The question is originated from Pedersen's book "C*-algebras and their automorphism groups" (P55 Def. 3.6.5).
If $M$ is a von Neumann algebra in $B(H)$, let the $T(H)$ denote the elements in $B(H)$ of trace class and the set $N=\{x\in T(H)|~ Tr(ux)=0, \forall u\in M \}$. Prove: $T(H)/N\cong M_*$ (isometric isomorphism), the $M_*$ denotes all the normal functional on $M$.
Proof. From the Theorem 3.6.4 of Pedersen's book, we can establish a natural map from $T(H)/N$ to $M_*$ by $$T(H)/N\longrightarrow M_*$$ $$x+N\longmapsto \phi$$ where $x$ is an operator of trace class such that $\phi(y)=Tr(xy)$ for $y\in M$. It is easy to see that this linear map is bijective. And I can verify $||x+N||_1\leq||\phi||$ by the definition of $||.||_1$ and polar decomposition of $M$. However, how to prove $||x+N||_1\geq||\phi||$? (Here, the $||.||_1:=Tr(|.|)$).
First, note that we have, for $a\in B(H)$ and $b\in T(H)$, the Hölder inequality $$\tag1 |\operatorname{Tr}(ab)|\leq\|a\|\,\operatorname{Tr}(|b|). $$ Indeed, writing $b=v|b|$ the polar decomposition, we have by Cauchy-Schwarz \begin{align} |\operatorname{Tr}(ab)|&=|\operatorname{Tr}(av|b|^{1/2}\,|b|^{1/2})| \leq\operatorname{Tr}(|b|^{1/2}v^*a^*av|b|^{1/2})^{1/2}\operatorname{Tr}(|b|)^{1/2}\\[0.3cm] &\leq\|v^*a^*av\|^{1/2}\,\operatorname{Tr}(|b|)=\|av\|\,\operatorname{Tr}(|b|)\\[0.3cm] &\leq\|a\|\,\operatorname{Tr}(|b|). \end{align}
Because the quotient map is a $*$-homomorphism, given $z\in N$ we have $|x+z|=|x|+w$ for some $w\in N$. Then if $\|y\|=1$ and $x=v|x|$ is the polar decomposition, $$ |\operatorname{Tr}(xy)|=\operatorname{Tr}(v^*|x|y)=\operatorname{Tr}((|x|+w)yv^*) =\operatorname{Tr}(|x+z|v^*y)\leq\|v^*y\|\,\operatorname{Tr}(|x+z|) \leq\operatorname{Tr}(|x+z|). $$ As this can be done for any $z\in N$, we get $$\|\phi\|=\sup\{|\operatorname{Tr}(xy)|:\ y\in M,\ \|y\|=1\}\leq\|x+N\|_1. $$