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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 $b=0$?

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    $\begingroup$ Robert Furber has given a nice commutative example, but it may also be worth noting a kind of canonical example. Take $S$ to be the convolution algebra of a (discrete) non-amenable group $G$, take $\Vert\cdot\Vert_1$ to be the norm induced by the left regular representation of $S$ acting on $\ell^2(G)$, and take $\Vert\cdot\Vert_2$ to be the full group $C^*$-norm $\endgroup$
    – Yemon Choi
    Oct 17, 2018 at 22:45

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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 $b \neq 0$.

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  • $\begingroup$ This is a nice answer! $\endgroup$
    – Nik Weaver
    Oct 17, 2018 at 15:45
  • $\begingroup$ @NikWeaver This is something I struggled a long time to convince myself about when I first learnt that completion does not preserve injectivity. $\endgroup$ Oct 17, 2018 at 15:46
  • $\begingroup$ @Robert do you mean injectivity of functions or of algebras? $\endgroup$
    – David Roberts
    Oct 17, 2018 at 20:45
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    $\begingroup$ @DavidRoberts Now you mention it, that is ambiguous. I meant functions. In this case it is that the completion functor applied to the identity map $(S,\|\cdot\|_2) \rightarrow (S,\|\cdot\|_1)$ produces a non-injective map $B \rightarrow A$ (in fact, it is the quotient map for the ideal of functions vanishing on $[0,1]$). $\endgroup$ Oct 17, 2018 at 20:54

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