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In Banach space theory, a super-property is a property of a Banach space that is preserved under ultrapowers. (Update (2015-09-28): The property must also be closed under isometric embeddings.) (Super-properties are also characterized--more commonly?--via finite representability).

What is the name for a property of a Banach space preserved under ultraproducts.

Update 3 (2015-09-28): Given that a super-property must also be closed under isometric embeddings (Thanks Bill Johnson!), I should update my question to the following version.

What is the name for a property of a Banach space preserved under ultraproducts and isometric embeddings.

My examples below satisfy this additional requirement.


For example, a Banach space $B$ is super-reflexive iff it is reflexive and all its ultrapowers are reflexive. However, super-reflexitivity is not preserved under ultraproducts. Indeed, the spaces $L^p$ for $1<p<\infty$ are super-reflexive, but if $\mathcal{U}$ is a non-principle ultrafilter on $\mathbb{N}$, then the ultraproduct $(\prod_{n=2}^\infty L^n)/\mathcal{U}$ is not reflexive.

Every super-reflexive space is isomorphic (in the Banach space sense) to a $q$-uniformly convex space for $2\leq q <\infty$. Such a space is said to have martingale cotype $q$. The property of having martingale cotype $q$ is preserved under ultraproducts (for a fixed $q$).

Update 1 (2015-09-27): As Bill Johnson pointed out, having $q$-martingale cotype is a bad example, since it is not actually preserved under ultraproducts.

For a better example, being a uniformly convex Banach space with modulus $\delta(\varepsilon)$ is preserved under ultraproducts. [Proof: Let $\{B_i\}_{i\in I}$ be uniformly convex spaces with common modulus $\delta(\varepsilon)$. Let $x=(x_i)_\mathcal{U}$ and $y=(y_i)_\mathcal{U}$ be elements in the interior of the unit ball of the ultraproduct $B = \prod_{i\in I} B_i/\mathcal{U}$ with $\|x-y\|_B>\varepsilon$. Then $\mathcal{U}$-a.s. $\|x_i\|_{B_i}, \|y_i\|_{B_i} \leq 1$ and $\|x_i - y_i\|_{B_i} \geq \varepsilon$. By uniform convexity, $\|(x + y)/2\|_B = \lim_\mathcal{U} \|(x_i + y_i)/2\|_{B_i} \leq 1-\delta(\varepsilon)$.]

For another example, a space $B$ is of martingale cotype $q$ iff it is isomorphic to a Banach space with modulus of uniform convexity $\delta(\varepsilon) = C \varepsilon^q$. In other words, there are constants $C_1$ and $C_2$ and a new norm $\| \|_0$ such that $(1/C_1) \|x\|_0 \leq \|x\|_B \leq C_1 \|x\|_0$ and $(B,\|\|_0)$ is uniformly convex with modulus $\delta(\varepsilon) = C_2 \varepsilon^q$. Call such a $B$ a space of martingale cotype $(C_1,C_2,q)$. Unlike martingale cotype $q$, martingale cotype $(C_1,C_2,q)$ is preserved by ultraproducts. [The proof is similar to the previous one.]

Update 2 (2015-09-27): The reason I am interested in this is because such properties have a lot of uniformity associated with them. For example, Avigad and I gave a variational norm on the mean ergodic theorem which holds for every Banach space of martingale cotype $(C_1,C_2,q)$. Our inequality only depends on $C_1$, $C_2$ and $q$.

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    $\begingroup$ Your last sentence is not correct. Finite dimensional spaces have cotype $q$ for all $q$, and every Banach space embeds isometrically into an ultraproduct of finite dimensional spaces. That explains why there is no name for properties that are preserved under arbitrary ultraproducts. Finite inequalities are approximately preserved under ultraproducts; see papers by Ward Henson to see how this elementary observation leads to a model theoretic characterization. $\endgroup$ Commented Sep 27, 2015 at 15:27
  • $\begingroup$ @BillJohnson, thanks for the correction! I spoke too soon. There is a constant $C$ that I forgot to account for. When that is accounted for, this property is preserved by ultraproducts. I fixed my question accordingly. However, I don't see why your example "explains why there is no name for properties that are preserved under arbitrary ultraproducts." Could you please elaborate? $\endgroup$
    – Jason Rute
    Commented Sep 27, 2015 at 20:20
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    $\begingroup$ Because any hereditary property possessed by all finite dimensional spaces and ultraproducts is possessed by all Banach spaces. Incidentally, James' original definition of super-Q was that every space that is finitely representable in such a space must have Q. In particular, the property must be hereditary. A space $Y$ is finitely representable in a space $X$ iff $Y$ embeds isometrically into some ultrapower of $X$. $\endgroup$ Commented Sep 27, 2015 at 21:43
  • $\begingroup$ Ok, @BillJohnson, thanks again! I see now that my initial characterization of super-property was missing the isometric embedding requirement. I've fixed that and also updated my question to include this requirement. (Notice, my two examples are closed under both ultraproducts and isometric embeddings.) $\endgroup$
    – Jason Rute
    Commented Sep 28, 2015 at 13:47

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