# well-defined inner product

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 $$b^\ast a-b'^\ast a'\in N$$...

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 $$x \in A$$.