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Let $A$ be a $C^\ast$-algebra and $f\colon A\to\mathbb{C}$ a positive linear functional. Take $N=\{a\in A:f(a^\ast a)=0\}$. Define an inner product on the quotient space $A/N$ by $$ \langle a+N,b+N\rangle:=f(b^\ast a). $$

Using the Cauchy-Schwarz Inequality, it turns out $N$ is a left ideal of $A$, which makes the inner product well-defined.


I understand that $N$ is a left ideal of $A$, but I don't understand how it makes the inner product well-defined. I provided my attempt below (apologies in advanced for the repetitive "rather"):

Suppose $a+N=a'+N$ and $b+N=b'+N$. We want to show $$ \langle a+N,b+N\rangle=\langle a'+N,b'+N\rangle. $$ Which means we want to show $$ f(b^\ast a)=f(b'^\ast a'). $$ Or rather we want to show $$ f(b^\ast a)-f(b'^\ast a')=0. $$ Or rather $$ f(b^\ast a-b'^\ast a')=0. $$ Or rather $b^\ast a-b'^\ast a'\in N$. So we know $a-a'\in N$ and $b-b'\in N$. Also $b^\ast,b'^\ast\in A$. Therefore $$ (b^\ast a-b'^\ast a') + (b'^\ast a - b^\ast a')= b^\ast a - b^\ast a' + b'^\ast a - b'^\ast a'= (b^\ast +b'^\ast)(a-a')\in N. $$ I just can't seem to get that $b^\ast a-b'^\ast a'\in N$...

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First note that $$ N = \{a \in A \mid f(b^*a) = 0 \text{ for all }b \in A\} $$ since by Cauchy-Schwarz we have $$ \lvert f(b^*a) \vert^2 \leq f(b^*b)f(a^*a) = 0 $$ for all $a \in N$, $b \in A$. Furthermore, $N$ is a left ideal as you already noted.

Now if $a + N = a' + N$ and $b + N = b' + N$ then \begin{align*} \langle a+N, b+N \rangle &= \langle a-a'+N, b+N \rangle + \langle a'+N,b+N \rangle\\ &= \underbrace{f(b^*\underbrace{(a-a')}_{\in N})}_{=0} + \langle a'+N, b+N \rangle\\ &= \langle a'+N, b-b'+N \rangle + \langle a'+N, b'+N \rangle\\ &= f((b-b')^*a') + \langle a'+N, b'+N \rangle\\ &= \overline{f((a')^*(b-b'))} + \langle a'+N, b'+N \rangle\\ &= \langle a'+N, b'+N \rangle \end{align*} since for positive linear functionals $f:A\rightarrow \mathbb C$ we have $f(x^*) = \overline{f(x)} $ for all $x \in A$.

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