Show subspace of H^1[0,1] is closed

[quasar987]

[SOLVED] Show subspace of H^1[0,1] is closed

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 $$||u_n-u||^2=(u_n-u,u_n-u)=(u_n,u_n)-2(u_n,u)+(u,u)$$,

$$\int_0^1[(u_n')^2+u_n^2] - 2\int_0^1[u_n'u'+u_nu]+\int_0^1[(u')^2+u^2]\rightarrow 0$$I don't see how to retrieve u(0)=u(1).

 

[taqseer khan]

solve-

how to construct an example of a non-separable Hilbert Space?

can anyone suggest me something please?