Consider the following theorem in Murphy's '$C^*$-algebras and operator theory':
Why is the marked equality true? This seems to boil down to showing that $$tr(uw^*) = tr(w^*u)$$
Myrphy already showed that $tr(uv) = tr(vu)$ when one of the operators $u,v$ is trace class or they both are Hilbert Schmidt operators, but neither of these conditions seem to apply here. I'm also a little bit confused why Murphy uses the trace for operators that are not necessarily trace-class (the trace was defined for such operators).
What am I missing?
As you have noted, that equality is no problem when $u$ is trace class. When $u$ is not trace class, then the statement $\Vert u \Vert_1 = \Vert u^* \Vert_1$ should be read as "neither is $u^*$". I.e. $u$ is trace class iff $u^*$ is. This is now clear, for example, by the characterisation of trace class operators as the product of Hilbert–Schmidt operators (Theorem 2.4.13), because the adjoint of a Hilbert–Schmidt operator is Hilbert–Schmidt (as can be seen from the explicit formula in Example 2.4.1.).