Unable to show the boundedness of the map $\varphi$ from $\hat M$ to the Hilbert space $H_\tau$

Question

Let $M$ be a von Neumann algebra and $\tau$ be a faithful normal state. Let $(\pi_\tau,H_\tau)$ be the GNS representation of $M$ with respect to $\tau$, that is, $\pi_\tau: M \to B(H_\tau)$ is a $*$-homomorphism. Since $\tau$ is faithful, we can view $M$ as a dense subspace of $H_\tau$. We use the notation $\hat x$ when we want to stress the point that $x$ is considered as an element of $H_\tau$.
Let $x \in M$. Now define $R_x: \hat M \to H_\tau$ by $R_x(\hat y)=\hat {yx}$ for all $y \in M$. Recall that the inner product $\langle,\rangle$ defined on $\hat M$ with respect to which $H_\tau$ is the completion is defined as $$\langle \hat x,\hat y\rangle=\tau(x^*y),~~\forall ~x,y \in M.$$ Now notice that $R_x$ is bounded. This is because, $$\|R_x(\hat y)\|^2=\|\hat{yx}\|^2=\langle \hat {yx},\hat {yx}\rangle=\tau(x^*y^*xy)=\tau(yxx^*y^*)\le \|x^*x\|\tau(y^*y)=\|x\|^2\|y\|^2.$$ Now since $R_x$ is bounded and $\hat M$ is dense in $H_\tau$, we can extend $R_x$ to $H_\tau$. Call it again $R_x$ and taking $R_x \in B(H_\tau)$. Let $\theta \in H_\tau$. Now define $\varphi: \hat M \to H_\tau$ by $$\varphi(\hat x)=R_x(\theta),~~\forall~ x \in M.$$ Now I am unable to show that $\varphi$ is bounded and also if not bounded I am unable to find counterexamples. Please help me to solve this. Thank you.

Answer 1

Suppose that $M$ is a II$_1$-factor with trace $\tau$, and let $\{p_n\}$ be a pairwise orthogonal sequence of projections with $\tau(p_n)=2^{-n}$. Define $$ \theta=\sum_k\sqrt k\,\hat p_k. $$ This sum converges because the $\hat p_k$ are pairwise orthgonal and $$ \|\theta\|_2^2=\sum_kk\,2^{-k}=2. $$ Let $$ x_n=2^{n/2}\,\hat p_n. $$ Then, for all $n$.
$$ \|\hat x_n\|^2_2=2^n\,\tau(p_n)=1. $$ And $$ \|\phi(\hat x_n)\|_2=\|R_{x_n}(\theta)\|_2=\sqrt n\,2^{n/2}\,\|\hat p_n\|_2 =\sqrt n\,2^{n/2}\,\tau(p_n)^{1/2}=\sqrt n. $$ Hence $\phi$ is unbounded.