Mixing
Almost everywhere convergent subsequence in a Sobolev space
$begingroup$
Let $1 < p < n$ and let $p^ast$ denote the Sobolev conjugate of $p$, i.e. let
$$
p^ast := frac{np}{n-p}.
$$
Denote by $mathcal{D}^{1,p}(mathbb{R}^n)$ the space
$$
mathcal{D}^{1,p}left(mathbb{R}^nright) := left{ u in L^{p^ast}(mathbb{R}^n) : nabla u in L^p(mathbb{R}^n) right}
$$
equipped with the norm
$$
leftVert u rightVert := leftVert nabla u rightVert_{L^p(mathbb{R}^n)}.
$$
Assume now that $v_m rightharpoonup v$ weakly in $mathcal{D}^{1,p}(mathbb{R}^n)$ and that $(v_m)$ is uniformly bounded in $L^{p^ast}(mathbb{R}^n)$. After passing to a subsequence, can we assume that $v_m to v$ almost everywhere point wise?
Naturally my first idea was to try and use results about compact embeddings, but I am unaware of any such result involving this space $mathcal{D}^{1,p}(mathbb{R}^n)$.
Edit: Note that in this case we are working in the space $mathcal{D}^{1,p}(mathbb{R}^n)$ which is not the same as $W^{1,p}(mathbb{R}^n)$. In particular, the embedding theorems for $W^{1,p}(mathbb{R}^n)$ do not seem to apply here.
real-analysis functional-analysis pde sobolev-spaces
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
$endgroup$
add a comment |
$begingroup$
Let $1 < p < n$ and let $p^ast$ denote the Sobolev conjugate of $p$, i.e. let
$$
p^ast := frac{np}{n-p}.
$$
Denote by $mathcal{D}^{1,p}(mathbb{R}^n)$ the space
$$
mathcal{D}^{1,p}left(mathbb{R}^nright) := left{ u in L^{p^ast}(mathbb{R}^n) : nabla u in L^p(mathbb{R}^n) right}
$$
equipped with the norm
$$
leftVert u rightVert := leftVert nabla u rightVert_{L^p(mathbb{R}^n)}.
$$
Assume now that $v_m rightharpoonup v$ weakly in $mathcal{D}^{1,p}(mathbb{R}^n)$ and that $(v_m)$ is uniformly bounded in $L^{p^ast}(mathbb{R}^n)$. After passing to a subsequence, can we assume that $v_m to v$ almost everywhere point wise?
Naturally my first idea was to try and use results about compact embeddings, but I am unaware of any such result involving this space $mathcal{D}^{1,p}(mathbb{R}^n)$.
Edit: Note that in this case we are working in the space $mathcal{D}^{1,p}(mathbb{R}^n)$ which is not the same as $W^{1,p}(mathbb{R}^n)$. In particular, the embedding theorems for $W^{1,p}(mathbb{R}^n)$ do not seem to apply here.
real-analysis functional-analysis pde sobolev-spaces
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
$endgroup$
$begingroup$
In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00
add a comment |
$begingroup$
Let $1 < p < n$ and let $p^ast$ denote the Sobolev conjugate of $p$, i.e. let
$$
p^ast := frac{np}{n-p}.
$$
Denote by $mathcal{D}^{1,p}(mathbb{R}^n)$ the space
$$
mathcal{D}^{1,p}left(mathbb{R}^nright) := left{ u in L^{p^ast}(mathbb{R}^n) : nabla u in L^p(mathbb{R}^n) right}
$$
equipped with the norm
$$
leftVert u rightVert := leftVert nabla u rightVert_{L^p(mathbb{R}^n)}.
$$
Assume now that $v_m rightharpoonup v$ weakly in $mathcal{D}^{1,p}(mathbb{R}^n)$ and that $(v_m)$ is uniformly bounded in $L^{p^ast}(mathbb{R}^n)$. After passing to a subsequence, can we assume that $v_m to v$ almost everywhere point wise?
Naturally my first idea was to try and use results about compact embeddings, but I am unaware of any such result involving this space $mathcal{D}^{1,p}(mathbb{R}^n)$.
