Let $A$ and $B$ be $C^\*$-algebras and let $A \otimes B$ their minimal tensor product and $M(A \otimes B)$ the associated multiplier algebra.


On $M(A \otimes B)$, we consider the strict topology which is the locally convex topology generated by the seminorms
$$M(A\otimes B)\ni x\mapsto \|x c\| $$
$$M(A \otimes B)\ni x\mapsto \|cx\| $$
for all $c \in A\otimes B$. This topology is weaker than the norm-topology on $M(A \otimes B)$.


It is easy to prove that the algebraic tensor product $A \odot B$ is strictly dense in $M(A \otimes B)$: simply observe that $A \odot B$ is norm-dense in $A \otimes B$ and $A \otimes B$ is strictly dense in $M(A \otimes B)$.



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> **Question**: Is the unit ball of $A \odot B$ dense in the unit ball of $M(A \otimes B)$?
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Is the unit ball of $A \odot B$ strictly dense in that of $M(A \otimes B)$?
Yes. This follows from the strict topology version of the Kaplansky Density Theorem (the proof is much the same as the usual proof of Kaplansky Density). See for example Proposition 1.4 in Lance's book about Hilbert $C^\*$-modules. In multiplier algebra language it says:


**Theorem:** Let $A$ be a $C^\*$-algebra and let $B$ be a strictly dense $C^\*$-subalgebra of $M(A)$. Then the unit ball of $B$ is strictly dense in the unit ball of $M(A)$.


It is easy to see that $B$ need not be closed, as in your example.


