# Terminology of Operator Topologies

Let $$X,Y$$ be normed vector spaces. In the case that $$X,Y$$ are Banach spaces, Folland defines the strong operator topology on the space of bounded linear maps $$X \to Y$$ to be the weak topology generated by the family of maps $$T \mapsto Tx$$ for each $$x \in X$$. Similarly he defines the weak operator topology to be generated by $$T \mapsto f(Tx)$$ where $$f \in Y^*$$, the space of bounded linear functionals and $$x \in X$$.

Is there any problem in vastly extending these definitions to define topologies on the space of all linear maps $$X \to Y$$ for arbitrary (not necessarily Banach) normed vector spaces? Would these be going against standard usage? It seems the restriction to bounded operators and Banach spaces is unnecessary. Is there anything wrong with this usage?

• The problem is that most (read: all that I know of) unbounded linear operators which anybody is interested in are not defined on all of $X$, but on dense subspaces that are part of the data of the operator. This means a bunch of things, for example unbounded operators cannot always be added or multiplied in meaningful ways. Further if you want to explicitly define a topology on the space of such operators pointwise convergence is not such a good idea since the domains vary so a sequence of operators can have no common vector in their domain, making pointwise convergence inadequate. Commented Apr 15, 2020 at 18:52
• That said pointwise convergence is the way most operations on unbounded operators are considered to be continuous. For example the measurable calculus from the spectral theorem is defined pointwise by the integral $f(A)\ v := \int_{\sigma(A)} f(\lambda) dP(\lambda)v$ for all $v$ so that $\int |f(\lambda) | \|dP(\lambda)v\|$ exists. Its just that its not very productive (in my opinion) to view all this as taking place in a special operator space equipped with a topology for the reasons mentioned above. Commented Apr 15, 2020 at 18:56
• @s.harp I guess that makes sense/is fair. I was mostly thinking about in the context of results like the following: suppose $T_n \to T$ pointwise and the $T_n$ are bounded operators, then $T$ is bounded (under sufficient assumptions like the domain is Banach). This can be phrased as the space of bounded operators is sequentially closed in the strong operator topology on the space of linear maps. But I see how it generally isn't too useful. If you want to put your comment in an answer, I'll accept it. Commented Apr 15, 2020 at 19:07

The definition of the strong topology as that of pointwise convergence may be extended without issue to $$\mathrm{End}(X)$$, the space of linear maps $$X\to X$$. The restriction of this topology to the bounded linear maps $$B(V)$$ will then agree with the usual definition of the strong operator topology on $$B(V)$$.
The problem with doing this is that unbounded operators defined on all of $$X$$ are not the kind of unbounded operators you are interested in. Usually an unbounded operator $$T$$ is defined on a dense sub-space $$D(T)$$ of the Banach space $$X$$, and when talking about or defining this operator the domain is taken as part of its data.
This means that the space of those unbounded operators which one is usually interested in is not a reasonable operator space. You will face difficulties if you want to add or multiply such operators, simply because the domains of two operators you wish to add can be disjoint. Giving this space the "topology" of pointwise convergence is also problematic, because if the domains of the individual operators change you will not be able to consistently consider the convergence of $$T_n x$$ for a sequence $$T_n$$ and $$x\in X$$.
That said the notion of pointwise convergence is the usual way in which operations on unbounded operators are considered continuous. As an example in the measurable calculus defined by the spectral theorem, both the defining $$f(T)=\int_{\sigma(T)}f(\lambda)\ dP(\lambda)$$ and the domain of $$f(T)$$ are declared by pointwise considerations.