Show that $(S^1)^*=B(\ell^2)$ knowing $(\ell^1)^*=l^\infty$

Question

Is there a way to show that dual of trace class operators, $S^1$, is $B(\ell^2)$, bounded operators on $\ell^2$, knowing that dual of $\ell^1$ is $\ell^\infty$?

Answer 1

The most you're going to get in that direction is a close analogy, with the rank one trace class operators $v\otimes w$ (= the map $u \mapsto \langle u,w\rangle v$), for unit vectors $v, w \in l^2$, corresponding to the standard basis vectors $e_n$ in $l^1$. We have a natural nonexpansive map from $l^\infty$ into $(l^1)^*$ (or from $B(l^2)$ into $TC(l^2)^*$), and you check that it's an isometry by looking at behavior on the $e_n$ (or the $v\otimes w$). Then you check that it's onto by starting with an arbitrary bounded linear functional and showing that it arises from pairing with something in $l^\infty$ (or in $B(l^2)$) by looking at its behavior on the $e_n$ (or the $v\otimes w$).

In the opposite direction, it is possible to deduce the $l^\infty$ result from the $B(l^2)$ result. Isometrically identify $l^1$ with the diagonal trace class operators; then by a standard result, the dual of $l^1 \subset TC(l^2)$ is the quotient of $TC(l^2)^* \cong B(l^2)$ by the annihilator of $l^1$. This annihilator is just the operators which are zero down the diagonal relative to the standard basis of $l^2$, so the quotient is just the bounded operators which are diagonalized, and that is isometrically identified with $l^\infty$. That's a sketch of a proof, but I think filling in the details is actually more work than just proving $(l^1)^* \cong l^\infty$ from scratch.

Answer 2

I think there is a way to at least relate the two results. Firstly, by a standard method you can reduce to the case of real sequences, resp. to self-adjoint operators. Then start with the finite dimensional cases. $\ell^1_n$ and $\ell^\infty_n$ are algebraically in duality and the fact that this also holds for the norms is a simple and standard computation. Analogously for the spacesof $n\times n$ matrices, again at first algebraically. It also holds for the trace and supremum norm. The only (slightly) non-trivial part is the equality of the supremum norm of a matrix and its norm as a functional and this can be reduced to the commutative case by looking at the matrix at which it attains its supremum and assuming, as we can by the spectral theorem, that it is diagonal. This is the point at which we deduce the non-commutative case from the commutative one—-at least at one step (which I think is the key one). To go from here to the infinite dimensional case is a standard approximation argument which I won‘t go into since it has nothing to do with your question. (However, I would be happy to do so if requested).