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Hilbert Space

  1. Oct 7, 2006 #1
    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 [tex]H^{1}[/tex] is a Hlbert space.

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

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

    Alternatively, is there a way to prove that the space [tex]V:=\{v \in H^{1}, v(1) = 0\}[/tex] 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 [tex]H^{1}[/tex]).

    Thanks.
     
    Last edited: Oct 7, 2006
  2. jcsd
  3. Oct 7, 2006 #2

    StatusX

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    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.
     
  4. Oct 8, 2006 #3
    thanks for the reply StatusX...

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

    thanks again.
     
  5. Oct 8, 2006 #4

    AKG

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    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.
     
  6. Oct 8, 2006 #5
    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
  7. Oct 8, 2006 #6

    AKG

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    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.
     
  8. Oct 8, 2006 #7
    if my understanding is correct, isn't the [tex]l^{2}[/tex] space a space of measurable functions (and hence the [tex]h^{1}[/tex] 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?
     
  9. Oct 8, 2006 #8

    StatusX

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    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
  10. Oct 8, 2006 #9
    the inner product of H1 is given by

    [tex](u, v)_{H^{1}} = \int u' v' (x) dx [/tex]

    the norm is

    [tex] \| \cdot \|_{H^{1}} = \sqrt{(\cdot, \cdot)_{H^{1}}[/tex]

    how exactly do you define a sequence of functions v_n?

    thanks.
     
    Last edited: Oct 9, 2006
  11. Oct 10, 2006 #10

    StatusX

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    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.
     
  12. Oct 10, 2006 #11

    matt grime

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    l^2 is the space of square summable series. L^2(X) is the space of square intergrable functions on X.
     
  13. Oct 11, 2006 #12
    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.
     
  14. Oct 11, 2006 #13

    AKG

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    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.
     
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