I'm reading Takesaki's second volume and I have a question:
Let $M\subseteq B(H)$ be a von Neumann algebra and let $\varphi: M_+ \to [0, \infty]$ a weight. We assume moreover that $\varphi$ is semifinite, which is defined as follows by Takesaki (p41):
The weight $\varphi$ is called-semifinite if $\mathfrak{p}_\varphi=\{x \in M_+: \varphi(x) <\infty\}$ generates $M$.
Question 1: What exactly does this mean? Is this generated as a von Neumann algebra? As a linear space? It feels natural to require that we first consider the $*$-algebra generated by $\mathfrak{p}_\varphi$ and we require density of this $*$-algebra in either of the weak/strong/strong$^*$-topologies or their $\sigma$-counterparts (the closures in all these topologies automatically coincide).
With respect to the weight $\varphi$, we can consider a GNS-construction. I'll briefly sketch how this can be done.
Basically, the first thing we do is consider $$\mathfrak{n}_\varphi = \{x \in M: \varphi(x^*x)< \infty\}$$ which is a left ideal of $M$. Then we consider the subspace $N_\varphi = \{x \in M: \varphi(x^*x) = 0\}$ and the quotient vector space $\mathfrak{n}_\varphi/N_\varphi$ becomes an inner product space for $$\langle x + N_\varphi, y + N_\varphi\rangle := \varphi(y^*x)$$ where the right hand side makes sense because we can define $\varphi$ on elements of the $*$-subalgebra $$\mathfrak{m}_\varphi = \mathfrak{n}_\varphi^*\mathfrak{n}_\varphi$$ in a way that agrees with how $\varphi$ is defined on $\mathfrak{p}_\varphi$. Then $H_\varphi$ is the Hilbert space completion of this quotient space and we have a unique $*$-representation $\pi_\varphi: M \to B(H_\varphi)$ uniquely given by $$\pi_\varphi(x): H_\varphi \to H_\varphi: y + N_\varphi \mapsto xy + N_\varphi.$$
Takesaki now claims that if $\varphi$ is faithful and semi-finite (and also normal, but I don't think this matters here), then $\pi_\varphi$ is also faithful.
Question 2: Why is this the case? If $\pi_\varphi(x)=0$ for some positive (without loss of generality, by the $C^*$-identity) $x \in M$, then $$0=\langle \pi_\varphi(x)(y+ N_\varphi), z+ N_\varphi\rangle = \varphi(z^*xy)$$ for all $z,y \in \mathfrak{n}_\varphi$.
In particular, if $z \in \mathfrak{p}_\varphi$, we get $$0 = \varphi(z^{1/2} x z^{1/2}) = \varphi((z^{1/2}x^{1/2})(z^{1/2}x^{1/2})^*)$$ so faithfulness of $\varphi$ gives $z^{1/2}x^{1/2}=0$. Hence, $zx=0$ for all $z \in p_\varphi$. Now, I guess to conclude I will have to know the answer to question $(1)$!
As already mentioned, semifinite means that $\mathfrak m_\varphi$ is $\sigma$-weak-dense (i.e. wot-dense, sot-dense, etc.).
Once you know that $zx=0$ for all $z\in p_\varphi$, you also get that $z^*z=x=0$, so $zx=0$ for all $z\in \mathfrak n_\varphi$. Similarly you also get that $zx=0$ for all $z\in\mathfrak m_\varphi$. As $\mathfrak m_\varphi$ is dense in $M$, you are done.