Periodic boundary solution forms a closed subspace in $H^1$
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I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
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add a comment |
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I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
$endgroup$
add a comment |
$begingroup$
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
$endgroup$
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
functional-analysis
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
asked Dec 2 '18 at 16:31
lxnllllxnlll
464
464
add a comment |
add a comment |
2 Answers
2
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oldest
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The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
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Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 '18 at 17:52
eddieeddie
525110
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
$endgroup$
add a comment |
$begingroup$
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
$endgroup$
add a comment |
$begingroup$
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
$endgroup$
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
answered Dec 3 '18 at 7:27
gerwgerw
19.5k11334
19.5k11334
add a comment |
add a comment |
$begingroup$
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 '18 at 17:52
eddieeddie
525110
$endgroup$
add a comment |
$begingroup$
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 '18 at 17:52
eddieeddie
525110
$endgroup$
add a comment |
$begingroup$
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 '18 at 17:52
eddieeddie
525110
$endgroup$
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 '18 at 17:52
eddieeddie
525110
answered Dec 2 '18 at 17:52
eddieeddie
525110
answered Dec 2 '18 at 17:52
eddieeddie
525110
answered Dec 2 '18 at 17:52
eddieeddie
525110
525110
add a comment |
add a comment |
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