The following question looks simple, but the answer is not obvious for me:


Let $S$ be a $\*$-algebra and $\left\Vert \cdot \right\Vert \_{1}$, $\left\Vert \cdot \right\Vert \_{2}$ $C^\*$-norms on $S$ with $\left\Vert \cdot \right\Vert \_{1} \leq \left\Vert \cdot \right\Vert \_{2}$. Denote by $A$, $B$ the generated $C^\*$-algebras. If a sequence $(x\_n)\_{n \in \mathbb{N}}$ is convergent in both norms with limit $0 \in A$ and $b\in B$, does this already imply $b=0$?



Sequence in *-algebra with different limits for two C*-norms?
The answer is no (to the main question, not the title). Consider the \*-algebra $S$ of \*-polynomials generated by one variable $z$ such that $zz^\* = z^\*z$, *i.e.* the free commutative \*-algebra on one generator. Each $a \in S$ can be considered to be a continuous function $\mathbb{C} \rightarrow \mathbb{C}$, so we can define
$$
\|a\|\_1 = \sup \{ a(x) \mid x \in [0,1] \} \\
\|a\|\_2 = \sup \{ a(x) \mid x \in [0,2] \}
$$
Since $[0,1] \subseteq [0,2]$, it is clear that $\|\cdot\|\_1 \leq \|\cdot\|\_2$. By the Stone-Weierstrass theorem, the completion of $S$ in $\|\cdot\|\_1$ is $C([0,1])$ and the completion of $\|\cdot\|\_2$ is $C([0,2])$. 


So we can take a continuous function $b : [0,2] \rightarrow \mathbb{C}$ that vanishes on $[0,1]$ but takes a non-zero value in $(1,2]$, and a sequence $(a\_i)\_i$ in $S$ such that $a\_i \to b$ with respect to $\|\cdot\|\_2$. Then
$$
\|a\_i\|\_1 = \|a\_i-b\|\_1 \leq \|a\_i - b\|\_2 \to 0
$$
So $(a\_i)\_i$ is convergent with respect to both norms and has limit $0$ in $\|\cdot\|\_1$, but its limit in $\|\cdot\|\_2$ is $b \neq 0$.