Edit: Note that in this case we are working in the space $mathcal{D}^{1,p}(mathbb{R}^n)$ which is not the same as $W^{1,p}(mathbb{R}^n)$. In particular, the embedding theorems for $W^{1,p}(mathbb{R}^n)$ do not seem to apply here.
real-analysis functional-analysis pde sobolev-spaces
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
$endgroup$
Let $1 < p < n$ and let $p^ast$ denote the Sobolev conjugate of $p$, i.e. let
$$
p^ast := frac{np}{n-p}.
$$
Denote by $mathcal{D}^{1,p}(mathbb{R}^n)$ the space
$$
mathcal{D}^{1,p}left(mathbb{R}^nright) := left{ u in L^{p^ast}(mathbb{R}^n) : nabla u in L^p(mathbb{R}^n) right}
$$
equipped with the norm
$$
leftVert u rightVert := leftVert nabla u rightVert_{L^p(mathbb{R}^n)}.
$$
Assume now that $v_m rightharpoonup v$ weakly in $mathcal{D}^{1,p}(mathbb{R}^n)$ and that $(v_m)$ is uniformly bounded in $L^{p^ast}(mathbb{R}^n)$. After passing to a subsequence, can we assume that $v_m to v$ almost everywhere point wise?
Naturally my first idea was to try and use results about compact embeddings, but I am unaware of any such result involving this space $mathcal{D}^{1,p}(mathbb{R}^n)$.
Edit: Note that in this case we are working in the space $mathcal{D}^{1,p}(mathbb{R}^n)$ which is not the same as $W^{1,p}(mathbb{R}^n)$. In particular, the embedding theorems for $W^{1,p}(mathbb{R}^n)$ do not seem to apply here.
real-analysis functional-analysis pde sobolev-spaces
real-analysis functional-analysis pde sobolev-spaces
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
edited Dec 31 '18 at 0:24
rolandcyp
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
asked Dec 30 '18 at 21:30
rolandcyprolandcyp
22029
22029
$begingroup$
In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00
add a comment |
$begingroup$
In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00
$begingroup$
In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $mathcal{D}^{1,p}(mathbb R^n) hookrightarrow L^p(mathbb R^n,mathbb R^n)$ given by $u mapsto nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ mathcal{D}^{1,p}(mathbb R^n)' cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
where
$$ Y = left{g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0 , forall u in D^{1,p}(mathbb R^n) right}.$$
(Note I do not claim that $mathcal{D}^{1,p}(mathbb R^n)$ is complete in this argument $(dagger)$.) From this it is easy to see that $v_m rightharpoonup v$ weakly in $D^{1,p}(mathbb R^n)$ if and only if for all $g in L^{p'}(mathbb R^n, mathbb R^n),$
$$ int_{mathbb R^n} g.nabla v_m rightarrow int_{mathbb R^n} g.nabla v. $$
Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $Omega subset mathbb R^n$ a bounded domain, then extending by zero we get for any $g in L^{p'}(Omega,mathbb R^n),$
$$ int_{Omega} g.nabla v_m rightarrow int_{Omega} g.nabla v. $$
Now the restriction $(v_n|_{Omega})$ defines a bounded sequence in $W^{1,p}(Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u in W^{1,p}(Omega)$ and a subsequence $(v_{n_k}|_{Omega})$ such that $v_{n_k} rightharpoonup u$ weakly in $W^{1,p}(Omega)$ and $v_{n_k} rightarrow u$ a.e. also. Then for any $g in L^{p'}(Omega,mathbb R^n)$ we have,
$$ 0 = lim_{mrightarrow 0} int_{mathbb R^n} g.nabla v_{n_k} - g.nabla v = int_{mathbb R^n} g.nabla(u-v). $$
Hence $nabla(u-v) = 0$ a.e. in $Omega,$ and so there is $lambda_{Omega} in mathbb R$ such that $v_{n_k} rightarrow (v + lambda_{Omega})$ a.e. in $Omega.$
Now take a compact exhaustion $mathbb R^n = bigcup_j Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} rightarrow v + lambda_i$ a.e. as $k rightarrow infty$ in $Omega_i$ for all $i leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |Omega_j| leq liminf_{k rightarrow infty} int_{Omega_j} |v_{k,j} - v|^{p^*} leq sup_{m} int_{mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < infty, $$
so $k_j rightarrow 0$ as $j rightarrow infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} rightarrow v$ a.e. in $mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$
