# Sequence in *-algebra with different limits for two C*-norms?

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 $$b=0$$?

• 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 Oct 17, 2018 at 22:45

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 $$b \neq 0$$.
• @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 $[0,1]$). Oct 17, 2018 at 20:54