Operator Topologies

A Hilbert space has two useful topologies (weak and strong); the space of operators on a Hilbert space has several. The metric topology induced by the norm is one of them; to distinguish it from the others, it is usually called the norm topology or the uniform topology. The next two are natural outgrowths for operators of the strong and weak topologies for vectors. A subbase for the strong operator topology is the collection of all sets of the form

correspondingly a base is the collection of all sets of the form

Here $k$ is a positive integer, $f_1, \ldots , f_k$ are vectors, and $\varepsilon$ is a positive number. A subbase for the weak operator topology is the collection of all sets of the form

where $f$ and $g$ are vectors and $\varepsilon > 0$; as above (as always) a base is the collection of all finite intersections of such sets. The corresponding concepts of convergence (for sequences and nets) are easy to describe: $A_n \to A$ strongly if and only if $A_n f \to Af$ strongly for each $f$ (i.e., $\left\| (A_n - A)f\right\| \to 0$ for each $f$), and $A_n \to A$ weakly if and only if $A_n f \to Af$ weakly for each $f$ (i.e., $\left\langle A_n f, g \right\rangle \to \left\langle Af, g \right\rangle$ for each $f$ and $g$).

The weak topology is smaller (weaker) than the strong topology, and the strong topology is smaller than the norm topology: every weak open set is a strong open set, and every strong open set is norm open. Norm convergence implies strong convergence, and strong convergence implies weak convergence. In the presence of uniformity on the unit sphere, the implications are reversible.

The norm topology (or uniform topology) is the metric topology induced by the operator norm. A subbase for the strong operator topology is the collection of all sets of the form $\left\{ A : \left\| (A - A_0)f\right\| < \varepsilon \right\}$; a subbase for the weak operator topology is the collection of all sets of the form $\left\{ A : \left| \left\langle (A - A_0)f, g \right\rangle \right| < \varepsilon \right\}$, where $f$ and $g$ are vectors and $\varepsilon > 0$.

The assertion means that if $\left\{ A_n\right\}$ is a net converging weakly to $A$, then $\left\| A\right\| \leq \liminf_n \left\| A_n\right\|$. Equivalently: for every $\varepsilon > 0$ there exists an $n_0$ such that $\left\| A\right\| - \left\| A_n\right\| \leq \varepsilon$ whenever $n \geq n_0$.

A product (unlike the norm and the adjoint) is a function of two variables, so a continuity statement about products has a "joint" and a "separate" interpretation. It is usual, when nothing is said to the contrary, to interpret such statements in the joint sense, i.e., to interpret them as referring to the mapping that sends an ordered pair $\left\langle A, B \right\rangle$ onto the product $AB$.

There is a familiar argument that can be used frequently to show that if the linear operations and squaring are continuous, then multiplication is continuous also. Since the argument depends on the identity $ab = \tfrac {1}{4}\left((a+b)^{2} - (a-b)^{2}\right)$, it can be used only when multiplication is commutative. Operator multiplication is not commutative; this problem asks whether there is nonetheless a hope for the weak sequential continuity of squaring, at least for some operators.

If a net $\left\{ P_n\right\}$ of projections converges weakly to a projection $P$, then it converges strongly to $P$. Proof: for each $f$,

and therefore Problem 20 is applicable.