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It seems that quite a standard trick of showing two $C^*$ algebras are as follows:

Let $A$ be a $C^*$ algebra $B$ another $C^*$ algebra. $A' \subseteq A$ be a subalgebra that is closed under $*$. (Not necessarily closed) and is dense in $A$. Let $L:A' \rightarrow B$ be a $*$-isometric homomoprhism which has dense image. Then $L$ extends to a $*$-isometric isomorphism .

So my aim is showing that $L$ actually extends to an injective map - since it is not necessarily case that $L$ is injective in its closure.

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Given $a\in A$ take $\{a_j\}\subset A'$ with $a_j\to a$. Since $L$ is isometric, we have $$ \|L(a_j)-L(a_k)\|=\|L(a_j-a_k)\|=\|a_j-a_k\|, $$ so $\{L(a_j)\}$ is Cauchy. As $B$ is complete, there exists $b_a=\lim L(a_j)$. We want to extend $L$ by defining $L(a)=b_a$. For this we need to show that $b_a$ is unique for each $a$; and this follows easily: if $c_j\to a$, then $$ \|L(c_j)-L(a_j)\|=\|c_j-a_j\|\leq\|c_j-a\|+\|a-a_j\|\to0, $$ so $L(c_j)\to b_a$. So the extension $L$ is well-defined. Also $$ \|L(a)\|=\|\lim_j L(a_j)\|=\lim_j\|L(a_j)\|=\lim_j\|a_j\|=\|a\|, $$ so $L$ is isometric.

If you don't require $L$ to be isometric, the result is not true. Let $A=C[0,1]$, $B=\mathbb C\oplus C[0,1]$, and take $A'\subset A$ to be the $*$-algebra of complex polynomials. Define $L:A'\to B$ by $$ L(p)=p(2)\oplus p. $$ This is clearly an injective $*$-homomorphism. And it has dense image: Given $(\lambda,f)\in B$, we can choose a sequence of polynomials $\{p_j\}$ such that $p_j\to f$ uniformly on $[0,1]$ and $p_j(2)=\lambda $ (simply use Stone-Weierstrass on a continuous function $g$ that agrees with $f$ on $[0,1]$ and with $g(2)=\lambda$).

So $L$ satisfies all the hypotheses; but it is not bounded. If you consider $p_n(x)=x^n$, then $\|p_n\|=1$, while $p_n(2)=2^n$, so $\|L(p_n)\|=2^n$.

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  • $\begingroup$ I suppose you mean $\pi(A')=L(A')$? The statement I know requires $\rho$ to be bounded is that its domain is a Banach $*$ algebra i.e. closed. It seems that the claim is: $L:A \rightarrow B$ is bounded, if $A$ is a $*$-algebra (with a norm such that $A$ is not necessarily complete), then $B$ is a $C^*$ algebra. Please correct me if I mistinterpret you. $\endgroup$ – CL. Apr 16 at 19:24
  • $\begingroup$ Martin, I have edited my problem, I noticed I have quite a lot of typos. $\endgroup$ – CL. Apr 16 at 19:34
  • $\begingroup$ Yes, I was glossing over the part that $*$-homomorphisms are bounded, and that doesn't work when the domain is not closed. I have rewritten the answer with a counterexample. $\endgroup$ – Martin Argerami Apr 16 at 20:09
  • $\begingroup$ Hi Martin, thanks a lot, I have added the condition that $L$ is an isometric isomorphism, I wonder if the argument holds from here. In fact, I am just trying to generalize the trick you did here: math.stackexchange.com/questions/1918086/… $\endgroup$ – CL. Apr 16 at 20:34
  • $\begingroup$ And please do keep the counter example in your edits : ) $\endgroup$ – CL. Apr 16 at 20:35
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The answer by Martin is of course correct, I just want to emphasise that this statement and the way to see it is essentially elementary functional analysis. In my opinion is very helpful to be able to differentiate when a statement is about the linear structure and when a statement needs the C*.

Central is the following statement: Let $A:V\to W$ be a continuous linear map between normed vector spaces, then there is a unique continuous linear extension $\overline A :\overline V\to\overline W$ between the completions. This can be done as an elementary exercise and is basically the first half of Martin's answer.

(As a remark: The above statement is itself an expression of a more general statement, namely that Lipschitz maps between metric spaces $A,B$ so that $B$ is complete and $A$ is dense in some other metric space $A'$ always have a unique continuous extension $A'\to B$.)

The rest of what you need follows from:

  1. If a continous linear map is isometric on a dense sub-set, then it is isometric on its entire domain.

  2. If a continuous linear map is multiplicative wrt continuous multiplications on a dense sub-set, then it is multiplicative on its entire domain.

  3. If a continuous linear map $A$ satisfies $A(v^*)= A(v)^*$ on a dense sub-set and the $*$ operation is continuous on domain and codomain then the above equality holds on the entire domain.

Note that this is basically the same statement $3$ times: Continuous linear maps that preserve a continuous structure on a dense sub-set automatically do so on the entire domain.

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