# Sz.-Nagy dilation for uniformly convex Banach spaces

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. $c\|x\| \leq \|T^n x\| \leq C\|x\|$).

• If you can prove it for all finite dimensional spaces with a given common modulus of uniform convexity with the dilation spaces all having a (possibly different) common modulus of uniform convexity, then you get the same result for all spaces that have the approximation property and have that given common modulus of uniform convexity. This follows from a standard Banach space ultraproduct argument (basically the same one I used to extend the finite dimensional Akcoglu-Sucheston theorem, which they already knew, to the infinite dimensional setting).... – Bill Johnson May 4 '13 at 20:10
• The point is that, modulo abstract nonsense, for spaces that possess the approximation property, all the difficulty is in the finite dimensional version of the problem. BTW, I doubt this has a positive answer even in the setting of $L_p$ spaces. – Bill Johnson May 4 '13 at 20:16

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

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.

• Cédric, thanks! I'll definitely look into this more. To be clear, in your definition of loose dilation, the isomorphism U must be power bounded from above, but can it also be power bounded from below ($\|U^n\| > C$ for all $n$)? Also, I am not sure if it is too late, but I noticed a typo in the PDF in Theorem 4.5. The second "is" should be an "if". – Jason Rute May 5 '13 at 11:14
• It seems to me that it is obvious. Indeed, you also have $||U^n||\leq C$ for negative integers. Thank you for the typo. – user33709 May 5 '13 at 15:14
• Cédric, thank you. I didn't notice the $\mathbb{Z}$. – Jason Rute May 5 '13 at 19:43

I would give this partial answer as a comment but it seems that I do not have earned enough credit points yet.

I do not know whether such a general dilation theorem holds. But on $L^p$-spaces there is an analogue for positive, contractive operators: the Akcoglu-Sucheston dilation theorem. I personally like the presentation in On Dilations and Transference for Continuous One-Parameter Semigroups of Positive Contractions on $\mathcal{L}^p$-spaces by G. Fendler and the lattice-theoretic approach of R. Nagel & G. Palm.

• Now we have the remarkable dilation theorem proved by you on general Banach spaces with mild assumptions. – A beginner mathmatician Jun 16 at 10:17