# bilinear form on Hilbert space

$H$ 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?

• Yes you are right. – Alexander Nov 11 '12 at 19:42

## 1 Answer

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 $a(u,v)=L(v)$.
• 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