarks

Equivalently, $(x_n)$ is a Riesz basis if and only if $x_n = T e_n$ for an orthonormal basis $(e_n)$ of $H$ and a bounded linear operator $T \colon H \to H$ with a bounded inverse. Then every $x \in H$ has a unique expansion $x = \sum_n c_n x_n$, the series converges unconditionally, and $A \sum \abs{c_n}^2 \le \norm{x}^2 \le B \sum \abs{c_n}^2$. The best constants are $A = \norm{T^{-1}}^{-2}$ and $B = \norm{T}^2$.

Example: let $(e_n)$ be an orthonormal basis of $H$ and $x_n = e_n + \frac12 e_{n+1}$. Then $x_n = T e_n$ with $T = I + \frac12 S$, where $S$ is the shift $e_n \mapsto e_{n+1}$, of norm $1$. Since $\norm{\frac12 S} < 1$, the Neumann series shows that $T$ is invertible with bounded inverse, so $(x_n)$ is a Riesz basis.

Non-example: $x_n = e_n / n$. Its closed span is $H$ and the upper bound holds with $B = 1$. The lower bound fails: for $c = (0, \dots, 0, 1)$ with the $1$ in place $n$, the middle term is $1/n^2$, which is not bounded below by a positive $A$.