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I've encountered this space in a book (see Proposition 1.2) and don't understand what is meant. It's a notation I only know from quotient spaces, but I can't make sense of that here.

0xbadf00d
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2 Answers2

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On page 11 in your link it is written that this is the space of all function orthogonal to the constant functions,

$$\{p\in L^2(\Omega) :\int_\Omega p\, dx = 0\}$$

Thomas
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  • This seems to be a very informal definition of a quotient space. What are the equivalence classes? My best guess is that it would be two functions are in the same equivalence class if they differ by a constant, so that you can always find one in the equivalence class aware that integral is 0. – Paul Sep 01 '16 at 16:01
  • @Paul I just copied what is written in the book. There they don't seem to view this as a quotient, but as a symbol denoting the set I wrote down (which is well defined). – Thomas Sep 01 '16 at 16:02
  • Yes. I'm just trying to understand the connection to a quotient space, to relate it to OP's previous knowledge. The notation is likely this way for a reason. – Paul Sep 01 '16 at 16:03
  • @Paul sure, look at the equivalence classes ${f + c:c \in \mathbb{R}}$ – Thomas Sep 01 '16 at 16:05
  • I didn't found that definition cause it occurs after its first usage. Thanks, Thomas. – 0xbadf00d Sep 01 '16 at 18:23
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Consider the Stokes (and similarly Navier-Stokes) problem in $\Omega$ \begin{equation} \begin{cases} -\mu\Delta u + \nabla p & = f,\\ \nabla \cdot u & = 0, \end{cases} \end{equation} with a presecribed boundary condition on $u$. For any solution $(u,p)$ of the problem and any $\bar{p} \in \mathbb{R}$, $(u, p+\bar{p})$ is also a solution. Therefore the solution (if exists) is unique up to an isomorphism in $L^2(\Omega)$. The solution space $L^2(\Omega)$ for pressure is thus decomposed into a direct sum $$L^2(\Omega) = \mathbb{R} \oplus (L^2(\Omega)/\mathbb{R}),$$ and the average-removed pressure is then uniquely determined in the space $$L^2(\Omega)/\mathbb{R} = \{q \in L^2(\Omega): (q,1)_{0,\Omega} = 0\}.$$

Empiricist
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  • While Thomas answer provides what I was looking for, I guess your answer is the main motivation for the introduction of $L^2(\Omega)/\mathbb R$ of the author of the referenced book. – 0xbadf00d Sep 01 '16 at 18:25