Let $N$ be a finite von Neumann algebra with a fixed faithful normal trace $\tau$. Let $L^1(N)$ be the completion of $N$ with respect to the norm $\|\cdot\|_1$ defined by $\|x\|_1=\tau (|x|)=\sup\{|\tau(xy)|:\|y\|\leq 1,\,y\in N\}$.
Problem: Prove that the Banach space dual of $L^1(N)$ is isometrically isomorphic to $N$.
I got stuck with this while reading Section $3.6$ from the book 'Finite von Neumann algebras and masas' by Sinclair and Smith. They says that it follows from a standard duality argument. But I did not get their argument. More specifically, how do I show that for any bounded linear functional $\Phi:L^1(N)\rightarrow\mathbb{C}$, there exists a unique $x\in N$ such that $\Phi(y)=\tau(xy)$ for each $y\in N$? Thanks in advance for any help.
Let $(H_\tau,\pi_\tau,\Omega_\tau)$ be the cyclic GNS representation associated with $\tau$. For $x,y\in N$ we have $$ |\Phi(yx^\ast)|\leq \|\Phi\|\tau(|yx^\ast|)\leq \|\Phi\|\tau(|x|^2)^{1/2}\tau(|y|^2)^{1/2}. $$ Hence there exists a unique bounded sesquilinear form $b$ on $H_\tau$ such that $b(\pi_\tau(x)\Omega_\tau,\pi_\tau(y)\Omega_\tau)=\Phi(yx^\ast)$. By the Riesz representation theorem there exists a unique $a\in B(H_\tau)$ such that $\Phi(yx^\ast)=\langle \pi_\tau(x)\Omega_\tau,a\pi_\tau(y)\Omega_\tau\rangle$.
It remains to show that $a\in \pi_\tau(N)$. For that purpose recall that $\pi_\tau(N)'=\{\pi_\tau'(z)\mid z\in N\}$, where $\pi_\tau'(z)$ acts on $H_\tau$ by $\pi_\tau'(z)\pi_\tau(x)\Omega_\tau=\pi_\tau(xz)\Omega_\tau$. We have \begin{align*} \langle \pi_\tau(x)\Omega_\tau,a\pi_\tau'(z)\pi_\tau(y)\Omega_\tau\rangle&=\langle\pi_\tau(x)\Omega_\tau,a\pi_\tau(yz)\Omega_\tau\rangle\\ &=\Phi(yzx^\ast)\\ &=\Phi(y(xz^\ast)^\ast)\\ &=\langle \pi_\tau(xz^\ast)\Omega_\tau,a\pi_\tau(y)\Omega_\tau\rangle\\ &=\langle\pi_\tau(x)\Omega_\tau,\pi_\tau'(z)a\pi_\tau(y)\Omega_\tau\rangle \end{align*} for all $x,y,z\in N$. Thus $a\pi_\tau'(z)=\pi_\tau'(z)a$ for all $z\in N$, that is, $a\in \pi_\tau(N)''=\pi_\tau(N)$.