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Density of a subspace
Question about limits of weakly convergent sequence in $H^1_0(Omega)$Solution regularity of the heat equation after $t>varepsilon$How to modify a $H^1$ weak convergence sequence so that I have the $L^2$ equi-integrability of gradient?Alternative characterisation of weak derivativesReal Analysis, Folland problem 5.5.63 Hilbert SpacesDemonstrating that constructed solution to PDE satisfies initial conditionSobolev Imbedding and Gagliardo Nirenberg Sobolev inequalityConstructing the limit of functions in $C[a,b]$ with maximum value normIn compact embedding Theorem, $u_0$ lies in $W^1,p(Omega)$?Prove that the space $C^k(barU)$ is Banach.
$begingroup$
Let $OmegasubsetmathbbR^d$ be open, bounded and simply connected with smooth boundary $partialOmega$. Define $mathcalH:=~Delta uin L^2(Omega)$ with norm
$$|u|_mathcalH^2:=|u|_H^1(Omega)^2+|Delta u|_L^2(Omega)^2.$$
I want to show that $H^2(Omega)$ is dense in $mathcalH$ under $|cdot|_mathcalH$.
This means, given $uinmathcalH$ and $varepsilon>0$ I need to find a $u_0in H^2(Omega)$ such that
$$|u-u_0|_H^1(Omega)^2+|Delta u-Delta u_0|_L^2(Omega)^2<varepsilon.$$
Alternatively it means that given a sequence $u_ninH^2(Omega)$ that is Cauchy in $|cdot|_mathcalH$ I need to show that its limit $u$ is in $mathcalH$.
So far I've attempted reasoning as follows:
If $u_n$ is $mathcalH$-Cauchy then also it is $H^1(Omega)$-Cauchy, i.e.,
$$|u_n-u_m|_H^1(Omega)rightarrow0qquad(n,mrightarrowinfty).$$
Hence by completeness there is an $H^1(Omega)$-limit $u$:
$$|u_n-u|_H^1(Omega)rightarrow0qquad(nrightarrowinfty).$$
The issue here is that $Delta u$ is not necessarily defined. One may of course remark that by the same reasoning in the complete space $L^2(Omega)$ the sequence $Delta u_n$ has a limit, say, $y$, in $L^2(Omega)$. But there is no immediate link between $y$ and $u$.
Could one attempt to solve the Dirichlet problem
$$left{beginarrayrclcc
Delta f &=&y&textin&Omega\
f&=&u|_partialOmega&texton&partialOmega
endarrayright.quad?$$
This seems like an incredibly convoluted way to do it, when I feel it should be extremely simple---$H^2(Omega)$ is trivially $H^1$-dense in $H^1(Omega)$.
functional-analysis hilbert-spaces sobolev-spaces
asked Mar 13 at 9:53
StromaelStromael
675615
$endgroup$
add a comment |
$begingroup$
Let $OmegasubsetmathbbR^d$ be open, bounded and simply connected with smooth boundary $partialOmega$. Define $mathcalH:=~Delta uin L^2(Omega)$ with norm
$$|u|_mathcalH^2:=|u|_H^1(Omega)^2+|Delta u|_L^2(Omega)^2.$$
I want to show that $H^2(Omega)$ is dense in $mathcalH$ under $|cdot|_mathcalH$.
This means, given $uinmathcalH$ and $varepsilon>0$ I need to find a $u_0in H^2(Omega)$ such that
$$|u-u_0|_H^1(Omega)^2+|Delta u-Delta u_0|_L^2(Omega)^2<varepsilon.$$
Alternatively it means that given a sequence $u_ninH^2(Omega)$ that is Cauchy in $|cdot|_mathcalH$ I need to show that its limit $u$ is in $mathcalH$.
So far I've attempted reasoning as follows:
If $u_n$ is $mathcalH$-Cauchy then also it is $H^1(Omega)$-Cauchy, i.e.,
$$|u_n-u_m|_H^1(Omega)rightarrow0qquad(n,mrightarrowinfty).$$
Hence by completeness there is an $H^1(Omega)$-limit $u$:
$$|u_n-u|_H^1(Omega)rightarrow0qquad(nrightarrowinfty).$$
The issue here is that $Delta u$ is not necessarily defined. One may of course remark that by the same reasoning in the complete space $L^2(Omega)$ the sequence $Delta u_n$ has a limit, say, $y$, in $L^2(Omega)$. But there is no immediate link between $y$ and $u$.
Could one attempt to solve the Dirichlet problem
$$left{beginarrayrclcc
Delta f &=&y&textin&Omega\
f&=&u|_partialOmega&texton&partialOmega
endarrayright.quad?$$
This seems like an incredibly convoluted way to do it, when I feel it should be extremely simple---$H^2(Omega)$ is trivially $H^1$-dense in $H^1(Omega)$.
functional-analysis hilbert-spaces sobolev-spaces
asked Mar 13 at 9:53
StromaelStromael
675615
$endgroup$
$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50
add a comment |
$begingroup$
Let $OmegasubsetmathbbR^d$ be open, bounded and simply connected with smooth boundary $partialOmega$. Define $mathcalH:=~Delta uin L^2(Omega)$ with norm
$$|u|_mathcalH^2:=|u|_H^1(Omega)^2+|Delta u|_L^2(Omega)^2.$$
I want to show that $H^2(Omega)$ is dense in $mathcalH$ under $|cdot|_mathcalH$.
