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A formulation of continuity for bilinear forms

  1. Feb 1, 2008 #1

    quasar987

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    [SOLVED] A formulation of continuity for bilinear forms

    1. The problem statement, all variables and given/known data
    My HW assignment read "Let H be a real Hilbert space and a: H x H-->R be a coninuous coersive bilinear form (i.e.
    (i) a is linear in both arguments
    (ii) There exists M>0 such that |a(x,y)|<M||x|| ||y||
    (iii) there exists B such tthat a(x,x)>a||x||^2"

    So apparently, condition (ii) is the statement about continuity. But I fail to see how this statement is equivalent to "a is continuous".

    I see how (ii) here implies continuous, but not the opposite.

    3. The attempt at a solution

    Let z_n = (x_n,y_n)-->0. Then x_n-->0 and y_n-->0. So |a(x,y)|<M||x|| ||y|| implies a(x,y)-->0. a is thus continuous at 0, so it is so everywhere, being linear.
     
    Last edited: Feb 1, 2008
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  3. Feb 1, 2008 #2

    Dick

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    I'm not quite sure what the question is. It seems to have been omitted. But just looking at your argument, z_n=(x_n,y_n)->0 doesn't imply x_n->0 or y_n->0. Does it?
     
  4. Feb 2, 2008 #3

    quasar987

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    Well, I'm making use of the fact that if (M,d) is metric space, then the product topology on M x M is generated by the metric

    D((x1,y1),(x2,y2))=[d(x1,x2)² + d(y1,y2)²]^½

    I conclude that the norm on H x H is

    ||(x,y)|| = [||x||² + ||y||²]^½

    And now if (x_n,y_n)-->0, this means that

    [||x_n||² + ||y_n||²]^½ --> 0,

    which can only happen if x_n-->0 and y_n-->0.

    ------------

    You're right, I have not actually typed the question entirely. It is because I was confused by the fact that they seem to imply that condition (ii) is equivalent to continuity. While (ii) implies continuity, does continuity implies (ii)?
     
    Last edited: Feb 2, 2008
  5. Feb 2, 2008 #4

    Hurkyl

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    Well, there is the theorem that a linear functional on a Hilbert space is continuous if and only if it's bounded...
     
  6. Feb 2, 2008 #5

    quasar987

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    You mean bounded in the unit ball?

    In this case, you're right, it works. Because ||(x,y)|| = [||x||² + ||y||²]^½ <1 ==> ||x||, ||y||<1 and (ii) implies|a(x,y)|<M for all (x,y) in H x H such that ||(x,y)||<1.
     
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