# Lax-Milgram Theorem Proof

1. Jun 5, 2013

### muzialis

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. Jun 5, 2013

### micromass

Staff Emeritus
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 $\|~\|_E$.

3. Jun 10, 2013

### muzialis

Micromass,

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