Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$H$ Hilbert space, $a:H\times H \to\mathbb R$ a bilinear form with $|a(x,y)|\le C\|x\| \|y\|$ and $a(x,x)\ge\alpha\|x\|^2$ $\forall x,y\in H$ , $C>0, \alpha>0$, $L$ continuous linear functional on $H$

I already could prove that there exists a linear operator S, such that: $a(x,y)=(Sx,y)$

How can it be followed that there exists an unique $u\in H$, such that $\forall v\in H: a(u,v)=L(v)$ ?

The second thing I am interested in is: for $a(\cdot,\cdot)$ symmetric I define $\phi(x)=\frac{1}{2}a(x,x)-L(x)\forall x\in H$. Now it should be possible to characterize $u$ through $\phi(u)=\min_{x\in H}\phi(x)$ But how?

share|improve this question
    
Yes you are right. –  Alexander Nov 11 '12 at 19:42

1 Answer 1

up vote 1 down vote accepted
  1. If $a$ is symmetric, then $(H,a)$ is Hilbert space (norm $p(x) := \sqrt{a(x,x)}$) and $L$ is also continuous with respect to the norm $p$. Apply Riesz representation theorem... If $a$ is not symmetric, the proof is a bit more complicated (see: Lax-Milgram-lemma, Proof)
  2. Assume that $u$ solves $a(u,v)=L(v)$ and set $\phi(v) := \frac{1}{2}a(v,v)-L(v)$. Then $$\phi(w) = \underbrace{\frac{1}{2} a(u,u)-L(u)}_{\phi(u)}+\underbrace{a(u,w-u)-L(w-u)}_{0}+\frac{1}{2}\underbrace{a(w-u,w-u)}_{\geq \alpha \cdot \|w-u\|^2>0}>\phi(u)$$ for all $w \not= u$. On the other hand: If $u= \min_v \phi(v)$, then $\mathbb{R} \ni J(u+t \cdot v)$ has a minimum in $t=0$ for all $v \in H$, thus $$\frac{dJ(u+t \cdot v)}{dt} \bigg|_{t=0} = 0$$ Straight-forward calculations yield $a(u,v)=L(v)$.
share|improve this answer
    
Thank you for your help. –  Alexander Nov 11 '12 at 20:20
    
where can I find the LaTeX command for the term groupings? –  user45793 Nov 11 '12 at 20:42
    
Do you mean \underbrace{x}{y}? –  saz Nov 11 '12 at 20:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.