Suppose $A\in L(X,Y)$. $||\cdot||$ denotes spectral norm and denotes the largest singular value of a matrix, i.e. the largest eigenvalue of $\sqrt{A^*A}$.
$||\cdot||_{tr}$ denotes trace norm. We have that $$||A||_{tr}=tr\sqrt{A^*A}$$ So I would like to prove the statement that $$||A||_{tr}=\max\{|tr(A^*B)|: B\in L(X,Y), ||B||=1 \}$$
I know that from Nielsen and Chuang lemma 9.5 that
$$|tr(AU)|\le tr |A|$$ and equality is achieved by a unitary.
We have by definition that $|A|=\sqrt{A^*A}$. So $||A||_{tr}=tr|A|$.
I think my question is if $B$ is not a unitary but has norm 1, can we have that
$$|tr(AB)|> tr |A|\ge |tr(AU)|$$ for any unitary? And if yes, why the maximum is still achieved by a unitary?