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Lax-Milgram Theorem Proof

  1. Jun 5, 2013 #1
    Hi All,

    I am going through the proof of Lax-Milgram's theorem.
    A lemma states, "H is a Hilbert space, let a(.,.) be a symmetric bilinear form, continous on H and coercive on W, subset of H. then, {W, a(.,.)} is a Hilbert Space".

    This is proved as follows:
    Being a(.,.) coercive, it is also an inner product for W.
    Let us define $$\parallel \cdot \parallel_{E} = \sqrt{a(w,w}$$ and take a Cauchy sequence $$w_n$$.
    This will also be a Cauchy sequence in H because coercitivity implies $$ a (w,w) \geq \alpha \parallel w \parallel ^{2} _{H}$$.
    H is a Hilbert space, then complete, so $$w_n \to w$$, with $$w \in H$$.
    Since W is closed in H, $$w \in W$$.
    Hence, $${ W, \parallel \cdot \parallel_{E} }$$ is complete.

    What use is mad then of the continuity assumption?

    Thanks
     
  2. jcsd
  3. Jun 5, 2013 #2

    micromass

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    So you took a Cauchy sequence for ##\|~\|_E##. You showed that it is convergent for the norm ##\|~\|_H## and that the limit is in ##W##. But you still need to show that the convergence is also for the norm ##\|~\|_E##.
     
  4. Jun 10, 2013 #3
    Micromass,

    your point is of course valid. Thank you very much for the help, as usual.
     
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