Let $A$ be an infinite dimensional faithful Banach algebra and let $\mathcal U$ be a free ultrafilter. Is the ultrapower $(A)_{\mathcal U}$ faithful?

(This merely expands on my comment above.) Let $A$ be a Banach algebra and let $L: A\to {\mathcal B}(A)$ be defined by $L(a)(x)=ax$. The map $L$ is a normdecreasing homomorphism but it need not have closed range. (For instance, take $A=\ell^p$ with pointwise multiplication, for any $1\leq p <\infty$.) So, let us suppose $L$ does not have closed range . Then there is a sequence $(a_n)$ in $A$ such that $\Vert a_n\Vert=1$ for all $n$ while $\Vert L(a_n) \Vert \to 0$. Let $[a]$ be the element of the ultrapower that corresponds to the sequence $(a_n)$. Then clearly $[a]$ is a nonzero, left annihilator of any element of the ultrapower. 