Added later $(dagger)$: Writing $X = mathcal{D}^{1,p}(mathbb R^n)$ and $lVert cdot rVert = lVert nabla cdot rVert_{L^p(mathbb R^n,mathbb R^n)},$ it is not clear whether $(X,lVertcdotrVert)$ is complete (which is equivalent to its image being closed in $L^p(mathbb R^n,mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(mathbb R^n,mathbb R^n)$ identified via this isometric embedding. Then the restriction map $overline X' ni f mapsto f|_X$ defines an isometric isomorphism $X' cong overline X'.$ Also we have,
begin{align*}
Y &= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in X right} \
&= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in overline X right}
end{align*}
Identifying $L^{p'}(mathbb R^n,mathbb R^n) cong L^p(mathbb R^n,mathbb R^n)',$ we have $Y = X^{perp} = overline X^{perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' cong overline X' cong L^p(mathbb R^n,mathbb R^n)' / overline{X}^{perp} cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
$endgroup$
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
add a comment |
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$begingroup$
First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $mathcal{D}^{1,p}(mathbb R^n) hookrightarrow L^p(mathbb R^n,mathbb R^n)$ given by $u mapsto nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ mathcal{D}^{1,p}(mathbb R^n)' cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
where
$$ Y = left{g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0 , forall u in D^{1,p}(mathbb R^n) right}.$$
(Note I do not claim that $mathcal{D}^{1,p}(mathbb R^n)$ is complete in this argument $(dagger)$.) From this it is easy to see that $v_m rightharpoonup v$ weakly in $D^{1,p}(mathbb R^n)$ if and only if for all $g in L^{p'}(mathbb R^n, mathbb R^n),$
$$ int_{mathbb R^n} g.nabla v_m rightarrow int_{mathbb R^n} g.nabla v. $$
Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $Omega subset mathbb R^n$ a bounded domain, then extending by zero we get for any $g in L^{p'}(Omega,mathbb R^n),$
$$ int_{Omega} g.nabla v_m rightarrow int_{Omega} g.nabla v. $$
Now the restriction $(v_n|_{Omega})$ defines a bounded sequence in $W^{1,p}(Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u in W^{1,p}(Omega)$ and a subsequence $(v_{n_k}|_{Omega})$ such that $v_{n_k} rightharpoonup u$ weakly in $W^{1,p}(Omega)$ and $v_{n_k} rightarrow u$ a.e. also. Then for any $g in L^{p'}(Omega,mathbb R^n)$ we have,
$$ 0 = lim_{mrightarrow 0} int_{mathbb R^n} g.nabla v_{n_k} - g.nabla v = int_{mathbb R^n} g.nabla(u-v). $$
Hence $nabla(u-v) = 0$ a.e. in $Omega,$ and so there is $lambda_{Omega} in mathbb R$ such that $v_{n_k} rightarrow (v + lambda_{Omega})$ a.e. in $Omega.$
Now take a compact exhaustion $mathbb R^n = bigcup_j Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} rightarrow v + lambda_i$ a.e. as $k rightarrow infty$ in $Omega_i$ for all $i leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |Omega_j| leq liminf_{k rightarrow infty} int_{Omega_j} |v_{k,j} - v|^{p^*} leq sup_{m} int_{mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < infty, $$
so $k_j rightarrow 0$ as $j rightarrow infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} rightarrow v$ a.e. in $mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$
Added later $(dagger)$: Writing $X = mathcal{D}^{1,p}(mathbb R^n)$ and $lVert cdot rVert = lVert nabla cdot rVert_{L^p(mathbb R^n,mathbb R^n)},$ it is not clear whether $(X,lVertcdotrVert)$ is complete (which is equivalent to its image being closed in $L^p(mathbb R^n,mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(mathbb R^n,mathbb R^n)$ identified via this isometric embedding. Then the restriction map $overline X' ni f mapsto f|_X$ defines an isometric isomorphism $X' cong overline X'.$ Also we have,
begin{align*}
Y &= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in X right} \
&= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in overline X right}
end{align*}
Identifying $L^{p'}(mathbb R^n,mathbb R^n) cong L^p(mathbb R^n,mathbb R^n)',$ we have $Y = X^{perp} = overline X^{perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' cong overline X' cong L^p(mathbb R^n,mathbb R^n)' / overline{X}^{perp} cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
$endgroup$
