When does the dual to the space $K(X)$ of compact operators consist of nuclear functionals?

Question

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 $T$.

Answer 1

This is the question of duality of injective and projective tensor products of Banach spaces, and the natural question would be whether the dual of $K(X)$ can be represented by the functionals in $N(X^*)$. A quick answer is: If $X^*$ or $X^{**}$ has the approximation property and if $X^*$ or $X^{**}$ has the Radon-Nikodym property, then this is so; see Chapter VIII in Diestel and Uhl's "Vector Measures'' or Ryan's "Introduction to Tensor Products of Banach Spaces'' or Defant and Floret's "Tensor Norms and Operator Ideals''. So if $X$ is reflexive with the approximation property things are ok.