Let $X$ be a Banach space and $B(X)$ be its space of all (bounded) operators. A *nuclear functional* on $B(X)$ 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 $N(X)$ the space of all nuclear functionals on $B(X)$.

If $X$ is a Hilbert space, then it is well known (see G.J.Murphy, C*-Algebras and Operator Theory, Theorem 4.2.1) that the dual space $K(X)^*$ to the space of all compact operators $K(X)$ coinsides with the space of all nuclear functionals: $$ K(X)^*=N(X) $$ (this is an isomorphism of Banach spaces, but for me it is important that this is an equality of sets).

Is the same true for all Banach spaces $X$? Or at least for all Banach spaces with the (classical) approximation property?

I am mostly interested in the case when $X=C(T)$, the space of continuous functions on a compact topological space $T$.

paperby Wolfgang Ruess that you have in mind: zbmath.org/?q=an%3A0573.46007 $\endgroup$ – Dirk Werner Apr 7 '19 at 21:12