Lemma 34.11.2.label Let $A$ be a $C^{*}$-algebra, $\phi \in S(A)$, and
\[N_{\phi} = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}\]
then:
- (1)
For any $x, y \in A$ with $x \in N_{\phi}$ or $y \in N_{\phi}$, $\dpn{x, y}{A}= 0$.
- (2)
$N_{\phi}$ is a closed left ideal of $A$.
Proof. (1): By the Cauchy-Schwarz inequality, for any $x, y \in A$,
\[|\dpn{x, y}{\phi}|^{2} \le \dpn{x, x}{\phi}\cdot \dpn{y, y}{\phi}\]
If $x \in N_{\phi}$ or $y \in N_{\phi}$, then the above inequality shows that $\dpn{x, y}{\phi}= 0$.
(2): As the zero set of a continuous function on $A$, $N_{\phi}$ is closed.
For any $x, y \in N_{\phi}$,
\begin{align*}\dpn{x + y, x + y}{\phi}&= \dpn{x, x}{\phi}+ \dpn{x, y}{\phi}+ \dpn{y, x}{\phi}+ \dpn{y, y}{\phi}\\&= \dpn{x, y}{\phi}+ \dpn{y, x}{\phi}\end{align*}
By (1), $\dpn{x, y}{\phi}= \dpn{y, x}{\phi}= 0$. Therefore $x + y \in N_{\phi}$.
Finally, for each $x \in N_{\phi}$ and $y \in A$,
\[\dpn{yx, yx}{\phi}= \dpn{x^*y^*yx, \phi}{A}= \dpn{x^*(y^*yx), \phi}{A}= \dpn{y^*yx, x}{\phi}=0\]
by (1).$\square$
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