Is $N_\phi = \{x \in E: \phi(\langle x,x\rangle)=0\}$ a Hilbert submodule of $E$?

Question

Let $E$ be a (right) Hilbert module over the $C^*$-algebra $B$. Let $\phi$ be a state on the $C^*$-algebra $B$. Then consider $$N_\phi:= \{x \in E: \phi(\langle x,x\rangle)=0\}.$$

I want to show that $N_\phi$ is a submodule of $E$, but for this I need to show that $$b \in B, x \in N_\phi \implies \phi(b^*\langle x,x\rangle b)=\phi(\langle xb,xb\rangle)= 0.$$

Why is this true?

Answer 1

It is not true. Take $B= M_2(\mathbb C)$ (with standard matrix units $e_{i,j}$), $E= B$ as a Hilbert $B$-module in the usual way, and let $\phi \in B^\ast$ be compression to the $(1,1)$-corner. Then $x = e_{2,2} \in E$ and $b = e_{2,1} \in B$ satisfy $x\in N_\phi$ and $xb \notin N_\phi$.