# Two versions of Lax-Milgram theorem

I'm having some troubles differentiating between two versions of Lax-Milgram theorem, one shown in my class and one that I saw is common on the internet.

Let $H$ be hilbert space, $B$ bilinear form on $H$ which is bounded and coercive.

1. (my class): There is a unique bounded linear operator $S$ with an invertible bounded operator such as $B(Sx,y)=<x,y>$.

2. (interenet): let $\phi$ be a linear functional on $H$. so there is a unique $x \in H$ that maked $\phi(y)=B(y,x)$ for every $y\in H$

hope you could show me at least why the first definition leads to the second. I really tryed and had absolutely no idea (I used riesz' theorem and took $x$ such that for every $y \in H$ $\phi (y)=<y,x>=\overline{<x,y>}=\overline{B(Sx,y)}$, and from here i can't unfortunately lose the conjuget and replace the order of $Sx$ and $y$, which would leed to the solution.

Thanks

• The second is not Lax-Milgram, but the Riesz representation theorem en.wikipedia.org/wiki/Riesz_representation_theorem. – gerw Nov 15 '14 at 17:51
• @gerw: you are right of course, i fixed my questions. It should be $\phi(y)=B(y,x)$ instead of $\phi(y)=<y,x>$ – user188400 Nov 15 '14 at 17:57

The problem is that the $B$ in the second formulation does not equal the $B$ in the first one.
Let us proof the second one by using the first one. Set $\hat B(x,y) = \overline{B(y,x)}$. By using the first theorem, you find an operator $\hat S$ associated to $\hat B$. Then, \begin{equation*} \phi(y) = \langle y,x\rangle = \overline{\langle x, y \rangle} = \overline{\hat B(\hat S x, y)} = B(y, \hat S x). \end{equation*}