Modern Fourier Analysis by Grafakos states in Proposition 2.3.1:
Let $0 < p,q < \infty$ and $\alpha \in \mathbb{R}$. The homogenous Triebel-Lizorkin space $\dot{F}^{\alpha,q}_p(\mathbb{R}^n) \ldots$ is complete.
Moreover, we further know that $\dot{F}^{s,2}_2 \approx \dot{H}^s$ so the above implies $\dot{H}^s(\mathbb{R}^n)$ is complete.
However, Fourier Analysis and Nonlinear PDE by Bahouri states in Proposition 1.34:
$\dot{H}^s(\mathbb{R}^d)$ is a Hilbert space if and only if $s < \frac{d}{2}$.
So where is the discrepancy? I don't think it's merely due to (pre-)Hilbertness. In fact the latter book proceeds by
If $s \geq d / 2$, observe that the function $$ N: u \longmapsto\|\widehat{u}\|_{L^1(B(0,1))}+\|u\|_{\dot{H}^s} $$ is a norm over $\dot{H}^s(\mathbb{R}^d)$ and that $(\dot{H}^s(\mathbb{R}^d), N)$ is a Banach space. Now, if $\dot{H}^s(\mathbb{R}^d)$ endowed with $\|\cdot\|_{\dot{H}^s}$ were also complete, then, according to Banach's theorem, there would exist a constant $C$ such that $N(u) \leq C\|u\|_{H^s}$...
So they are in-fact concluding $\dot{H}^s$ cannot be complete when $s \geq d/2$, in contradiction to the claims from Grafakos' book. So where is my line of reasoning going wrong?
Edit:
Here are the respective definitions each book uses: For Grafakos: $$ \dot{L}^{p}_{s} := \{u \in \mathcal{S}'/\mathcal{P} : ( \lvert \xi \rvert^s \hat{u})^\vee \in \dot{L}^p\} $$ where $\dot{L}^p := \{u \in \mathcal{S}'/\mathcal{P} : \exists Q \in \mathcal{P} : u + Q \in L^p\}$ where $ \mathcal{P}$ is the space of polynomials. Here $\dot{H}^s = \dot{L}^2_s$.
For Bahouri: $$ \dot{H}^s := \{u \in \mathcal{S}': \hat{u} \in L^1_{\rm loc}(\mathbb{R}^d)\, \land \, \lvert \xi \rvert^s \hat{u} \in L^2\}. $$
DISCLAIMER: This anwser despite being quite lengthy is far from being exhaustive: the topic is quite heavy and there ARE A LOT of technicalities involved here and there I won't adress or at best partially.
Essential points are written in bold and some times in CapsLock. Sorry, if it looks like agressive. I really want to emphasize on key points.
I] The various definitions of homogeneous Sobolev and Besov spaces over different (sub-)spaces of tempered distributions.
Before we start, I recall briefly that the main concern and issue about defining homogeneous function spaces over $\mathcal{S}'(\mathbb{R}^n)$ is that (semi-)norms like $$\lVert u \rVert_{\dot{\mathrm{H}}^{m,2}(\mathbb{R}^n)}:= \lVert \nabla^m u\rVert_{{\mathrm{L}}^{2}(\mathbb{R}^n)}$$ does not separate polynomials and that non-zero polynomials $P\in\mathcal{P}(\mathbb{R}^n)$ are fully localized on $0$ in the frequency side, in the sense that, in $\mathcal{S}'(\mathbb{R}^n)$, $$\mathcal{F}P = \sum_{|\alpha|=0}^k a_\alpha (-i)^{|\alpha|}\partial^\alpha\delta_0$$ where $P=\sum_{|\alpha|=0}^k a_\alpha x^\alpha$, and $\mathcal{F}$ is the Fourier transform.
Therefore, the same goes with homogeneous Besov and Triebel-Lizorkin (semi-)norms \begin{align*} \lVert u \rVert_{\dot{\mathrm{B}}^{s}_{p,q}(\mathbb{R}^n)} = \lVert (2^{js}\dot{\Delta}_j u)_{j\in\mathbb{Z}}\rVert_{\ell^q(\mathbb{Z},\mathrm{L}^p(\mathbb{R}^n))} \text{ and } \lVert u \rVert_{\dot{\mathrm{F}}^{s}_{p,q}(\mathbb{R}^n)} = \lVert (2^{js}\dot{\Delta}_j u)_{j\in\mathbb{Z}}\rVert_{\mathrm{L}^p(\mathbb{R}^n,\ell^q(\mathbb{Z}))}. \end{align*} where $(\dot{\Delta}_j)_{j\in\mathbb{Z}}$ is a homogeneous Littlewood-Paley decomposition.
The main goal is then to try to get rid of non-zero polynomials in the definition of the ambient space $\mathcal{S}'(\mathbb{R}^n)$. I will give the details of each main constructions so that one can compare them easily.
