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Equivalence of norms

  1. Aug 24, 2009 #1

    quasar987

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    Let H be a separable Hilbert space and let {e_k} be a Hilbert basis (aka total orthonormal sequence) for H. Then

    [tex]|||u|||_1:=\sum_{k=1}^{+\infty}\frac{1}{2^k}|(e_k,u)|[/tex]

    is a norm. If {f_k} is another Hilbert basis, we get another norm by setting

    [tex]|||u|||_2:=\sum_{k=1}^{+\infty}\frac{1}{2^k}|(f_k,u)|[/tex]

    How to show that these two norm are equivalent???
     
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  3. Aug 24, 2009 #2

    morphism

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    Try using the fact that

    [tex]f_k = \sum_l (f_k,e_l) e_l.[/tex]
     
  4. Aug 24, 2009 #3

    quasar987

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    I did:

    [tex]|||u|||_2 =\sum_k\frac{1}{2^k}|(f_k,u)|=\sum_k\frac{1}{2^k}\left|\left(\sum_l(f_k,e_l)e_l,u\right)\right| = \sum_k\frac{1}{2^k}\left|\sum_l(f_k,e_l)(e_l,u)\right|[/tex]
    [tex]\leq\sum_k\frac{1}{2^k}\sum_l|(f_k,e_l)||(e_l,u)|[/tex]

    and then I'm stuck...
     
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