I'm trying to understand $(W_0^{1,p} (Ω))^*=W_0^{-1,p^*} (Ω)$, and what a proper representation of its elements is. I understand the basics such as: if $f∈L^{p^*} (Ω)⇒f∈W_0^{-1,p^*}(Ω)$ and the duality is given by $(f,u)=∫_Ωfudx$ $∀ u∈W_0^{1,p^*}(Ω). $
Now my question is the following: assume that $F∈ (W_0^{1,p} (Ω))^*$ but $F∉L^{p^*}(Ω)$ what is a proper representation of $F$ ?
here is my attempt:
$∀ F∈(W_0^{1,p} (Ω))^* ∃ f_0...f_n ∈L^{p^*} (Ω)$ such that $∀ u∈W_0^{1,p}(Ω)$
$F(u)=lim_{n→∞} ⟨f_0-∑_{i=1}^n∂_i f_i,ϕ_n ⟩_{D^*×D}$ (1)
Where:
$∂_i f_i$ is the distribution of derivative of $f_i$ with respect to the $i$ variable.
$⟨*,*⟩_{D^*×D}$ is the dual bracket of $D^* (Ω)×D(Ω)$
for this to be correct I have to justify the expression itself, and then prove that the particular choice of sequence is irrelevant
Here is what I think proves my claim:
Step 1 we can find an isometry between $W_0^{1,p}(Ω)$ and $(L^p (Ω))^{n+1}$. Without going too much into detail using this isometry we conclude that $∀F∈(W_0^{1,p}(Ω))^*∃ f_0...f_n ∈L^{p^*}(Ω)$ such that $∀ u∈W_0^{1,p}(Ω):$
$F(u)=∫_Ω uf_0dx+∑_{i=1}^N∫_Ωf_i ∂_i udx$ (2)
Step 2. this is in my opinion the tricky part. Firstly since we are working in $W_0^{1,p}(Ω)$ we can always choose a sequence of test functions $ϕ_n$ that converges to $u$ in $W_0^{1,p}(Ω)$.Now by continuity of F we have:
$F(u)=lim_{n→∞} ∫_Ω ϕ_n f_0dx+∑_{i=1}^N∫_Ω f_i ∂_i ϕ_n dx.$
However since these are test functions, we can consider the integration to be the duality bracket in $D^* (Ω)×D(Ω)$, thus we get.
$F(u)=lim_{n→∞} ⟨f_0 ,ϕ_n⟩_{D^*×D}+∑_{i=1}^N⟨f_i,∂_i ϕ_n ⟩_{D^*×D}$ (3)
by using the derivative of distribution we get (1) directly from (3). As far as I can tell the important part is that: convergence in$W_0^{1,p}(Ω)$ does not imply convergence in $D(Ω)$ even so the limit in (3) exists by the calculations in step 1 and step 2; namely we know that the limit it is equal to some $F$ acting on some $u$.
Step 3. we prove that the choice of sequence is irrelevant. Let $ψ_n$ be another sequence converging to $u$ in $W_0^{1,p}(Ω)$ by using (3) we get:
$lim_{n→∞} ⟨f_0,ϕ_n-ψ_n ⟩_{D^*×D}+∑_{i=1}^N ⟨f_i,∂_i ϕ_n-ψ_n ⟩_{D^*×D} =0$
Is my approach correct ? And if not can you please correct some mistakes.