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Do complex-normed spaces exist? Is there an extension of $p$-norms to $p\in\Bbb C\setminus\Bbb R$?

A while ago I thought of extending $L^p$-spaces to the complex-normed setting. After some discussions, we came up with variations involving $$\int_S|f|^{|p|}\arg f^p\,d\mu$$ and exponentiated versions. The argument would convey a sense of "spin", however the choice of terms seems a bit arbitrary and there wasn't any natural interpretation of it. Note that, for this specific variation, $p=i$ would get us very close to the Shannon entropy.

The only information I could find on this extension was an unresolved MSE question.

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    $\begingroup$ Since you mention "spin": is it possible that you're not necessarily/exactly wanting a "norm" with complex index, but, rather, computing a (Fourier-like) component with respect to a (compact?) group action? $\endgroup$ Commented Jun 16, 2023 at 19:08
  • $\begingroup$ @paulgarrett It's probably not the best word choice, but I am interested in extending the norm to a complex index. Also, that was just one variant we considered. I did a fair bit of searching but was unable to find anything on this topic. $\endgroup$ Commented Jun 16, 2023 at 19:23
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    $\begingroup$ For what it's worth, I can confirm that I myself have absolutely never seen a literal complex-indexed "norm"... $\endgroup$ Commented Jun 16, 2023 at 19:24
  • $\begingroup$ Perhaps it makes sense to start with Hölder's inequality. What would it say for complex $p,q$ and some simple functions? $\endgroup$ Commented Jun 16, 2023 at 19:33
  • $\begingroup$ If there are some things about the L^p spaces that vary real-analytically with p, then one pathway might be first to analytically continue those things to a complex neighbourhood of (0,oo), and then to infer from that how the p in L^p might be non-real. [Written before reading the answer by Dmitri Pavlov.] $\endgroup$ Commented Jun 17, 2023 at 2:33

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Not only do $\def\L{{\sf L}}\L^p$-spaces make sense for all complex $p$, but their noncommutative generalizations play a crucial role in the Tomita–Takesaki modular theory.

One construction is described in the answer to the question Is there an introduction to probability theory from a structuralist/categorical perspective?.

The input data is a measurable space $(X,M)$ together with a σ-ideal $N$ of negligible (alias measure 0) sets. This is less data than a traditional measure space, since $\L^p$-spaces do not depend on the choice of a measure, only on the σ-ideal of negligible sets.

Below, it is convenient to replace $p$ by its reciprocal $1/p$. Thus, what used to be $\L^2$ is now denoted by $\L^{1/2}$ and $\L^∞$ is denoted by $\L^0$.

Now the space $\L^p(X,M,N)$ can be constructed as the quotient of the set of pairs $(f,μ)$ by an equivalence relation $\sim$. Here $f$ is an equivalence class of bounded measurable functions on $(X,M,N)$ (modulo equality almost everywhere), $μ$ is a measure on $(X,M,N)$ (required to vanish on elements of $N$), and $μ$ is required to be finite if $\Re p>0$. The equivalence relation is generated by $$(fg^p,μ)\sim (f,gμ),$$ where $g$ is an equivalence class of a strictly positive real measurable function on $(X,M,N)$ and $gμ$ denotes the multiplication of a measure by a function, as in the Radon–Nikodym theorem.

Equivalently, if $\Re p>0$, we can drop the boundedness condition on $f$ and finiteness condition on $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This construction produces the same $\L^p$-space because $$(f,μ)=(\arg f\cdot (|f|^{1/\Re p})^{-\Im p}(|f|^{1/\Re p})^p,μ)\sim(\arg f\cdot (|f|^{1/\Re p})^{-\Im p},|f|^{1/\Re p}μ),$$ and in the last pair, the first component has absolute value at most 1 (hence is bounded), whereas $|f|^{1/\Re p}μ$ is a finite measure. Here and below we write $p=\Re p+\Im p$ for the real and purely imaginary parts of $p$. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$ observe that for any faithful measure $μ$ the map $$f↦(f,μ)$$ yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$ where the right side does not depend on the choice of $μ$.

As a special case, $\L^0(X,M,N)$ is the commutative von Neumann algebra of equivalence classes of bounded measurable functions on $(X,M,N)$, whereas $\L^1(X,M,N)$ is the Banach space of finite complex-valued measures on $(X,M,N)$.

