Relation between bochner space $L^1(I,X)$ and $C(I,X)$
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I am new in Bochner spaces and I have the following problem. I am not able to prove, that if $u in L^1((0,1),X),u' in L^1((0,1),X) $ then $u in C([0,1],X)$, where $X$ is Banach space and $L^1((0,1),X) $ Bochner space.
bochner-spaces
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
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add a comment |
$begingroup$
I am new in Bochner spaces and I have the following problem. I am not able to prove, that if $u in L^1((0,1),X),u' in L^1((0,1),X) $ then $u in C([0,1],X)$, where $X$ is Banach space and $L^1((0,1),X) $ Bochner space.
bochner-spaces
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
$endgroup$
$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
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– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49
add a comment |
$begingroup$
I am new in Bochner spaces and I have the following problem. I am not able to prove, that if $u in L^1((0,1),X),u' in L^1((0,1),X) $ then $u in C([0,1],X)$, where $X$ is Banach space and $L^1((0,1),X) $ Bochner space.
bochner-spaces
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
$endgroup$
I am new in Bochner spaces and I have the following problem. I am not able to prove, that if $u in L^1((0,1),X),u' in L^1((0,1),X) $ then $u in C([0,1],X)$, where $X$ is Banach space and $L^1((0,1),X) $ Bochner space.
bochner-spaces
bochner-spaces
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
edited Dec 11 '18 at 10:48
lojdmoj
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
asked Dec 11 '18 at 1:52
lojdmojlojdmoj
877
877
$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49
add a comment |
$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49
$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49
add a comment |
1 Answer
1
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You are mostly asking about the embedding
$$W^{1,1}(I;X) hookrightarrow C(I;X),$$
which is quite standard and can be seen in most books that deal with Bochner spaces and PDES, see e.g. Roubicek or Cazenave&Haraux.
Let $I=[a,b]$ be some interval and let $u in W^{1,1}(I;X)$, then $u' in L^1(I;X)$ and we set $v(t):=int_a^t u'(s) , text{d}s$. Then
$$| v(t_2)-v(t_1)|_X = left| int_{t_1}^{t_2} u'(s) , text{d}s right|_X leq int_a^b |u'(s)|_X , text{d}s$$
for all $a leq t_1 leq t_2 leq b$, showing $t mapsto v(t):I to X$ is continuous. Since $u'=v'$, we have $v=u+c$, $c in X$, and therefore $u$ is continuous, too.
Now, you can simply prove the continuity of the embedding, i.e. $|u|_{C(I;X)} leq C |u|_{W^{1,1}(I;X)}$, you can look at the given references.
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
You are mostly asking about the embedding
$$W^{1,1}(I;X) hookrightarrow C(I;X),$$
which is quite standard and can be seen in most books that deal with Bochner spaces and PDES, see e.g. Roubicek or Cazenave&Haraux.
Let $I=[a,b]$ be some interval and let $u in W^{1,1}(I;X)$, then $u' in L^1(I;X)$ and we set $v(t):=int_a^t u'(s) , text{d}s$. Then
$$| v(t_2)-v(t_1)|_X = left| int_{t_1}^{t_2} u'(s) , text{d}s right|_X leq int_a^b |u'(s)|_X , text{d}s$$
for all $a leq t_1 leq t_2 leq b$, showing $t mapsto v(t):I to X$ is continuous. Since $u'=v'$, we have $v=u+c$, $c in X$, and therefore $u$ is continuous, too.
Now, you can simply prove the continuity of the embedding, i.e. $|u|_{C(I;X)} leq C |u|_{W^{1,1}(I;X)}$, you can look at the given references.
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
$endgroup$
add a comment |
$begingroup$
You are mostly asking about the embedding
$$W^{1,1}(I;X) hookrightarrow C(I;X),$$
which is quite standard and can be seen in most books that deal with Bochner spaces and PDES, see e.g. Roubicek or Cazenave&Haraux.
Let $I=[a,b]$ be some interval and let $u in W^{1,1}(I;X)$, then $u' in L^1(I;X)$ and we set $v(t):=int_a^t u'(s) , text{d}s$. Then
$$| v(t_2)-v(t_1)|_X = left| int_{t_1}^{t_2} u'(s) , text{d}s right|_X leq int_a^b |u'(s)|_X , text{d}s$$
for all $a leq t_1 leq t_2 leq b$, showing $t mapsto v(t):I to X$ is continuous. Since $u'=v'$, we have $v=u+c$, $c in X$, and therefore $u$ is continuous, too.
Now, you can simply prove the continuity of the embedding, i.e. $|u|_{C(I;X)} leq C |u|_{W^{1,1}(I;X)}$, you can look at the given references.
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
$endgroup$
add a comment |
$begingroup$
You are mostly asking about the embedding
$$W^{1,1}(I;X) hookrightarrow C(I;X),$$
which is quite standard and can be seen in most books that deal with Bochner spaces and PDES, see e.g. Roubicek or Cazenave&Haraux.
Let $I=[a,b]$ be some interval and let $u in W^{1,1}(I;X)$, then $u' in L^1(I;X)$ and we set $v(t):=int_a^t u'(s) , text{d}s$. Then
$$| v(t_2)-v(t_1)|_X = left| int_{t_1}^{t_2} u'(s) , text{d}s right|_X leq int_a^b |u'(s)|_X , text{d}s$$
for all $a leq t_1 leq t_2 leq b$, showing $t mapsto v(t):I to X$ is continuous. Since $u'=v'$, we have $v=u+c$, $c in X$, and therefore $u$ is continuous, too.
Now, you can simply prove the continuity of the embedding, i.e. $|u|_{C(I;X)} leq C |u|_{W^{1,1}(I;X)}$, you can look at the given references.
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
$endgroup$
You are mostly asking about the embedding
$$W^{1,1}(I;X) hookrightarrow C(I;X),$$
which is quite standard and can be seen in most books that deal with Bochner spaces and PDES, see e.g. Roubicek or Cazenave&Haraux.
Let $I=[a,b]$ be some interval and let $u in W^{1,1}(I;X)$, then $u' in L^1(I;X)$ and we set $v(t):=int_a^t u'(s) , text{d}s$. Then
$$| v(t_2)-v(t_1)|_X = left| int_{t_1}^{t_2} u'(s) , text{d}s right|_X leq int_a^b |u'(s)|_X , text{d}s$$
for all $a leq t_1 leq t_2 leq b$, showing $t mapsto v(t):I to X$ is continuous. Since $u'=v'$, we have $v=u+c$, $c in X$, and therefore $u$ is continuous, too.
Now, you can simply prove the continuity of the embedding, i.e. $|u|_{C(I;X)} leq C |u|_{W^{1,1}(I;X)}$, you can look at the given references.
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
answered Dec 23 '18 at 10:43
MarvinMarvin
2,5363920
2,5363920
add a comment |
add a comment |
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$begingroup$
How is $u'$ defined when you don't know that $u$ is continuous?
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 6:01
$begingroup$
@KaviRamaMurthy Now it is correct task. I am sorry for wrong interpretation
$endgroup$
– lojdmoj
Dec 11 '18 at 10:49