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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,

Conjecture: 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 $B_{co}(H)$.

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 $B_{co}(H)$.

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,

Conjecture: 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 seems to be not sufficient for being equicontinuous on $B_{co}(H)$.

For me any answer, positive or negative, will be satisfactory.

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,

Conjecture: 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 $B_{co}(H)$.

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,

Conjecture: 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 $$ \sup_{x\in T}||Ax||\le 1\quad \Rightarrow\quad \sup_{f\in F}|f(A)|\le 1. $$

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 $B(H)$. From the Banach-Steinhauss theorem for $H$ it follows 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 seems to be not sufficient for being equicontinuous on $B_{co}(H)$.

For me any answer, positive or negative, will be satisfactory.

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,

Conjecture: 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 seems to be not sufficient for being equicontinuous on $B_{co}(H)$.

For me any answer, positive or negative, will be satisfactory.