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Let $\cal A$ be a Banach algebra, $\cal U$ be a free ultrafilter, and $\phi$ be a character. Let ${(w_{\alpha})}_{\alpha}$ be a net in $(\cal A)_{\cal U}$, and suppose that for every $(a_i)\in (\cal A)_{\cal U}$ we have $$\lim_{\alpha}\|a_i w_{\alpha}-\phi(a_i)w_{\alpha}\|=0,$$ so$$\lim_{\cal U} \lim_{\alpha}\|a_i w_{\alpha}-\phi(a_i)w_{\alpha}\|=0.$$

When do we have that $$\lim_{\alpha}\lim_{\cal U}\|a_i w_{\alpha}-\phi(a_i)w_{\alpha}\|=0 ?$$

When can we "displace" an ultrafilter limit with another limit?

Thank you so much!

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Doing ultra-character amenability, are we? – Yemon Choi Jun 27 '13 at 6:36
What reason do you have to believe that one can interchange limits? In general, in analysis, one cannot do this, so you should give some evidence or special cases which show why this might be true. – Yemon Choi Jun 27 '13 at 6:38
Before asking a question on MO, you should demonstrate that you have made a serious attempt to answer the question on your own. We are not in the business of writing people's master's theses for them – Yemon Choi Jun 27 '13 at 7:43
By the way, if you used your real name, I would be more willing to help with what is clearly an attempt to write a paper/thesis on ultra-character amenability. Hiding behind a pseudonym does you no favours in this particular case; I don't see why we should do the hard work and risk not being credited in the final product. – Yemon Choi Jun 27 '13 at 18:27
Yes, I want to study ultra $\phi$-amenability and ultra character amenability. I want to show that if $\phi\in \Delta_{\cal A}$, and $\cal A$ is $\phi$-amenable, then $(\cal A)_{\cal U}$ is $(\phi)_{\cal U}$-amenable, for every ultrafilter $\cal U$. But I confront to interchanging limits. – Albert harold Jun 27 '13 at 20:25
up vote 2 down vote accepted

With hindsight, this question is really too elementary for MO, in my opinion. However, since I am not on MSE, here is a sketch of what I think is a counter-example (but I think there is some merit to the argument that questions at this level should not be encouraged on MO). Also, I will not be happy if I learn later that this example is used in a paper without proper citation.

Note that the counter-example has nothing to do with ultraproducts, as I commented above you are trying to interchange limits and this hope is unfortunately far too naive.

$\newcommand{\norm}[1]{{\Vert#1\Vert}}$ Let $A=C[0,1]$ with usual multiplication and the supremum norm. Let $\phi\in\Delta_A$ be the character defined by $\phi(f)=f(0)$.

Let $(w_n) \subset A$ be a bounded sequence with the following properties: $w_n(0)=1=\norm{w_i}$ and $\operatorname{supp}(w_n)\subseteq [0,1/n]$ for all $n\geq 1$. (For instance $w_n(t) = \max(0,1-nt)$ would do.)

Claim 1: for any $a\in A$, $\lim_n\norm{aw_n - \phi(a)w_n} = 0$.

Now let $(a_k)\subset A$ be any bounded sequence with the following properties: for each $k\geq 1$, we have $a_k(0)=0$ and $a_k(t)=1=\norm{a_k}$ for all $t\in [1/k, 1]$. (For instance, $a_k(t) = \min(1, kt)$ would do.)

Claim 2: for each $n$, $\lim_k \norm{a_k w_n - \phi(a_k)w_n} = 1$.

Since $\lim_{k\in \mathcal U} \equiv \lim_k$ for any sequence of convergent complex numbers, this shows that the interchange of limits which you desire, is false even for a very well-behaved commutative Banach algebra. I leave the proofs of the two claims to you, the crucial point is of course that elements of $A$ are continuous functions on $[0,1]$.

On the other hand, note that every ultrapower of $C[0,1]$ is a commutative $C^*$-algebra, hence is amenable. So the result you actually want to prove, concerning $\phi_{\mathcal U}$-amenability of $A_{\mathcal U}$, may be true for some special cases of $A$ - it might even be true for all $A$, or it might not - but you have to come up with a better argument than the one you suggest in your question.

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Is the rambling really necessary? – Michael Greinecker Jun 28 '13 at 0:31
@MichaelGreinecker perhaps not, but the injunctions are, in my opinion. I have a good idea who the OP is, and that their level of questions will not be appropriate to MO. But email me if you would like to know the backstory. – Yemon Choi Jun 28 '13 at 0:41
I will. But first I need some sleep. – Michael Greinecker Jun 28 '13 at 0:42
@MichaelGreinecker OK - I have reworded some of the more subjective stuff at the beginning – Yemon Choi Jun 28 '13 at 0:45

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