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 $f\in F$ $$f(A)=\sum_{n\geq0} f(A(P_{n+1}-P_n))\le \sum_{n\geq0} \|f(\,\cdot\,(1-P_n))\| \|AP_{n+1}\|\le 1.$$

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  • $\begingroup$ 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. $T:=\{ v : \exists n\in{\bf N} \ \ v \in P_nH\ \ \& \ \ \| v \|\le 2^{-n+2}\}$)? $\endgroup$ – Sergei Akbarov Mar 18 '19 at 7:34
  • $\begingroup$ 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?) $\endgroup$ – Sergei Akbarov Mar 18 '19 at 8:06
  • $\begingroup$ @Sergei Akbarov: Yes to the first question. As to the second, it's norm as a linear functional $B(H)\ni B\mapsto f(B(1-P_n)) \in {\bf C}$. $\endgroup$ – Narutaka OZAWA Mar 18 '19 at 11:50
  • $\begingroup$ Narutaka, yes, it's OK, I would only replace $\limsup_n$ by $\lim_n$, because it's a bit more complicated than necessary. $\endgroup$ – Sergei Akbarov Mar 19 '19 at 14:08

This is a comment on the above answer but I am not entitled. The result is just a special case of the celebrated Banach-Dieudonné theorem (which gives a rather stronger statement).

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  • $\begingroup$ Could you please, give some details? I know only the form of the Banach-Dieudonné theorem from J.Horvath's book (Theorem 3.10.1). I don't see how it implies what I need. $\endgroup$ – Sergei Akbarov Mar 18 '19 at 16:26

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