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](https://rads.stackoverflow.com/amzn/click/com/0125113609), 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 $T$.



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


