What kind of universal property does the strong dual topology on $X'$ have, for $X$ being a locally convex space. Is it possible to define $X'$ as the projective limit of the normed spaces $\mathcal{L}(B,\mathbb{C})$ where $B\in \mathfrak{B}_X$ an element of the directed set of bounded subsets of X? What kind of universal property does the uniform norm have?

EDIT: What I actually want to know is, in which sense the (bounded) uniform convergence topologies are categorical

EDIT: It is known that there is an adjoint pair of functors $(F,G)$ between the symmetric monoidal categories of locally convex spaces and convex bornological spaces. The latter is closed, i.e. for $X,Y \in \mathsf{cbs}$ there is an internal hom object $[X,Y] \in \mathsf{cbs}$, which is just the set of bounded maps together with the bornology of equibounded sets of linear maps. For $A,B \in \mathsf{lcs}$ we obtain a topological space $G(F(A),F(B))$ and since the adjoint pair is the identity on set level, we have $G(F(A),F(B))=B(A,B)$. Moreover, we know that $\mathcal{L}(A,B) \subseteq B(A,B)$. Thus, we can endow $\mathcal{L}(A,B)$ with the initial topology with respect to this inclusion. My guess is, that this coincides with the bounded uniform convergence topology.

  • $\begingroup$ What is $\mathcal L(B,\mathbb C)$? $\endgroup$ – Jochen Mar 2 '16 at 7:38
  • $\begingroup$ Restrictions of continuous linear functions on $X$ to $B$ $\endgroup$ – Bipolar Minds Mar 2 '16 at 7:50

The strong topology $\beta(X',X)$ on the dual $X'$ is given by the system of semi-norms $$p_B(f)=\sup\lbrace |f(x)|: x\in B\rbrace $$ and these are the norms of your spaces $\mathcal L(B,\mathbb C)$. Hence, $(X',\beta(X',X))$ is contained (as a topological subspace) in the projective limit of all $\mathcal L(B,\mathbb C)$. This limit consists of all linear maps $f:X\to\mathbb C$ which are bounded on all bounded sets and, in general, it is strictly bigger than $X'$.

They are equal for so-called ''bornological'' locally convex spaces, that is, every absolutely convex set which absorbs all bounded sets is a $0$-neighbourhood. Metrisable spaces are bornological (this is quite elementary) and a beautiful theorem of Laurent Schwartz states that the strong dual of every complete Schwartz space is bornological (you can find a proof in the book Inroduction to Functional Analysis by Meise and Vogt).

On the other hand, there are Frechet spaces (constructed by Köthe and Grothendieck) whose strong duals are not bornological.

  • $\begingroup$ Thx! I clarified my second edit, this seems to be compatible with what you said, since bornological locally convex spaces are just the objects on which $GF$ is the identity, right? $\endgroup$ – Bipolar Minds Mar 2 '16 at 21:41

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