5
$\begingroup$

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 $B_{co}(H)$.

$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.