Norm in Sobolev Spaces.Is the Sobolev Space $H^k(0,1)$ a banach algebra?Trace theorem, Sobolev spaceDual space of Sobolev functions with homogenous neumann BCIntuitive question about the trace operator (Sobolev spaces)Extending Sobolev functions by zeroSobolev spaces embeddings & isomorphismsCompactness of Sobolev spacesSobolev embeddings; continuous representativesCounterexample of Sobolev tracesCompactly supported functions are dense in local Sobolev spaces
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Norm in Sobolev Spaces.
Is the Sobolev Space $H^k(0,1)$ a banach algebra?Trace theorem, Sobolev spaceDual space of Sobolev functions with homogenous neumann BCIntuitive question about the trace operator (Sobolev spaces)Extending Sobolev functions by zeroSobolev spaces embeddings & isomorphismsCompactness of Sobolev spacesSobolev embeddings; continuous representativesCounterexample of Sobolev tracesCompactly supported functions are dense in local Sobolev spaces
$begingroup$
In an exercise, i would like to use the next afirmation.
Si $L=delta_0$ y $vin H^1_0$ entonces
$|<L,v>| =|v(0)| le ||v||_infty le K_1 ||v||_H^1 le K_2 ||v||_H^1_0.$
But, i do not know if Its true that $||v||_infty le K_1 ||v||_H^1$. Is it a consequence of v is, in particular, continous ($H^1_0subset C^0)$?
All of this in a domain $Omega$ one dimensional.
functional-analysis pde sobolev-spaces
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
add a comment |
$begingroup$
In an exercise, i would like to use the next afirmation.
Si $L=delta_0$ y $vin H^1_0$ entonces
$|<L,v>| =|v(0)| le ||v||_infty le K_1 ||v||_H^1 le K_2 ||v||_H^1_0.$
But, i do not know if Its true that $||v||_infty le K_1 ||v||_H^1$. Is it a consequence of v is, in particular, continous ($H^1_0subset C^0)$?
All of this in a domain $Omega$ one dimensional.
functional-analysis pde sobolev-spaces
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
add a comment |
$begingroup$
In an exercise, i would like to use the next afirmation.
Si $L=delta_0$ y $vin H^1_0$ entonces
$|<L,v>| =|v(0)| le ||v||_infty le K_1 ||v||_H^1 le K_2 ||v||_H^1_0.$
But, i do not know if Its true that $||v||_infty le K_1 ||v||_H^1$. Is it a consequence of v is, in particular, continous ($H^1_0subset C^0)$?
All of this in a domain $Omega$ one dimensional.
functional-analysis pde sobolev-spaces
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
In an exercise, i would like to use the next afirmation.
Si $L=delta_0$ y $vin H^1_0$ entonces
$|<L,v>| =|v(0)| le ||v||_infty le K_1 ||v||_H^1 le K_2 ||v||_H^1_0.$
But, i do not know if Its true that $||v||_infty le K_1 ||v||_H^1$. Is it a consequence of v is, in particular, continous ($H^1_0subset C^0)$?
All of this in a domain $Omega$ one dimensional.
functional-analysis pde sobolev-spaces
functional-analysis pde sobolev-spaces
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
edited Mar 29 at 21:55
Rodrigo Moron Sanz
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Mar 29 at 20:36
Rodrigo Moron SanzRodrigo Moron Sanz
415
415
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it is true that $H_0^1(Omega) subset C^0(Omega)$ when $Omega subset mathbbR$. This follows from the (second part of the) Sobolev embedding theorem as written here. In fact, listed there is the stronger result that
$$H_0^1(Omega) subseteq C^0,frac12(Omega)$$
(the space of $1/2$-Holder continuous functions) with the embedding being continuous which gives the desired inequality.
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
$endgroup$
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, it is true that $H_0^1(Omega) subset C^0(Omega)$ when $Omega subset mathbbR$. This follows from the (second part of the) Sobolev embedding theorem as written here. In fact, listed there is the stronger result that
$$H_0^1(Omega) subseteq C^0,frac12(Omega)$$
(the space of $1/2$-Holder continuous functions) with the embedding being continuous which gives the desired inequality.
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
$endgroup$
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
add a comment |
$begingroup$
Yes, it is true that $H_0^1(Omega) subset C^0(Omega)$ when $Omega subset mathbbR$. This follows from the (second part of the) Sobolev embedding theorem as written here. In fact, listed there is the stronger result that
$$H_0^1(Omega) subseteq C^0,frac12(Omega)$$
(the space of $1/2$-Holder continuous functions) with the embedding being continuous which gives the desired inequality.
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
$endgroup$
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
add a comment |
$begingroup$
Yes, it is true that $H_0^1(Omega) subset C^0(Omega)$ when $Omega subset mathbbR$. This follows from the (second part of the) Sobolev embedding theorem as written here. In fact, listed there is the stronger result that
$$H_0^1(Omega) subseteq C^0,frac12(Omega)$$
(the space of $1/2$-Holder continuous functions) with the embedding being continuous which gives the desired inequality.
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
$endgroup$
Yes, it is true that $H_0^1(Omega) subset C^0(Omega)$ when $Omega subset mathbbR$. This follows from the (second part of the) Sobolev embedding theorem as written here. In fact, listed there is the stronger result that
$$H_0^1(Omega) subseteq C^0,frac12(Omega)$$
(the space of $1/2$-Holder continuous functions) with the embedding being continuous which gives the desired inequality.
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
answered Mar 29 at 22:10
Rhys SteeleRhys Steele
7,8001930
7,8001930
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
add a comment |
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
Thanks. I understand it, but, why gives it the inequality? I don’t see it sorry.
$endgroup$
– Rodrigo Moron Sanz
Mar 29 at 22:49
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
The inequality comes from the fact the embedding is continuous. Continuity of the embedding exactly means that $|v|_C^0,1/2 leq K |v|_H_0^1$ for some $K$. The inequality you want then follows since $|v|_infty leq |v|_C^0,1/2$.
$endgroup$
– Rhys Steele
Mar 29 at 22:51
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
$begingroup$
Okei! Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Mar 30 at 0:10
add a comment |
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