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

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Yes we can. Presumably you meant to ask: "is an ultrapower of a faithful Banach algebra always faithful?" and the answer is no. – Yemon Choi Oct 18 '12 at 19:20
You say "free ultrafilter", but probably intend to consider only ultrafilters concentrating on a countable index set, or at least non-$\sigma$-complete ultrafilters. After all, if there is a measurable cardinal, then one can have countably-complete ultrafilters, whose ultrapowers leave all small structures invariant, including your Banach algebra $A$, so that $A=(A)_U$. – Joel David Hamkins Oct 19 '12 at 2:07
up vote 7 down vote accepted

(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 norm-decreasing 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 non-zero, left annihilator of any element of the ultrapower.

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