I will try to focus on the use of Bahouri-Chemin-Danchin and Grafakos' books to detail my answer but I will probably need additional and external references.
a) The standard construction of homogeneous function spaces $\dot{\mathrm{A}}^{s}_{p,q}(\mathbb{R}^n)$, $A\in\{B,F\}$, $s\in\mathbb{R}$, $p,q\in[1,+\infty]$ following [Gra14, Chapter 2, Section 2.2] :
The first natural idea is to consider the quotient structure $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ instead of $\mathcal{S}'(\mathbb{R}^n)$ which turns out to be good for many reasons such as :
I want to really emphasize that $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ is about equivalence classes of tempered distribution modulo polynomials. In particular, even for $u\in\mathcal{Z}(\mathbb{R}^n)$, one cannot distinguish in $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ the two elements $$u+P \text{ and } u+Q$$ for any $P,Q\in\mathcal{P}$.
b) Homogeneous Sobolev space over $\mathrm{L}^2$ : $\dot{\mathrm{H}}^{s}(\mathbb{R}^n)$, $s\in\mathbb{R}$ following [BCD11, Chapter 1, Section 1.3] :
Now, we introduce the homogeneous Sobolev space over $\mathrm{L}^2(\mathbb{R}^n)$, for $s\in\mathbb{R}$ : $$ \dot{\mathrm{H}}^{s}(\mathbb{R}^n):=\{\,u\in\mathcal{S}'(\mathbb{R}^n)\,|\,\mathcal{F}u \in\mathrm{L}^1_{\text{loc}}(\mathbb{R}^n)\text{ and } \int_{\mathbb{R}^n}|\xi|^{2s}|\mathcal{F}u(\xi)|^2\mathrm{d}\xi <+\infty\,\}. $$
c) Homogeneous Besov spaces $\dot{\mathrm{B}}^{s}_{p,q}(\mathbb{R}^n)$, $s\in\mathbb{R}$, $p,q\in[1,+\infty]$ following [BCD11, Chapter 2, Section 2.3] :
We introduce the subspace of tempered distributions \begin{align*} \mathcal{S}'_h(\mathbb{R}^n) &:= \left\{ u\in \mathcal{S}'(\mathbb{R}^n)\,\Big{|}\,\forall \Theta \in \mathrm{C}_c^\infty(\mathbb{R}^n),\, \left\lVert \Theta(\lambda D) u \right\rVert_{\mathrm{L}^\infty(\mathbb{R}^n)} \xrightarrow[\lambda\rightarrow+\infty]{} 0\right\}\text{,} \end{align*} where $\Theta(\lambda D)= \mathcal{F}^{-1}\Theta(\lambda \cdot)\mathcal{F}$. Notice that the condition could be strengthen (or weakened depending on the point of view) asking only for $\Theta$ to be such that $\Theta(0)=1$. And them again only asking for $\Theta(2^{-j} D) u$ to go to zero as $j$ goes to $-\infty$.
The point is that for any $P\in\mathcal{P}$, any $\Theta \in \mathrm{C}_c^\infty(\mathbb{R}^n)$ such that $\Theta(0)=1$, due to considerations on the support of their Fourier transform, we have for all $\lambda>0$ $$\Theta(\lambda D)P = P.$$ Asking then $\Theta(\lambda D)P$ to be in $\mathrm{L}^\infty$ implies that first $P$ must be a constant, but then the only polynomial that vanishes at infinity is the zero polynomial. Therefore $\mathcal{S}'_h(\mathbb{R}^n)$ does not contain any non-zero polynomial.
However, $\mathcal{S}'_h(\mathbb{R}^n)$ IS NOT isomorphic to $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ since $\mathcal{S}'_h(\mathbb{R}^n)\oplus\mathcal{P}\subsetneq \mathcal{S}'(\mathbb{R}^n)$. See [Cob21, Appendix B, Proposition B.1]. (We only have a canonical injection $\mathcal{S}'_h(\mathbb{R}^n)\hookrightarrow \mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$, and moreover, $\mathcal{S}'_h(\mathbb{R}^n)$ is a dense subspace of $\mathcal{S}'(\mathbb{R}^n)$).
With $\mathcal{S}'_h(\mathbb{R}^n)$ as an ambient space :
Notice that, with the two definitions available in [BCD11]:
Other definitions of $\mathcal{S}'_h(\mathbb{R}^n)$ are available, and in the complete case all the underlying definitions of homogeneous function spaces coincides and are equal. In this case, the definition given by $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ can also be identified. ${}{}$
II] Why different constructions ? Pro and cons for each one
One may wonder the reason about such variety of constructions for homogeneous function spaces, and moreover why accepting the loss of completeness is (maybe) a fair game instead of counter-intuitive non-sense. I will follow the recent exposure of the preprint [Gau22, Section 1] with additional comments.
The main issue is the following : The function spaces built over $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ are definitively not suitable for nonlinear Partial Differential Equation.
