I think, nuclear functionals on the space of operators $B(X)$ (on a Banach space $X$) must have a characterization in terms of some special continuity. I would be grateful if somebody could help me with the following hypothesis.
First, a nuclear functional on $B(X)$ (where $X$ is a Banach space) is a linear functional $u:B(X)\to{\mathbb C}$ that can be represented in the form $$ u(A)=\sum_{n=1}^\infty \lambda_n\cdot f_n(Ax_n),\qquad A\in B(X), $$ where $\lambda_n\in{\mathbb C}$, $x_n\in X$, $f_n\in X^*$ are such that $$ \sum_{n=1}^\infty |\lambda_n|<\infty,\quad \sup_{n}||x_n||\le 1,\quad \sup_{n}||f_n||\le 1. $$ Let us denote by $B_{pc}(X)$ the space $B(X)$ of operators $A:X\to X$ endowed with the topology of pointwise convergence (in the case when $X$ is a Hilbert space it is usually called the strong operator topology).
Note that a set of operators $T\subseteq B_{pc}(X)$ is totally bounded in $B_{pc}(X)$ iff for each $x\in X$ the set $\{Ax;\ A\in T\}$ is totally bounded in $X$.
And note that each nuclear functional $u:B(X)\to{\mathbb C}$ is continuous on each totally bounded set $T\subseteq B_{pc}(X)$ with respect to the topology induced from $B_{pc}(X)$ (since $T$ is bounded with respect to the norm in $B(X)$ by the Banach-Steinhauss theorem, and using this we can "cut off the tail of the series up to arbitrary $\varepsilon>0$").
It seems to me, this can be turned back:
Conjecture: a linear functional $u:B(X)\to{\mathbb C}$ is nuclear if and only if it is continuous on each totally bounded set $T\subseteq B_{pc}(X)$ with respect to the topology induced from $B_{pc}(X)$.
This is true when $X$ is a Hilbert space. Is it true for arbitrary Banach space $X$? At least for $X$ with the approximation property?
Of course, we can assume in addition that $u$ is contunuous on $B(X)$ with respect to the usual norm topology, since this follows from the continuity of $u$ on each totally bounded set $T\subseteq B_{pc}(X)$.
