Is it possible to characterize the Hausdorff locally convex spaces on which all bounded linear functionals are continuous?

It is known that a space is bornological if and only if the space is Mackey and every bounded linear functional is continuous. It follows that, in a way, I am interested in bornological spaces that are not Mackey, but this is not a usable characterization. (I am especially interested in the space of continuous functions on a Hausdorff completely regular space endowed with various topologies.)

  • $\begingroup$ What kind of "characterisation" are you thinking of? I mean it is always possible to write down the coarsest locally convex vector space topology $\tau_0$ on $V$ that makes all bounded functionals $V\to\mathbb C$ continuous. A vector space topology $\tau$ on $V$ makes all bounded linear functionals $V\to\mathbb C$ continuous if and only if $\tau_0\subseteq\tau$. Is that the kind of characterisation you are looking for? $\endgroup$ – Johannes Hahn Dec 23 '17 at 20:42
  • $\begingroup$ @JohannesHahn: Such a characterization, while correct, is not easily usable in practice. Anyway, I am open to anything. Think about this: does the space of bounded continuous functions on a Hausdorff completely regular space, endowed with the strict topology, have the property that I am asking about? Can you use your characterization to answer my question? (The strict topology is given by the seminorms $p_f (F) = \sup f |F|$ with $f$ in the set of positive bounded functions that vanish at infinity.) $\endgroup$ – Alex M. Dec 23 '17 at 20:58
  • $\begingroup$ Spaces with strict topologies are the worst possible candidates for this behaviour. This is because they have the same bounded sets as the finer norm topology but not the same dual, in general. By the way, don't you mean continuous functions in your definition of the strict topology? This, in the locally compact case - semi-continuity in the general one. $\endgroup$ – mühl Dec 24 '17 at 8:14

Your strict topology ($\beta_t$) is the strongest locally convex topology on $ C_b(T)$ coinciding with the compact-open topology on the unit ball $ \{ x \in C_b(T): \sup_{t \in T} | x(t) |\leq 1 \}$.

If each bounded linear functional on $ C_b(T)$ is $\beta_t$-continuous, then $T$ is compact. Suppose $T$ is not compact. Let $p\in\beta T\setminus T$ ($\beta T$ is the Stone-Čech compactification of $T$). Define the subset $N$ of $C(\beta T)$ by $N=\{f:0\le f\le 1, ~f(p)=1\}$. Labelling $N$ as $\{f_i : i\in I\}$ we make $I$ into a directed set by saying that $i\ge j$ iff $f_i\le f_j$. Let $g_i$ is the restriction of $f_i$ to $T$. Then the net $\{g_i : i\in I\}$ decreases pointwise to $0$. By the Dini theorem $g_i\rightarrow0$ uniformly on compact sets. By the above characterization of the strict topology we have that $g_i\rightarrow0$ in $\beta_t$. The linear functional $\delta_p(g)=f(p)$, where $f$ is the continuous extension of $g\in C_b(T)$ to $\beta T$, is bounded, so $\delta_p(g_i)\rightarrow 0$. This is in contradiction with $\delta_p(g_i)=f_i(p)=1$.

  • $\begingroup$ This answers my question in the interesting case of the strict topology, but does there exist an answer for general locally convex spaces? $\endgroup$ – Alex M. Dec 24 '17 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.