TITLE: Sz.-Nagy dilation for uniformly convex Banach spaces
QUESTION [5 upvotes]: The Sz.-Nagy dilation theorem says that for a Hilbert space $H$ with nonexpansive operator $T$, there is a larger space $H'$ containing $H$ and a unitary operator $U$ on $H'$ such that for all $x \in H$ and all $n$, $T^n x  = P U^n x$ where $P$ is the projection from $H'$ to $H$.

Is there anything analogous to the Sz.-Nagy dilation theorem except for uniformly convex Banach spaces?


A few more details:
I have a result for linear isometries on uniformly convex Banach spaces with "power type" modulus of convexity, $\eta(\varepsilon) = C \varepsilon^p$. I want to extend the result to linear nonexpansive operators on the same class of spaces.  I need the larger space to be uniformly convex, ideally with the same power type.
It would be a bonus if this also holds for power-bounded operators $T$ (i.e. $\|T^n x\| \leq C\|x\|$).  I don't actually need the other map to be an isometry, just something that is power bounded from above and below (i.e. $c\|x\| \leq \|T^n x\| \leq C\|x\|$).

REPLY [6 votes]: The Akcoglu-Sucheston-Peller dilation theorem gives indeed a characterization of operators on a $L^p$-space with an isometric dilation on a $L^p$-space. These operators are the contractively regular operators. And it is well-known that we can find contractive operators without this property (some 2x2 matrix on $\ell^p_2$ with $p\not=$ 1,2,$\infty$)
Moreover, you cannot obtain your "bonus". Indeed, the existence of an isometric dilation fo $T$ imply that $T$ is a contraction. However, it seems to me that you need the notion of "loose dilation" with a power-bounded isomorphism. See the paper "Dilation of Ritt operators on Lp-spaces" on
https://sites.google.com/site/cedricarhancet/publications-1
With the methods of the paper, it is not very difficult to obtain a dilation result for $R$-Ritt operators (a nice class of power-bounded operators) on UMD spaces. I warn you that there exist power-bounded operators on a $L^p$-space without loose dilation on a $L^p$-space.
Finally, if you really need a isometry instead of a power-bounded isomorphism, it is possible with a ultraproduct argument.