Does an Automorphism on a C*-Algebra Induce a *-Isomorphism?

[Oxymoron]

If n\in N then \alpha(a)(n) \in N for all a \in A. But we have

f((an)^*(an)) = f(n^*a^*an) = 0

which is in N. So \alpha\,:\,N\rightarrow N gives L\,:\,A/N\rightarrow A/N such that

L(a)(b+N) = \alpha(a)(b) + N = ab+N

Then, L is linear because alpha is. To show that it is bounded we have

\|L(a)(b+N)\|^2 = (ab+N\,|\,ab+N) = f(b^*a^*ab) = f((cb)^*cb) \geq 0

Thus

\|a\|^2(b+N\,|\,b+N) = \|a\|^2f(b^*b) \geq f(b^*a^*ab) = \|L(a)(b+N)\|^2

and L(a) is bounded with \|L(a)\|\leq \|a\|. Therefore L is an isometry.

 

[matt grime]

If n\in N then \alpha(a)(n) \in N for all a \in A.

Why? I see no reason for N to be an ideal.

But we have

f((an)^*(an)) = f(n^*a^*an) = 0

again, why would that be true? N is not an ideal, f is not multiplicative.

which is in N. So \alpha\,:\,N\rightarrow N gives L\,:\,A/N\rightarrow A/N such that

L(a)(b+N) = \alpha(a)(b) + N = ab+N

again, this doesn't make sense. What are you doing?

Then, L is linear because alpha is. To show that it is bounded we have

\|L(a)(b+N)\|^2 = (ab+N\,|\,ab+N) = f(b^*a^*ab) = f((cb)^*cb) \geq 0

Thus

\|a\|^2(b+N\,|\,b+N) = \|a\|^2f(b^*b) \geq f(b^*a^*ab) = \|L(a)(b+N)\|^2

and L(a) is bounded with \|L(a)\|\leq \|a\|. Therefore L is an isometry.

What has multiplication by L(a) got to do with anything? I don't understand what you're trying to accomplish by looking at this.

f is just a linear functional, it does not have any of these properties you are assigning to it.

I already explained why L is an isometry, and this isn't it.L maps A/N to A/N. So, take b+N in A/N, what does L map it to? Lb+N, that is all. It does not require you to pick and a and work out L(a)(b+N).