# Hilbert Space

1. Oct 7, 2006

### island-boy

Fact 1: we know that a closed subspace of a Hilbert Space is also a Hilbert Space.

Fact 2: we know that the Sobolev Space $$H^{1}$$ is a Hlbert space.

How do I show that the space $$V:=\{v \in H^{1}, v(1) = 0\}$$ is a Hilbert space?

Is V automatically a closed subspace of $$H^{1}$$? How do I show this? cause I can't see how it is so.

Alternatively, is there a way to prove that the space $$V:=\{v \in H^{1}, v(1) = 0\}$$ is a Hilbert space, without using facts 1 and 2? that is how do I show that V is complete? (it is automatically an inner product space because it is in $$H^{1}$$).

Thanks.

Last edited: Oct 7, 2006
2. Oct 7, 2006

### StatusX

Show the function H_1->R given by evalutation at a point is continuous. Then the preimage of a point, which is a closed set, must be closed.

3. Oct 8, 2006

### island-boy

how about for the space $$V:= \{v \in H^{1}, v(0) = v(1) \}$$?
How do I prove that this is a Hilbert Space?...more specifically, how do I prove that this is a closed subspace of $$H^{1}$$?

thanks again.

4. Oct 8, 2006

### AKG

If you understand post 2, you can do this question. He asked you to find a function H1 to R with some certain properties. Try explicitly writing down a function with those properties.

5. Oct 8, 2006

### island-boy

okay, lets see if I understood this correctly...

for the first one, I have to find a function that maps from H1 to v(1) (which is equal to 0 and hence is in R).

For the second, I have to find a function that maps from H1 to v(0) - v(1) (which is also equal to 0 and hence is in R).

And since 0 is a closed set, its preimage, which is in H1, is also a closed set, and therefore also a Hilbert Space.

is this correct?

also, do I have to specifically define a function, like f(x) = something.... where x is in H1?

Last edited: Oct 8, 2006
6. Oct 8, 2006

### AKG

That doesn't really make sense.
You should, yes.

And prove the functions you find are continuous, otherwise you can't argue that the preimages are closed.

0 is not a closed set, it's not even a set. {0} is, however.

And just in case it matters (I don't know): it doesn't have to map into R, it can map into any T1 space, hence into any Hausdorff space.

7. Oct 8, 2006

### island-boy

if my understanding is correct, isn't the $$l^{2}$$ space a space of measurable functions (and hence the $$h^{1}$$ space also a space of measruable functions).

My confusion lies is this: I have to find a function that maps the space of measurable function (h1) to the real space. Is this correct?
or is the V-space(the space of v in h1 where v(0) = 0) the function in question?

8. Oct 8, 2006

### StatusX

The function sends v, an element of H1, (which is a function itself) to v(1) (or v(1)-v(0)), which is an element of R (or C). To show it is continuous, show that if a sequence of functions v_n has limit v in H1, then the sequence in R (or C) given by v_n(1) has limit v(1). First you'll have to be more specific about what you mean by H1 (eg, what is the norm?).

Last edited: Oct 8, 2006
9. Oct 8, 2006

### island-boy

the inner product of H1 is given by

$$(u, v)_{H^{1}} = \int u' v' (x) dx$$

the norm is

$$\| \cdot \|_{H^{1}} = \sqrt{(\cdot, \cdot)_{H^{1}}$$

how exactly do you define a sequence of functions v_n?

thanks.

Last edited: Oct 9, 2006
10. Oct 10, 2006

### StatusX

You don't have to explicitly define a sequence. Just show that if {v_n} is a sequence such that ||v_n-v||->0, then |v_n(1)-v(1)|->0.

11. Oct 10, 2006

### matt grime

l^2 is the space of square summable series. L^2(X) is the space of square intergrable functions on X.

12. Oct 11, 2006

### island-boy

hi guys,

just submitted the work today...am not sure if what I did was 100% correct, but at least I know it was generally correct.

I did try to show that v_n(1) -> v(1) if v_n conerges to v...something like that :)

just want to say thanks for the help.

13. Oct 11, 2006

### AKG

H1 is a set of functions from X to Y (I don't know what X and Y are). Shouldn't you have defined a function f : H1 -> Y defined by: f(v) = v(1). It's clear that V = f-1({0}) so you would only need to prove two things: a) that {0} is closed in Y, and b) that f is continuous. For the second problem, you would define a function g : H1 -> Y by g(v) = v(0) - v(1). The V here is g-1({0}), so assuming you've already proved {0} to be closed in Y, you now just have to show g is continuous.