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
add a comment |
$begingroup$
First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $mathcal{D}^{1,p}(mathbb R^n) hookrightarrow L^p(mathbb R^n,mathbb R^n)$ given by $u mapsto nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ mathcal{D}^{1,p}(mathbb R^n)' cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
where
$$ Y = left{g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0 , forall u in D^{1,p}(mathbb R^n) right}.$$
(Note I do not claim that $mathcal{D}^{1,p}(mathbb R^n)$ is complete in this argument $(dagger)$.) From this it is easy to see that $v_m rightharpoonup v$ weakly in $D^{1,p}(mathbb R^n)$ if and only if for all $g in L^{p'}(mathbb R^n, mathbb R^n),$
$$ int_{mathbb R^n} g.nabla v_m rightarrow int_{mathbb R^n} g.nabla v. $$
Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $Omega subset mathbb R^n$ a bounded domain, then extending by zero we get for any $g in L^{p'}(Omega,mathbb R^n),$
$$ int_{Omega} g.nabla v_m rightarrow int_{Omega} g.nabla v. $$
Now the restriction $(v_n|_{Omega})$ defines a bounded sequence in $W^{1,p}(Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u in W^{1,p}(Omega)$ and a subsequence $(v_{n_k}|_{Omega})$ such that $v_{n_k} rightharpoonup u$ weakly in $W^{1,p}(Omega)$ and $v_{n_k} rightarrow u$ a.e. also. Then for any $g in L^{p'}(Omega,mathbb R^n)$ we have,
$$ 0 = lim_{mrightarrow 0} int_{mathbb R^n} g.nabla v_{n_k} - g.nabla v = int_{mathbb R^n} g.nabla(u-v). $$
Hence $nabla(u-v) = 0$ a.e. in $Omega,$ and so there is $lambda_{Omega} in mathbb R$ such that $v_{n_k} rightarrow (v + lambda_{Omega})$ a.e. in $Omega.$
Now take a compact exhaustion $mathbb R^n = bigcup_j Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} rightarrow v + lambda_i$ a.e. as $k rightarrow infty$ in $Omega_i$ for all $i leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |Omega_j| leq liminf_{k rightarrow infty} int_{Omega_j} |v_{k,j} - v|^{p^*} leq sup_{m} int_{mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < infty, $$
so $k_j rightarrow 0$ as $j rightarrow infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} rightarrow v$ a.e. in $mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$
Added later $(dagger)$: Writing $X = mathcal{D}^{1,p}(mathbb R^n)$ and $lVert cdot rVert = lVert nabla cdot rVert_{L^p(mathbb R^n,mathbb R^n)},$ it is not clear whether $(X,lVertcdotrVert)$ is complete (which is equivalent to its image being closed in $L^p(mathbb R^n,mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(mathbb R^n,mathbb R^n)$ identified via this isometric embedding. Then the restriction map $overline X' ni f mapsto f|_X$ defines an isometric isomorphism $X' cong overline X'.$ Also we have,
begin{align*}
Y &= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in X right} \
&= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in overline X right}
end{align*}
Identifying $L^{p'}(mathbb R^n,mathbb R^n) cong L^p(mathbb R^n,mathbb R^n)',$ we have $Y = X^{perp} = overline X^{perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' cong overline X' cong L^p(mathbb R^n,mathbb R^n)' / overline{X}^{perp} cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
$endgroup$
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
add a comment |
$begingroup$
First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $mathcal{D}^{1,p}(mathbb R^n) hookrightarrow L^p(mathbb R^n,mathbb R^n)$ given by $u mapsto nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ mathcal{D}^{1,p}(mathbb R^n)' cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
where
$$ Y = left{g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0 , forall u in D^{1,p}(mathbb R^n) right}.$$
(Note I do not claim that $mathcal{D}^{1,p}(mathbb R^n)$ is complete in this argument $(dagger)$.) From this it is easy to see that $v_m rightharpoonup v$ weakly in $D^{1,p}(mathbb R^n)$ if and only if for all $g in L^{p'}(mathbb R^n, mathbb R^n),$
$$ int_{mathbb R^n} g.nabla v_m rightarrow int_{mathbb R^n} g.nabla v. $$
Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $Omega subset mathbb R^n$ a bounded domain, then extending by zero we get for any $g in L^{p'}(Omega,mathbb R^n),$
$$ int_{Omega} g.nabla v_m rightarrow int_{Omega} g.nabla v. $$
Now the restriction $(v_n|_{Omega})$ defines a bounded sequence in $W^{1,p}(Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u in W^{1,p}(Omega)$ and a subsequence $(v_{n_k}|_{Omega})$ such that $v_{n_k} rightharpoonup u$ weakly in $W^{1,p}(Omega)$ and $v_{n_k} rightarrow u$ a.e. also. Then for any $g in L^{p'}(Omega,mathbb R^n)$ we have,
$$ 0 = lim_{mrightarrow 0} int_{mathbb R^n} g.nabla v_{n_k} - g.nabla v = int_{mathbb R^n} g.nabla(u-v). $$
Hence $nabla(u-v) = 0$ a.e. in $Omega,$ and so there is $lambda_{Omega} in mathbb R$ such that $v_{n_k} rightarrow (v + lambda_{Omega})$ a.e. in $Omega.$
Now take a compact exhaustion $mathbb R^n = bigcup_j Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} rightarrow v + lambda_i$ a.e. as $k rightarrow infty$ in $Omega_i$ for all $i leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |Omega_j| leq liminf_{k rightarrow infty} int_{Omega_j} |v_{k,j} - v|^{p^*} leq sup_{m} int_{mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < infty, $$
so $k_j rightarrow 0$ as $j rightarrow infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} rightarrow v$ a.e. in $mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$
Added later $(dagger)$: Writing $X = mathcal{D}^{1,p}(mathbb R^n)$ and $lVert cdot rVert = lVert nabla cdot rVert_{L^p(mathbb R^n,mathbb R^n)},$ it is not clear whether $(X,lVertcdotrVert)$ is complete (which is equivalent to its image being closed in $L^p(mathbb R^n,mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(mathbb R^n,mathbb R^n)$ identified via this isometric embedding. Then the restriction map $overline X' ni f mapsto f|_X$ defines an isometric isomorphism $X' cong overline X'.$ Also we have,
begin{align*}
Y &= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in X right} \
&= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in overline X right}
end{align*}
Identifying $L^{p'}(mathbb R^n,mathbb R^n) cong L^p(mathbb R^n,mathbb R^n)',$ we have $Y = X^{perp} = overline X^{perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' cong overline X' cong L^p(mathbb R^n,mathbb R^n)' / overline{X}^{perp} cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
$endgroup$
First of all we need to understand what weak convergence in this space means. Observe that we have an isometric embedding $mathcal{D}^{1,p}(mathbb R^n) hookrightarrow L^p(mathbb R^n,mathbb R^n)$ given by $u mapsto nabla u.$ By standard results about dual spaces of subspaces, we get that
$$ mathcal{D}^{1,p}(mathbb R^n)' cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
where
$$ Y = left{g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0 , forall u in D^{1,p}(mathbb R^n) right}.$$
(Note I do not claim that $mathcal{D}^{1,p}(mathbb R^n)$ is complete in this argument $(dagger)$.) From this it is easy to see that $v_m rightharpoonup v$ weakly in $D^{1,p}(mathbb R^n)$ if and only if for all $g in L^{p'}(mathbb R^n, mathbb R^n),$
$$ int_{mathbb R^n} g.nabla v_m rightarrow int_{mathbb R^n} g.nabla v. $$
Now the idea is that a.e. convergence is a local property, so it suffices to restrict to a bounded domain where we get compact embeddings into $L^p.$ For this fix $Omega subset mathbb R^n$ a bounded domain, then extending by zero we get for any $g in L^{p'}(Omega,mathbb R^n),$
$$ int_{Omega} g.nabla v_m rightarrow int_{Omega} g.nabla v. $$
Now the restriction $(v_n|_{Omega})$ defines a bounded sequence in $W^{1,p}(Omega),$ where we have used Hölder and the uniform bound of $(v_n)$ in $L^{p^*}(mathbb R^n).$ We know by the Rellich-Kondrachov compactness theorem that there exists $u in W^{1,p}(Omega)$ and a subsequence $(v_{n_k}|_{Omega})$ such that $v_{n_k} rightharpoonup u$ weakly in $W^{1,p}(Omega)$ and $v_{n_k} rightarrow u$ a.e. also. Then for any $g in L^{p'}(Omega,mathbb R^n)$ we have,
$$ 0 = lim_{mrightarrow 0} int_{mathbb R^n} g.nabla v_{n_k} - g.nabla v = int_{mathbb R^n} g.nabla(u-v). $$
