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Fixed some typos.
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Bokuan Li
2026-08-08 15:06:50 -04:00
parent bc1aea01b3
commit 8efc71c1a1
1 changed files with 5 additions and 5 deletions
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10
src/op/c-star/quotients.tex
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@@ -23,7 +23,7 @@
\]
\end{lemma}
\begin{proof}[Proof, {{\cite[Lemma 15.6]{Zhu}}}. ]
For each $y \in I$, $e_\beta y \to y$. Thus
For each $y \in I$, $ye_\beta \to y$. Thus
\[
\limsup_{\beta \in B}\norm{xe_\beta - x}_A = \limsup_{\beta \in B}\norm{(x - y)(1 - e_\beta)}_A \le \norm{x - y}_A
\]
@@ -44,11 +44,11 @@
For each $x \in A$ and $y \in I$,
\begin{align*}
\norm{x + I}_{A/I}^2 &= \lim_{\beta \in B}\norm{x - e_\beta x}_A^2 = \lim_{\beta \in B}\norm{(x - e_\beta x)(x - e_\beta x)^*}_A \\
&= \lim_{\beta \in B}\norm{(1_A - e_\beta)xx^*(1_A - e_\beta)}_A \\
&= \lim_{\beta \in B}\norm{(1_A - e_\beta)(xx^* + y)(1_A - e_\beta)}_A \le \norm{xx^* + y}_{A}
\norm{x + I}_{A/I}^2 &= \lim_{\beta \in B}\norm{x - xe_\beta }_A^2 = \lim_{\beta \in B}\norm{(x - xe_\beta )^*(x - x e_\beta)}_A \\
&= \lim_{\beta \in B}\norm{(1_A - e_\beta)x^*x(1_A - e_\beta)}_A \\
&= \lim_{\beta \in B}\norm{(1_A - e_\beta)(x^*x + y)(1_A - e_\beta)}_A \le \norm{x^*x + y}_{A}
\end{align*}
As the above holds for all $y \in I$, $\norm{x + I}_{A/I}^2 \le \norm{xx^* + I}_{A}$. Thus $A/I$ equipped with the quotient norm is a $C^*$-algebra.
As the above holds for all $y \in I$, $\norm{x + I}_{A/I}^2 \le \norm{x^*x + I}_{A/I}$. Thus $A/I$ equipped with the quotient norm is a $C^*$-algebra.
\end{proof}