Is the left regular representation of an algebra, always faithful?

Let $\mathcal{A}$ be a unital associative algebra with a countable basis $\mathcal{b}$ over $\mathbb{C}$.
Let $H=l^2(b)$ be the Hilbert space generated by $\mathcal{b}$.
Let $H_0 = \{v \in H \ \vert \ a.v \in H \ \forall a \in \mathcal{A} \}$ and $\rho$ the left regular representation of $\mathcal{A}$ on $H_0$.

Question: Is $H_0$ a dense subspace of $H$ and $\rho$ faithful? Else what are the first counter-examples?
Also what would be the (minimal) additional assumptions for having $H_0$ dense and $\rho$ faithful?

Remark: the problem seems reduce to the type of closure on $\mathcal{A}$ (see the comments below).

Remark: for the Heisenberg algebra $\mathcal{A} = \langle a,b \ \vert \ [a,b]=1 \rangle$, then $H_0$ is dense and $\rho$ faithful, whereas $\mathcal{A}$ can't have a Banach structure (see here).

• Unless I'm misunderstanding your definitions, it seems obvious that $H_0$ contains all sequences of finite support and thus the answer is trivially yes. – Eric Wofsey Dec 7 '14 at 22:00
• @EricWofsey: Why $H_0$ should contain all sequences of finite support? This means: $\forall v \in b$ and $\forall a \in \mathcal{A}$, $a.v$ should be square-summable regarding $b$, why? – Sebastien Palcoux Dec 8 '14 at 3:25
• I was assuming $b$ was a basis for $\mathcal{A}$ in the algebraic sense, so every element of $\mathcal{A}$ has finite support over $b$. If this isn't what you meant, you should clarify what additional structure $\mathcal{A}$ has that makes a different definition of basis meaningful. – Eric Wofsey Dec 8 '14 at 8:17
• @EricWofsey: You're right, if we add the assumption "every element of $\mathcal{A}$ has finite support over $b$" then $\mathcal{A} = <b>_{fin}$ and $H_0$ is dense. This assumption is not the minimal one; a weak assumption could be "every element of $\mathcal{A}$ is square-summable over $b$", but it's also not minimal: for example if we take $\mathcal{A} = B(V)$ with $V$ an infinite dimensional Hilbert space, and $b$ the canonical basis $(e_{ij})$, the identity operator itself is not square-summable over $b$. The problem seems to be around the topology closure of $\mathcal{A}$. – Sebastien Palcoux Dec 8 '14 at 9:39