# Universal property of topology of uniform convergence

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.

• What is $\mathcal L(B,\mathbb C)$? – Jochen Mar 2 '16 at 7:38
• Restrictions of continuous linear functions on $X$ to $B$ – 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).
• 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? – Bipolar Minds Mar 2 '16 at 21:41