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[SOLVED] Show subspace of H^1[0,1] is closed
I have an assignment that deals with some Sobolev spaces but I have never worked with them before. Only the definitions are given.
Consider the Sobolev space
W^{1,2}([0,1])=H^1([0,1])=\{u\in C([0,1]): \mbox{ there exists } u'\in L^2[0,1] \mbox{ such that } u(t)-u(0)=\int_0^tu'(s)ds \ \ \forall t \in [0,1]\}
It is a Hilbert space with the inner product
(u,v)=\int_0^1[u'v'+uv]
I am trying to show that the following subspace of H^1 is closed:
E=\{u\in H^1([0,1]):u(0)=u(1)\}
So I say, let u_n-->u (in H^1); we must show that u(0)=u(1).
u_n-->u means that ||u_n - u||-->0 in the norm induced by the above inner product. So it means
\int_0^1[(u_n-u)'^2+(u_n-u)^2] \rightarrow 0
or, using the fact that ||u_n-u||^2=(u_n-u,u_n-u)=(u_n,u_n)-2(u_n,u)+(u,u),
\int_0^1[(u_n')^2+u_n^2] - 2\int_0^1[u_n'u'+u_nu]+\int_0^1[(u')^2+u^2]\rightarrow 0I don't see how to retrieve u(0)=u(1).
Homework Statement
I have an assignment that deals with some Sobolev spaces but I have never worked with them before. Only the definitions are given.
Consider the Sobolev space
W^{1,2}([0,1])=H^1([0,1])=\{u\in C([0,1]): \mbox{ there exists } u'\in L^2[0,1] \mbox{ such that } u(t)-u(0)=\int_0^tu'(s)ds \ \ \forall t \in [0,1]\}
It is a Hilbert space with the inner product
(u,v)=\int_0^1[u'v'+uv]
I am trying to show that the following subspace of H^1 is closed:
E=\{u\in H^1([0,1]):u(0)=u(1)\}
The Attempt at a Solution
So I say, let u_n-->u (in H^1); we must show that u(0)=u(1).
u_n-->u means that ||u_n - u||-->0 in the norm induced by the above inner product. So it means
\int_0^1[(u_n-u)'^2+(u_n-u)^2] \rightarrow 0
or, using the fact that ||u_n-u||^2=(u_n-u,u_n-u)=(u_n,u_n)-2(u_n,u)+(u,u),
\int_0^1[(u_n')^2+u_n^2] - 2\int_0^1[u_n'u'+u_nu]+\int_0^1[(u')^2+u^2]\rightarrow 0I don't see how to retrieve u(0)=u(1).
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