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$$\lim\_{\cal U} \lim\_{\alpha}\|a\_i w\_{\alpha}-\phi(a\_i)w\_{\alpha}\|=0.$$



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> When do we have that $$\lim\_{\alpha}\lim\_{\cal U}\|a\_i w\_{\alpha}-\phi(a\_i)w\_{\alpha}\|=0 ?$$
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> 


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


Thank you so much!



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




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$\newcommand{\norm}[1]{{\Vert#1\Vert}}$
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.