Indeed, for $[u]\in\mathcal{S}'(\mathbb{R}^n)$, and $u+P, u+Q\in\mathcal{S}'(\mathbb{R}^n)$ two representatives of $[u]$, we have \begin{align*} (u+P)^2-(u+Q)^2 = (P-Q)u +P^2-Q^2 \end{align*} Therefore, even if we have product law that makes sense, although $P^2-Q^2$ is a polynomial, this is not the case for $(P-Q)u$ so that the product $[u]^2$ depends on the choice of representatives! The problem remains even if $P$ and $Q$ were constants. Products are then not compatible with quotient structure (as well as composition with smooth maps, whenever one can make it meaningful.)
Here, this is really the ambiant structure that makes everything fail.Notice that moreover, if one wants to use this structure and choose a representative, then the given representative would depend on the function space.
More precisely, as an example, when $s'\geqslant n/2$, for $[u]\in\dot{\mathrm{H}}^{s}(\mathbb{R}^n)\cap \dot{\mathrm{H}}^{s'}(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$, one may have representatives $u_s\in \mathcal{S}'(\mathbb{R}^n)\cap\dot{\mathrm{H}}^{s}(\mathbb{R}^n)$ and $u_{s'}\in \mathcal{S}'(\mathbb{R}^n)\cap\dot{\mathrm{H}}^{s'}(\mathbb{R}^n)$ can be such that $$[u]=[u_s]=[u_{s'}] \text{ in } \mathcal{S}'(\mathbb{R}^n)/\mathcal{P} \text{, but } u_s\neq u_{s'}\text{ in } \mathcal{S}'(\mathbb{R}^n).$$ This happens even if $s<n/2$, the issue truly relies on $s'\geqslant n/2$. See [Bou88, Theorem 1].
From this point, using $\mathcal{S}'_h(\mathbb{R}^n)$ prevents us from such kind of issues since there would be only one consitent representative whatever would be the (semi-)norms. But the price to pay seems quite high : the loss of completeness.
But do we really have to lost completeness ? In fact one would think, "okay anyway let's perform the nonlinear estimates, and consider the completion of Schwartz function for norm and let's extend it by density". It seems that this approach is not really suitable either.
Proceeding this way won't give you a space made of distributions, not even elements of $\mathcal{D}'(\mathbb{R}^n)$. In fact, one can check that by choosing $\varphi\in\mathrm{C}^\infty_c(\mathbb{R}^n,[0,+\infty))$ such that $\int \varphi =1$, we have $$\lVert \varphi \rVert_{\dot{\mathrm{H}}^{-\frac{n}{2}}(\mathbb{R}^n)}=+\infty.$$
Therefore the (pre-)dual space of $\dot{\mathrm{H}}^{\frac{n}{2}}(\mathbb{R}^n)$ DOES NOT CONTAIN $\mathrm{C}^\infty_c(\mathbb{R}^n)$. This tells us that as a Hilbert space, this yields that any completion of Schwartz functions with respect to $\lVert\cdot\rVert_{\dot{\mathrm{H}}^{\frac{n}{2}}}$ CANNOT BE INCLUDED (THEN A FORTIORI, CANNOT BE EMBEDDED) in $\mathcal{D}'(\mathbb{R}^n)$.
Thereby, if one wants to deal with homogeneous function spaces one can only have two out of the following three properties:
(i) spaces whose elements are distributions, in a reasonable sense;
(ii) well-defined product laws;
(iii) all spaces are complete.
We have the following matches
Finally, few comments :
The major issue for complete realization arise from the morphism for the identification of function spaces built over $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ as subspaces of $\mathcal{S}'(\mathbb{R}^n)$. The morphism will actually depend on the regularity of the function spaces giving a new morphism for each interval $(n/p+k,n/p+k+1)$, $k\in\mathbb{N}$, see [Bou13, Theorem 4.2].
This also tells us that one cannot perform complex and real interpolation in a legitimate way between those two spaces with different regularity indices $s_0\in (n/p+k_0,n/p+k_0+1)$ and $s_1\in (n/p+k_1,n/p+k_1+1)$, $k_0\neq k_1$ : $$ \dot{\mathrm{B}}^{s_0}_{p,q_0} \text{ and } \dot{\mathrm{B}}^{s_1}_{p,q_1} $$ The function spaces cannot "directly talk" to each other : to do so they need to be embedded in the bigger ambient structure $\mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$.
About Interpolation theory : Real interpolation does not require completeness to be defined and still holds for $\mathcal{S}'_h$-realization as well as $\mathcal{S}'/\mathcal{P}$ (for all three constructions 1), 2 and 3) in fact). Complex interpolation requires completeness to be defined (since it deal with holomorphic function with values in normed vector spaces) then requires to deal with indices that makes every involved spaces complete (so it fully applies to 1) and 3), but only partially on 2)).
I leave everything there. I hope I am clear enough and that I didn't make too much mistakes (weird English, grammar, missing words; I reread myself but probably it is not sufficient.)
I you have any question or comments, feel free.
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