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

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 $$T$$.

• Look at the book of Diestel-Uhl or Ruess' book Duality and Geometry of Spaces of Compact Operators. For $X=C(T)$, $T$ compact Hausdorff, $K(X)^*= N(X^*)$ if and only if $T$ is dispersed (has no perfect subsets). Apr 7, 2019 at 19:10
• @Bill, probably it's this paper by Wolfgang Ruess that you have in mind: zbmath.org/?q=an%3A0573.46007 Apr 7, 2019 at 21:12
• Right, @Dirk Werner. Apr 10, 2019 at 19:14
• For Diestel and Uhl it is Corollary 6 on page 175 and the discussion immediately preceding it (as well as the fact, not stated there, that a compact Hausdorff space is scattered iff it admits no surjection onto $[0,1]$). Apr 11, 2019 at 8:59

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.