The condition $f≥0$ singles out the positive subspace $\L^p_{≥0}⊂\L^p$, and the map $$μ↦(1,μ),\qquad \L^1_{≥0}→\L^p_{≥0}$$ is a bijection. Here $\L^1_{≥0}$ can be identified with finite (positive real) measures on $(X,M,N)$. Furthermore, given a complex number $q$, we can define the power operation $$\L^p_{≥0}(X,M,N)→\L^{pq}(X,M,N),\qquad (1,μ)↦(1,μ),$$ using which one can show that $(f,μ)=fμ^p$, justifying the above construction and the equivalence relation $\sim$.

The Hölder inequality yields a canonical multiplication map $$\L^p(X,M,N)⊗\L^q(X,M,N)→\L^{p+q}(X,M,N)$$ given by $$((f,μ),(g,μ))↦(fg,μ),$$ where we used the equivalence relation $\sim$ to pick representatives for elements of $\L^p$ and $\L^q$ with the same measure $μ$ as the second component. The induced map $$\L^p(X,M,N)⊗_{\L^0(X,M,N)}\L^q(X,M,N)→\L^{p+q}(X,M,N)$$ is an isomorphism, where $⊗$ denotes the usual algebraic tensor product, which happens to be automatically complete.

Thus, $$\L^p(X,M,N)≅\L^{\Re p}(X,M,N)⊗_{\L^0(X,M,N)}\L^{\Im p}(X,M,N),$$ so the study of $\L^p$-spaces for complex $p$ reduces to the case of real $p$, which is well-known, and the case of purely imaginary $p$.

For a purely imaginary $p$, given a choice of a faithful measure $μ$, the map $$\L^0(X,M,N)→\L^p(X,M,N),\qquad f↦fμ^p$$ is an isomorphism, with its inverse being the map $f↦fμ^{-p}$. Thus, for traditional measure spaces, we do not see something radically new, although one can find some satisfaction in the fact that the above constructions of $\L^p$-spaces are independent of the choice of a measure, whereas the isomorphism $\L^0→\L^p$ requires such a noncanonical choice, so from a conceptual point of view, there is some value in $\L^p$-spaces for purely imaginary $p$.

All of the above generalizes to the case of noncommutative von Neumann algebras. For a purely imaginary $p$, the isomorphism $\L^0→\L^p$ is now only an isomorphism of left modules over a (noncommutative) von Neumann algebra. Indeed, the bimodules $\L^0$ and $\L^p$ are not isomorphic for a type III factor. This essentially the Tomita–Takesaki modular theory of von Neumann algebra, see the answer to Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"? for more information.

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  • $\begingroup$ In the "observe for any faithful finite measure... $f\mapsto(f,\mu)$ is an isomorphism" I don't believe this is well-defined. For $f$ in the traditional $L^p$ or $L^{1/p}$ space needn't be (equivalent to) a bounded measurable function, which you took to be a hypothesis in the construction of this new $\mathsf{L}^p$ space $\endgroup$
    – FShrike
    Commented Jun 16, 2023 at 20:08
  • $\begingroup$ Also what algebra structure / norm structure do you impose on this new $\mathsf{L}^p$ space? You refer to the algebraic tensor products but, for pure imaginary $q$ (say) we don't have a traditional $L^q$ space to refer back to. At first I thought you were using the isomorphism $L^p\cong\mathsf{L}^p$ (isomorphism of sets?) to get some norm/algebra structure on $\mathsf{L}^p$ but I can't see what you're doing for $q=i$ (say), a pure imaginary $\endgroup$
    – FShrike
    Commented Jun 16, 2023 at 20:13
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    $\begingroup$ @FShrike: Concerning the first comment, I added a paragraph explaining this. I also added the definition of the multiplication map and the L^p-norm. $\endgroup$ Commented Jun 16, 2023 at 20:29
  • $\begingroup$ Thank you for the clarifications. I like the new definition in terms of a quotient by pairs $(f;\mu)$ where $|f|^{1/\Re p}$ is $\mu$-integrable. Though I dont understand how your original formulation is equivalent to this new definition, I see no reason why that would be true as we still have a problem with functions not being replaceable with ae bounded functions (and it is not true in general that $g\mu$ is still finite and $fg^p$ is still bounded). I still feel the old version is not well defined (this is not critical, I’m a student after all, just trying to see how it could make sense) $\endgroup$
    – FShrike
    Commented Jun 16, 2023 at 21:16
  • $\begingroup$ @FShrike: I added details for the equivalence of two constructions. $\endgroup$ Commented Jun 16, 2023 at 21:39

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