Editing Trace class (section)

=== Trace class as the dual of compact operators ===

The [[dual space]] of <math>c_0</math> is <math>\ell^1(\N).</math> Similarly, we have that the dual of compact operators, denoted by <math>K(H)^*,</math> is the trace-class operators, denoted by <math>B_1.</math> The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let <math>f \in K(H)^*,</math> we identify <math>f</math> with the operator <math>T_f</math> defined by
<math display="block">\langle T_f x, y \rangle = f\left(S_{x,y}\right),</math>
where <math>S_{x,y}</math> is the rank-one operator given by
<math display="block">S_{x,y}(h) = \langle h, y \rangle x.</math>

This identification works because the finite-rank operators are norm-dense in <math>K(H).</math> In the event that <math>T_f</math> is a positive operator, for any orthonormal basis <math>u_i,</math> one has
<math display="block">\sum_i \langle T_f u_i, u_i \rangle = f(I) \leq \|f\|,</math>
where <math>I</math> is the identity operator:
<math display="block">I = \sum_i \langle \cdot, u_i \rangle u_i.</math>

But this means that <math>T_f</math> is trace-class. An appeal to [[polar decomposition]] extend this to the general case, where <math>T_f</math> need not be positive.

A limiting argument using finite-rank operators shows that <math>\|T_f\|_1 = \|f\|.</math> Thus <math>K(H)^*</math> is [[isometrically isomorphic]] to <math>B_1.</math>