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.
Best Answer
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 $\|L(p_n)\|=2^n$.