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*](https://en.wikipedia.org/wiki/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](https://www.encyclopediaofmath.org/index.php/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 $B\_{co}(H)$.



Is a set of nuclear functionals equicontinuous in compact-open topology if it is equicontinuous on each compact set?
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 $f\in F$
$$f(A)=\sum\_{n\geq0} f(A(P\_{n+1}-P\_n))\le \sum\_{n\geq0} \|f(\,\cdot\,(1-P\_n))\| \|AP\_{n+1}\|\le 1.$$


