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Let $Ω⊂R^d$ be a smooth open bounded set with boundary $Γ$. Define $D(Ω)$ the space of test function in $Ω$

$U=(D(Ω))^d∩Ker(div)=ϕ∈D(Ω)×…×D(Ω)$ and $div (ϕ)=0$

$V=$ the closure of $U$ in $H^1$ so that $V⊂(H_0^1 (Ω))^d$ close subset

$H=$ the closure of $U$ in $L^2$ so that $H⊂(L^2 (Ω))^d$ close subset

We also have: $V↪H=H'↪V'$ with dense inclusions. $V'$ is the dual of $V$

It is stated in “Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models” Franck Boyer, Pierre Fabrie that: “ the space $V'$ is not a distribution space. “ picture here

I can’t make sense of this statement other then: $V'⊄(D'(Ω))^d$ however, this seems to be a contradiction.

Let $ϕ=(ϕ_1…ϕ_d )∈U$ with suport in a compact $K⊂Ω$. Let $f∈V'$ by Riesz representation $∃ u_f∈V$ such that:

$(f,ϕ)_{V'×V}=(u_f,ϕ)_V=(u_f,ϕ)_{H_0^1}≤max_{|α|≤1}⁡‖∂^αϕ‖_{L^∞ (Ω)}∑_{i,j=1}^d‖\frac{∂u_i}{∂x_j}‖_{L^1 (Ω)}$

Now since $Ω$ is bounded and $\frac{∂u_i}{∂x_j}∈L^2 (Ω)⇒\frac{∂u_i}{∂x_j}∈L^1 (Ω)$ and since $ϕ$ is a test function it is bounded so all the quantities make sence and are finite. In adition notice that we did not identify $V=V'$ so the calculation are valid. In conclusion:

$∀ϕ$ test function with suport in $K⊂Ω$ we have: $|(f,ϕ)_{V'×V} |≤c max_{|α|≤1}⁡‖∂^α ϕ‖_{L^∞ (Ω)}$

This means that $f∈(D'(Ω))^d$ so its clear that I’m doing something wrong. One last Idea is that the author meant $V'⊄(H^{-1})^d$ but that is already stated above.

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Since you only consider maps in $U\subsetneq D(\Omega)^d$, you are not showing that $f$ is a distribution, you are only showing that there is a map $f \mapsto (u_f,\cdot)_{H^1_0}$ from $V'$ to $D'(\Omega)^d$.

The difference is a bit subtle. The main problem is that this map is not unique. In particular for any distribution $\Psi \in D'(\Omega)$, the map $$f \mapsto (u_f,\cdot)_{H^1_0} + \Psi( \operatorname{div}\cdot)$$ does the same job, as it has the same behavior as the previous map on functions $\phi$ for which $\operatorname{div}\phi=0$.

To explain the rest of the quote, note that in the context of the Stokes-equation the behavior on test functions with $\operatorname{div} \phi \neq 0$, is exactly the job of the pressure-term, which here is open to arbitrary choice.

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    $\begingroup$ The subtle difference is also the difference between the question in the title of this post and the question actually asked. :-) $\endgroup$ Commented Jan 10 at 19:43
  • $\begingroup$ @mlk Let me see if I understand this correctly. In my attempt I only shown the desired property for all functions in $U$ BUT to be right I had to show for all functions in $D(Ω)^d$and since you can define f to be anything (not a distribution) on functions that are not divergence-free I can’t prove that f is a distribution right? $\endgroup$ Commented Jan 10 at 23:48
  • $\begingroup$ @mlk And one more questions to make sense of $V'$ si larger then$ H^{-1} (Ω)^d$ : you notice that if $g∈H^{-1} (Ω)^d$ then $g∈V'$ so that $H^{-1} (Ω)^d⊂V'$ but $H^{-1} (Ω)^d$ is a space of distribution and as such you have $V'≠H^{-1} (Ω)^d$ thus $V'$ si larger then $H^{-1} (Ω)^d$ ? $\endgroup$ Commented Jan 10 at 23:52
  • $\begingroup$ @Alucard-oMing Yes, I think you have this essentially correct. There are of course some constraints on how you can define the map on non divergence free functions due to the linearity, but these are not strong enough to narrow it down to some canonical choice. $\endgroup$ Commented Jan 14 at 5:42

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