Is a set of nuclear functionals equicontinuous in compact-open topology if it is equicontinuous on each compact set?

Let $$H$$ be a Hilbert space and $$B(H)$$ be its space of all (bounded) operators. A nuclear functional on $$B(H)$$ is a linear functional $$f:B(H)\to{\mathbb C}$$ that can be represented in the form $$f(A)=\sum_{n=1}^\infty \lambda_n\cdot \langle Ax_n,y_n\rangle,\qquad A\in B(H),$$ where $$\lambda_n\in{\mathbb C}$$, $$x_n,y_n\in H$$ are such that $$\sum_{n=1}^\infty |\lambda_n|<\infty,\quad \sup_{n}||x_n||\le 1,\quad \sup_{n}||y_n||\le 1.$$ If we endow $$B(H)$$ with compact-open topology (what is a bit unusual), and denote by $$B_{co}(H)$$ this space with this new topology, then it is easy to show that nuclear (and only nuclear) functionals are continuous on $$B_{co}(H)$$. Let us denote by $$N(H)$$ the set of all nuclear functionals on $$B(H)$$ (or, what is the same, linear continuous functionals on $$B_{co}(H)$$).

I wonder if $$B_{co}(H)$$ satisfies the following weakened version of the Banach-Steinhauss theorem:

Conjecture: if a set of nuclear functionals $$F\subseteq N(H)$$ is equicontinuous on each compact set $$K\subseteq B_{co}(H)$$, then $$F$$ is equicontinuous on $$B_{co}(H)$$.

In other words,

If $$F\subseteq N(H)$$ and for each compact set $$K\subseteq B_{co}(H)$$ there is a compact set $$T\subseteq H$$ such that $$(A\in K\ \&\ \sup_{x\in T}||Ax||\le 1)\quad \Rightarrow\quad \sup_{f\in F}|f(A)|\le 1$$ then there is a compact set $$T\subseteq H$$ such that $$\sup_{x\in T}||Ax||\le 1\quad \Rightarrow\quad \sup_{f\in F}|f(A)|\le 1.$$

From the Banach-Steinhauss theorem for $$H$$ it follows that the compact sets $$K\subseteq B_{co}(H)$$ are the same as compact sets in what is called the strong operator topology (i.e. the topology of pointwise convergence) on $$B(H)$$. One can show also that if $$F\subseteq N(H)$$ is equicontinuous on every such a set $$K$$, then $$F$$ is bounded with respect to the usual nuclear norm: $$\sup_{f\in F}||f||<\infty$$ where $$||f||=\inf\sum_{n=1}^\infty|\lambda_n|$$ and the infimum is over all the representations of $$f$$ as a nuclear functional. But having bounded nuclear norm is not sufficient for being equicontinuous on $$B_{co}(H)$$.

The answer is YES. First of all, your $$N(H)$$ is nothing but the predual of $$B(H)$$ (the ultraweakly continuous linear functionals) and the nuclear norm is nothing but the norm as linear functionals.
We assume $$\sup_{f\in F}\| f\|\le1$$. Let $$P_n \in B(H)$$ be finite rank orthogonal projections such that $$P_n \nearrow 1$$ in SOT. (The result holds true for nonseparable case as well, but I assume $$H$$ is separable for simplicity.) I claim that the first condition implies that $$\limsup_n\sup_{f\in F}\|f(\,\cdot\,(1-P_n))\| = 0.$$ Indeed, if this were not the case, then there are $$\epsilon>0$$, $$n(k)\nearrow\infty$$, $$f_k\in F$$, and $$A_k\in B(H)$$ such that $$A_k=A_k(P_{n(k)}-P_{n(k-1)})$$, $$\| A_k\|\le1$$, and $$f_k(A_k)\geq\epsilon$$. The set $$\{ \epsilon^{-1}A_k \}\cup\{0\}$$ is SOT-compact and satisfies $$\limsup_k\sup_{v \in T} \|\epsilon^{-1}A_k v\|=0$$ for any compact subset $$T\subset H$$ and $$f_k(\epsilon^{-1}A_k)\geq1$$. Thus, after passing to a subsequence, we may assume that $$P_0=0$$ and $$\sup_{f\in F}\|f(\,\cdot\,(1-P_n))\| < 4^{-n}$$ for every $$n\geq1$$. Put $$T:=\{ v : n\in{\bf N},\,v \in P_nH,\,\| v \|\le 2^{-n+2}\}$$. Then $$T$$ is pre-compact in $$H$$. If $$\|Av\| \le 1$$ for all $$v\in T$$, then $$\|AP_n\|\le 2^{n-2}$$ and so for any $$f\in F$$ $$f(A)=\sum_{n\geq0} f(A(P_{n+1}-P_n))\le \sum_{n\geq0} \|f(\,\cdot\,(1-P_n))\| \|AP_{n+1}\|\le 1.$$
• Narutaka, thank you! It's monday, it will take me some time to understand this. In your choice of $T$ do you mean the quantor of existence for $n$ (i.e. $T:=\{ v : \exists n\in{\bf N} \ \ v \in P_nH\ \ \& \ \ \| v \|\le 2^{-n+2}\}$)? – Sergei Akbarov Mar 18 '19 at 7:34
• And, excuse me, what does the multiplication $\cdot$ mean in this phragment: $\|f(\,\cdot\,(1-P_n))\|$? (Where I suppose $\|\dots\|$ is not a norm but the absolute value?) – Sergei Akbarov Mar 18 '19 at 8:06
• @Sergei Akbarov: Yes to the first question. As to the second, it's norm as a linear functional $B(H)\ni B\mapsto f(B(1-P_n)) \in {\bf C}$. – Narutaka OZAWA Mar 18 '19 at 11:50
• Narutaka, yes, it's OK, I would only replace $\limsup_n$ by $\lim_n$, because it's a bit more complicated than necessary. – Sergei Akbarov Mar 19 '19 at 14:08