Hence $nabla(u-v) = 0$ a.e. in $Omega,$ and so there is $lambda_{Omega} in mathbb R$ such that $v_{n_k} rightarrow (v + lambda_{Omega})$ a.e. in $Omega.$
Now take a compact exhaustion $mathbb R^n = bigcup_j Omega_j$ and iteratively choose subsequences $v_{k,j}$ such that $v_{k,j} rightarrow v + lambda_i$ a.e. as $k rightarrow infty$ in $Omega_i$ for all $i leq j.$ Then observe by Fatou's lemma that,
$$ k_j^{p^*} |Omega_j| leq liminf_{k rightarrow infty} int_{Omega_j} |v_{k,j} - v|^{p^*} leq sup_{m} int_{mathbb R^n} 2^{p^*}(|v_m|^{p^*} + |v|^{p^*}) < infty, $$
so $k_j rightarrow 0$ as $j rightarrow infty.$ Hence the diagonal sequence $v_{n_k} = v_{k,k}$ satisfies $v_{n_k} rightarrow v$ a.e. in $mathbb R^n.$
Now as the limit is unique, we see that every subsequence of $(v_m)$ has a further subsequence which converges a.e. to $v.$ Hence the entire sequence converges a.e. to $v.$
Added later $(dagger)$: Writing $X = mathcal{D}^{1,p}(mathbb R^n)$ and $lVert cdot rVert = lVert nabla cdot rVert_{L^p(mathbb R^n,mathbb R^n)},$ it is not clear whether $(X,lVertcdotrVert)$ is complete (which is equivalent to its image being closed in $L^p(mathbb R^n,mathbb R^n)$). However I claim this is not necessary for the argument to work.
Let $overline X$ be the completion of $X$ with respect to this norm, which can be identified with the closure of $X$ in $L^p(mathbb R^n,mathbb R^n)$ identified via this isometric embedding. Then the restriction map $overline X' ni f mapsto f|_X$ defines an isometric isomorphism $X' cong overline X'.$ Also we have,
begin{align*}
Y &= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in X right} \
&= left{ g in L^{p'}(mathbb R^n,mathbb R^n) middle| int_{mathbb R^n} g.nabla u = 0, forall u in overline X right}
end{align*}
Identifying $L^{p'}(mathbb R^n,mathbb R^n) cong L^p(mathbb R^n,mathbb R^n)',$ we have $Y = X^{perp} = overline X^{perp}.$ Now the result about duality of subspaces is generally is stated for closed subspaces (see e.g. proposition 3.67 in section 3.16 of Introduction to Banach Spaces and Algebras by Allan & Dales), but in light of above we have,
$$ X' cong overline X' cong L^p(mathbb R^n,mathbb R^n)' / overline{X}^{perp} cong L^{p'}(mathbb R^n,mathbb R^n) / Y, $$
so it also holds even if $X$ is not complete.
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
edited Jan 1 at 11:04
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
answered Dec 31 '18 at 13:11
ktoiktoi
2,3611616
2,3611616
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
add a comment |
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
1
1
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
In the beginning, when you discuss the dual space of the subspace $D^{1,p}(mathbb{R}^n)$, don't we first need to show that $D^{1,p}(mathbb{R}^n)$ is closed as a subspace of $L^p(mathbb{R}^n;mathbb{R}^n)$?
$endgroup$
– Quoka
Dec 31 '18 at 18:48
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
$begingroup$
@Quoka I don't think closedness is needed for the argument to work. See my edit, where I've added more details showing what to do if it isn't.
$endgroup$
– ktoi
Jan 1 at 11:07
add a comment |
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In this post they seem to be using the compactness of the embedding $W^{1,2}(Omega) hookrightarrow L^p(Omega)$ for sufficiently nice $Omega$. However, I am not aware of any such embedding for the space $mathcal{D}^{1,p}(Omega)$. I do not see how this helps.
$endgroup$
– rolandcyp
Dec 31 '18 at 0:27
$begingroup$
Could you explain what you mean by $W^{1,p}$? I've seen it used to denote what you call $mathcal{D}^{1,p}$, and seen $mathcal{D}^{1,p}$ used to denote the closure of $C_c^{infty}$ in that space.
$endgroup$
– Sambo
Dec 31 '18 at 4:41
$begingroup$
In my case $W^{1,p}(Omega)$ denotes the space of functions in $L^p(Omega)$ having weak derivatives of the first order in $L^p(Omega)$.
$endgroup$
– rolandcyp
Dec 31 '18 at 6:08
$begingroup$
I see. I expect similar embeddings still hold. While I cannot point you directly to a reference, I do know of this paper which gives an embedding of a similar space (you'll see my confusion with the notation!)
$endgroup$
– Sambo
Dec 31 '18 at 7:00