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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 $T$.

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    $\begingroup$ 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). $\endgroup$ – Bill Johnson Apr 7 at 19:10
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    $\begingroup$ @Bill, probably it's this paper by Wolfgang Ruess that you have in mind: zbmath.org/?q=an%3A0573.46007 $\endgroup$ – Dirk Werner Apr 7 at 21:12
  • $\begingroup$ Right, @Dirk Werner. $\endgroup$ – Bill Johnson Apr 10 at 19:14
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    $\begingroup$ 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 $[0,1]$). $\endgroup$ – Robert Furber Apr 11 at 8:59
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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.

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