This means, given $uinmathcalH$ and $varepsilon>0$ I need to find a $u_0in H^2(Omega)$ such that
$$|u-u_0|_H^1(Omega)^2+|Delta u-Delta u_0|_L^2(Omega)^2<varepsilon.$$
Alternatively it means that given a sequence $u_ninH^2(Omega)$ that is Cauchy in $|cdot|_mathcalH$ I need to show that its limit $u$ is in $mathcalH$.
So far I've attempted reasoning as follows:
If $u_n$ is $mathcalH$-Cauchy then also it is $H^1(Omega)$-Cauchy, i.e.,
$$|u_n-u_m|_H^1(Omega)rightarrow0qquad(n,mrightarrowinfty).$$
Hence by completeness there is an $H^1(Omega)$-limit $u$:
$$|u_n-u|_H^1(Omega)rightarrow0qquad(nrightarrowinfty).$$
The issue here is that $Delta u$ is not necessarily defined. One may of course remark that by the same reasoning in the complete space $L^2(Omega)$ the sequence $Delta u_n$ has a limit, say, $y$, in $L^2(Omega)$. But there is no immediate link between $y$ and $u$.
Could one attempt to solve the Dirichlet problem
$$left{beginarrayrclcc
Delta f &=&y&textin&Omega\
f&=&u|_partialOmega&texton&partialOmega
endarrayright.quad?$$
This seems like an incredibly convoluted way to do it, when I feel it should be extremely simple---$H^2(Omega)$ is trivially $H^1$-dense in $H^1(Omega)$.
functional-analysis hilbert-spaces sobolev-spaces
asked Mar 13 at 9:53
StromaelStromael
675615
$endgroup$
Let $OmegasubsetmathbbR^d$ be open, bounded and simply connected with smooth boundary $partialOmega$. Define $mathcalH:=~Delta uin L^2(Omega)$ with norm
$$|u|_mathcalH^2:=|u|_H^1(Omega)^2+|Delta u|_L^2(Omega)^2.$$
I want to show that $H^2(Omega)$ is dense in $mathcalH$ under $|cdot|_mathcalH$.
This means, given $uinmathcalH$ and $varepsilon>0$ I need to find a $u_0in H^2(Omega)$ such that
$$|u-u_0|_H^1(Omega)^2+|Delta u-Delta u_0|_L^2(Omega)^2<varepsilon.$$
Alternatively it means that given a sequence $u_ninH^2(Omega)$ that is Cauchy in $|cdot|_mathcalH$ I need to show that its limit $u$ is in $mathcalH$.
So far I've attempted reasoning as follows:
If $u_n$ is $mathcalH$-Cauchy then also it is $H^1(Omega)$-Cauchy, i.e.,
$$|u_n-u_m|_H^1(Omega)rightarrow0qquad(n,mrightarrowinfty).$$
Hence by completeness there is an $H^1(Omega)$-limit $u$:
$$|u_n-u|_H^1(Omega)rightarrow0qquad(nrightarrowinfty).$$
The issue here is that $Delta u$ is not necessarily defined. One may of course remark that by the same reasoning in the complete space $L^2(Omega)$ the sequence $Delta u_n$ has a limit, say, $y$, in $L^2(Omega)$. But there is no immediate link between $y$ and $u$.
Could one attempt to solve the Dirichlet problem
$$left{beginarrayrclcc
Delta f &=&y&textin&Omega\
f&=&u|_partialOmega&texton&partialOmega
endarrayright.quad?$$
This seems like an incredibly convoluted way to do it, when I feel it should be extremely simple---$H^2(Omega)$ is trivially $H^1$-dense in $H^1(Omega)$.
functional-analysis hilbert-spaces sobolev-spaces
functional-analysis hilbert-spaces sobolev-spaces
asked Mar 13 at 9:53
StromaelStromael
675615
asked Mar 13 at 9:53
StromaelStromael
675615
asked Mar 13 at 9:53
StromaelStromael
675615
asked Mar 13 at 9:53
StromaelStromael
675615
asked Mar 13 at 9:53
StromaelStromael
675615
675615
$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50
add a comment |
$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50
$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50
add a comment |
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$begingroup$
Isn't $C^infty(Omega)$ dense in $mathcal H$? If that is the case then you're done, since $C^inftysubset H^2(Omega)$.
$endgroup$
– Giuseppe Negro
Mar 13 at 10:14
$begingroup$
$C^infty(Omega)$ is dense in $H^1(Omega)$ under the $H^1$-norm. I believe it is dense in $mathcalH$ under the $mathcalH$-norm, but that Laplacian keeps getting in the way of proving it; there's no obvious way to control $|Delta f|_L^2(Omega)$, even if $f$ is smooth.
$endgroup$
– Stromael
Mar 13 at 10:25
$begingroup$
Hum. If $Omega=mathbb R^n$, then $mathcalH$ is exactly equal to $H^2(mathbb R^n)$. The same if we consider functions that vanish on $partial Omega$ with their first derivatives (i.e., $H^2_0(Omega)$). In the general case, there's a boundary term: $$lVert Delta frVert_L^2(Omega)^2=lVert D^2 frVert_L^2(Omega)^2 + sum_ijint_partial Omega(partial_j f partial_i^2 f n_j - partial_j f partial_ipartial_q f n_q), dS.$$You probably know these things already but that's all I can say
$endgroup$
– Giuseppe Negro
Mar 13 at 10:50