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I think most physicists mostly model physical systems as some kind of Hilbert space.

Hilbert spaces are a strict subset of Banach spaces.


  • Can physical systems really have non-compact topologies, as a Banach space has?

  • Does anyone have an example of physics which requires a physical space which is Banach and not Hilbert?

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If you look close enough, you can associate Banach spaces to virtually every physical system. –  Qmechanic Jan 3 '13 at 21:36

3 Answers 3

Sobelev spaces (in which the norm of a function is a weighted sum of $L^p$ norms of the function and its derivatives) are Banach spaces, and they do come up in the analysis of PDE. Some physicists might regard the use of such spaces as intrinsically mathematics rather than physics.

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I learned about Sobelev spaces briefly in a functional analysis class years ago, but I'm no expert. I seem to remember, however, that although Sobelev spaces in general are not equal to $L_2$, they are still Hilbert spaces with some special inner product definition...maybe not all Sobolev spaces and just a subset? I don't recall exactly... –  daaxix Jan 3 '13 at 4:45
@daaxix: Just a subset. Many Sobelev spaces are not Hilbert. –  user1504 Jan 3 '13 at 14:59
But often one restricts oneself to the Sobolev spaces $W^{k, 2} = \{ \psi \in L^2 | D^\alpha \psi \in L^2 \text{ for } |\alpha| < k \}$, which for all k are Hilbert! –  Tobias Diez Jan 3 '13 at 21:26
Yes, in the special case $p=2$, $W^{k,p}$ is Hilbert. Otherwise not iirc. For theories with interactions in their Lagrangians, the higher powers need to be integrable as well. –  user1504 Jan 3 '13 at 22:59

(Classical) field theory lives naturally in more general spaces than Hilbert (and even more general than Banach). The space of smooth sections of a fiber bundle is a Fréchet manifold (if the basespace is compact). If this fiber bundle thing is new to you, you can consider your fields just as smooth functions $\psi: \text{spacetime} \rightarrow R$. In field theory you want to say something about the behavior of any field, so you are more or less forced to look at the space $C^\infty(M)$ (which is neither a Hilbert- nor a Banachspace)

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But is the $C^\infty(M)$ the structure of the space of all fields, or an actual physical space? –  daaxix Jan 9 '13 at 16:04
It is also very interesting that the set of infinitely differentiable functions is not a Banach space. This includes functions that are unbounded (and not integrable), however, correct? –  daaxix Jan 9 '13 at 16:09
@daaxix: You can require that the smooth functions vanish exponentially fast at infinity. The natural topology on this vector space of smooth functions is the nuclear topology, which is never Banach. –  user1504 Jan 9 '13 at 16:55
As far as I know, there are many useful topologies on $C^\infty (M)$ (e.g. uniform convergence on compacta of all partial derivatives), but none of these are Banach. Note, that $C^k$ for finite $k$ is a Banachspace (for compact M?). Futhermore, you are right, there are no restrictions on the behavior of the functions at infinity. –  Tobias Diez Jan 9 '13 at 17:28
Yes, I had path integral applications in mind. I should have been more clear about that. Sorry! –  user1504 Jan 9 '13 at 18:52

Hilbert spaces are occur everywhere where the Lagrangian\Hamiltonian is quadratic in derivatives. If the Lagrangian is non-quadratic then the Hilbert spaces are no longer so convenient. In particular in analysis of Navier–Stokes equations the Banach spaces(not Hilbert spaces) are active used.

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And C*-algebra is used quite often. –  user_user Jan 20 '13 at 16:48
What do you mean "Can physical systems really have non-compact topologies?" –  user_user Jan 20 '13 at 19:17

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