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Let $L$ be an $L^p$-space, $1

I would like to know if the ultraproduct $W$ of the $L_\alpha$ with respect to $\mathcal U$ is isometric isomorphic to $L$ itself. If $A$ contains a largest element this is true of course, but my intuition says it should be also true otherwise.

Here is what I have tried:

Let $\phi: L\to W, f\mapsto (E_\alpha f)_\mathcal U$, where $(x_\alpha)_\mathcal U$ denotes the equivalence class of $(x_\alpha)_{\alpha\in A}$ with respect to $\mathcal U$.

The mapping is isometric: $\|(E_\alpha f)_\mathcal U\| = \lim_\mathcal U\|E_\alpha f\|$. Since $\|E_\alpha f\|$ net-converges to $\|f\|$ this is also true for the ultrafilter limit due to the property we demanded of our ultrafilter.

I have problems with the surjectivity. Let $(f_\alpha)_\mathcal U\in W$ be arbitrary. My idea was to define a functional on $\varphi$ on $L^q(X)$ via $\varphi(g) = \lim_\mathcal U\int_{L_\alpha}(E_\alpha g)f_\alpha$. The limit exists because $\int_{L_\alpha}(E_\alpha g)f_\alpha$ is bounded. This also shows that $\varphi$ is bounded, hence there is $f\in L$ which induces $\varphi$. I think that we should have $(E_\alpha f)_\mathcal U = (f_\alpha)_\mathcal U$, but I can't show that $\lim_\mathcal U\|E_\alpha f-f_\alpha\| = 0$. If $(f_\alpha)_\mathcal U$ is already of the form $(E_\alpha f)_\mathcal U$ then this construction with the functional indeed gives back $f$, but in the general case I don't see it. Can anybody help me?

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    It is difficult to see that "this makes A a directed set for any measure space $X$"? – 2017-01-27
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    @Zouba No, this is not difficult to see. If you have two semi-partitions $\alpha, \beta$ then look at arbitrary intersections of sets in $\alpha$ and $\beta$. This is no problem because there are only finitely many of them. The semi-partition given by those intersections is an upper bound for $\alpha,\beta$. – 2017-01-29
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    The text is a part of a paper of Akcoglu/Sucheston. I read the paper some years ago. Out of curiosity, what is your Goal ? – 2017-01-30
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    @Zouba As you know, Akcoglu/Sucheston construct a dilation of a positive contraction $T$ on some $L^p$-space $L$ in this paper. We get a larger $L^p$-space $B$, a positive invertible isometry $Q:B\to B$, an isometric lattice embedding $D:L\to B$ and a positive projection $Q:B\to B$ such that $DT^n = PQ^n D$ for all $n\in\mathbb N$. This is proven for finite-dimensional $L$ first and then generalized via ultraproducts. However, in the finite-dimensional case, there is a much nicer version of the dilation, where we get $T^n = PQ^nD$ and $P:B\to L$ is a positive contraction instead. – 2017-01-30
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    I does not know if this is useful but if you have a ultrapower (L^p)^U, we can consider the conditional expectation $E : (L^p)^U \to L^p$, $(x_i)^U \mapsto weak-lim_{i \to U} x_i$. – 2017-01-30
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    I will try if this helps. Thanks for the hint. @Zouba – 2017-01-30
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    If you write a paper or a master thesis on this subject, let me know when you finish your work. This subject interests me. – 2017-01-31
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    I think about your problem, I believe that it is not possible. Moreover, I am a bit surprised by the fact that the isometric dilation on B is not sufficient for your goal. Out of curiosity, do you know a result on isometries which cannot be used with Ackoglu dilation on L? I would like understand your problem. – 2017-02-04

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I think that the answer is NO since an ultraproduct of $L^p$-spaces $L^p(\Omega_i)$ is an $L^p$-space on a measure space $\Omega^{\mathcal{U}}$ and such a measure space is a "bigger" measure space and generally no $\sigma$-finite even if the $\Omega_i$'s are $\sigma$-finite.