# Reflexive operator algebra

In functional analysis, a reflexive operator algebra A is an operator algebra that has enough invariant subspaces to characterize it. Formally, A is reflexive if it is equal to the algebra of bounded operators which leave invariant each subspace left invariant by every operator in A.
This should not be confused with a reflexive space.

## Examples

Nest algebras are examples of reflexive operator algebras. In finite dimensions, these are simply algebras of all matrices of a given size whose nonzero entries lie in an upper-triangular pattern.
In fact if we fix any pattern of entries in an n by n matrix containing the diagonal, then the set of all n by n matrices whose nonzero entries lie in this pattern forms a reflexive algebra.
An example of an algebra which is not reflexive is the set of 2 × 2 matrices
    $\left\{{\begin{pmatrix}a&b\\0&a\end{pmatrix}}\ :\ a,b\in \mathbb {C} \right\}.$
This algebra is smaller than the Nest algebra
    $\left\{{\begin{pmatrix}a&b\\0&c\end{pmatrix}}\ :\ a,b,c\in \mathbb {C} \right\}$
but has the same invariant subspaces, so it is not reflexive.
If T is a fixed n by n matrix then the set of all polynomials in T and the identity operator forms a unital operator algebra. A theorem of Deddens and Fillmore states that this algebra is reflexive if and only if the largest two blocks in the Jordan normal form of T differ in size by at most one. For example, the algebra
    $\left\{{\begin{pmatrix}a&b&0\\0&a&0\\0&0&a\end{pmatrix}}\ :\ a,b\in \mathbb {C} \right\}$
which is equal to the set of all polynomials in
    $T={\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}}$
and the identity is reflexive.

## Hyper-reflexivity

Let ${\mathcal {A}}$ be a weak*-closed operator algebra contained in B(H), the set of all bounded operators on a Hilbert space H and for T any operator in B(H), let
    $\beta (T,{\mathcal {A}})=\sup \left\{\left\|P^{\perp }TP\right\|\ :\ P{\mbox{ is a projection and }}P^{\perp }{\mathcal {A}}P=(0)\right\}.$
Observe that P is a projection involved in this supremum precisely if the range of P is an invariant subspace of ${\mathcal {A}}$.
The algebra ${\mathcal {A}}$ is reflexive if and only if for every T in B(H):
    $\beta (T,{\mathcal {A}})=0{\mbox{ implies that }}T{\mbox{ is in }}{\mathcal {A}}.$
We note that for any T in B(H) the following inequality is satisfied:
    $\beta (T,{\mathcal {A}})\leq {\mbox{dist}}(T,{\mathcal {A}}).$
Here ${\mbox{dist}}(T,{\mathcal {A}})$ is the distance of T from the algebra, namely the smallest norm of an operator T-A where A runs over the algebra. We call ${\mathcal {A}}$ hyperreflexive if there is a constant K such that for every operator T in B(H),
    ${\mbox{dist}}(T,{\mathcal {A}})\leq K\beta (T,{\mathcal {A}}).$
The smallest such K is called the distance constant for ${\mathcal {A}}$. A hyper-reflexive operator algebra is automatically reflexive.
In the case of a reflexive algebra of matrices with nonzero entries specified by a given pattern, the problem of finding the distance constant can be rephrased as a matrix-filling problem: if we fill the entries in the complement of the pattern with arbitrary entries, what choice of entries in the pattern gives the smallest operator norm?

## Examples

* Every finite-dimensional reflexive algebra is hyper-reflexive. However, there are examples of infinite-dimensional reflexive operator algebras which are not hyper-reflexive.
* The distance constant for a one-dimensional algebra is 1.
* Nest algebras are hyper-reflexive with distance constant 1.
* Many von Neumann algebras are hyper-reflexive, but it is not known if they all are.
* A type I von Neumann algebra is hyper-reflexive with distance constant at most 2.