On the Space of Continuous Linear Operators on LCTVS

Question

Suppose that $X$ is a locally convex topological vector space (LCTVS) and that $L(X)$ denotes the space of all continuous linear operators on $X$.

Question. How can we construct a topology on $L(X)$ which compatible with the vector space structure of $L(X)$?

I need help on this. Thanks in advance....

Answer 1

There is no natural topology on the dual space of a locally convex space unless the original space has a norm. In this case, if the original space is complete, then the $\sup_{B}$ norm is a norm on the dual space. Here $B$ is the unit ball in the original space.

I think the conventional way to endow a topology on the dual space is as follows. Assume $H$ has a topology defined by a family of seminorms such that $\bigcap \{x:p(x)=0\}=\{0\}$. The subbase in this case is all sets of the form $\{x:p(x-x_{0})<\epsilon\}$. So a set $U$ in $H$ is open if and only if at every point $x_{0}\in H$, we have $p_{i},\epsilon_{i},i=\{1\cdots n\}$ such that $\bigcap^{n}_{j=1}\{x\in H:p_{i}(x-x_{0})<\epsilon_{i}\}\subset U$.

Assuming this topology already well defined on $H$, we define a "weak-star topology" on $H^{*}$ by considering the family of seminorms given by $$\mathcal{P}=\{p_{x}:x\in H\}$$where $p_{x}(f)=f(x)$. Since $H^{*}$ is also a topological vector space, the above $p_{x}$ make it into a locally convex space in the same way as $H$. Some author claim that this is the only "natural" topology for the dual space of a locally convex space in the absence of a norm (See Conway). There is also a related personal note by Tao at here.

Answer 2

Although I am not an expert in this area, my guess would be that there is not a best or standard way of doing this. But I might give some suggestions with my shallow knowledge.

Let's first consider a normed vector space $(X,\|\cdot\|)$, which is a special kind of locally convex topological vector spaces. Then there are some very natural topologies on the space of continuous linear operators $\mathcal{L}(X)$, and we might study how these three topologies can be generalized to a LCTVS.

First, we have the norm topology on $\mathcal{L}(X)$ by \begin{equation} \|T\|=\sup_{\|x\|=1}\|Tx\|. \end{equation} Unfortunately I do not know how to generalize this to LCTVS. The problem is that there might be too many seminorms (note that the topology for a LCTVS always comes from a family of seminorms) for the $\sup$ to make sense.

However, if there are only countably many seminorms, then $X$ itself is a metric space with a metric $d$. Then of course we might just give $\mathcal{L}(X)$ the sup-norm, viewing it as a subset of $C(X)$, and the norm topology still makes sense.

Secondly, we have the strong operator topology on $\mathcal{L}(X)$ if $X$ is normed, which has a local base consisting of sets like \begin{equation} V(x_1,\dots,x_n;\epsilon_1,\dots,\epsilon_n):=\{T\in\mathcal{L}(X):\|Tx_j\|<\epsilon_j\}, \end{equation}or in terms of convergence of nets, $T_{\alpha}$ converges to $T$ in this topology if and only if $T_{\alpha}x$ converges to $Tx$ for all $x\in X$. Luckily this still makes sense for a general LCTVS.

Similarly, you might consider other topologies on $\mathcal{L}(X)$, like weak operator topology and weak$^*$ operator topology, when $X$ is a normed space, and see how they generalize to a LCTVS.