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?



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


