Let $b:H \times H \to \mathbb{R}$ be a bounded bilinear form on Hilbert space $H$. Fix $u \in H$. Then $b(u, \cdot):H \to \mathbb{R}$ is bounded so $b(u,\cdot) \in H^*.$ Then $b(u,\cdot) = F_u(\cdot)$ where $F_u \in H^*$.
Right? Isn't this a kind of Riesz representation theorem without coercivity?
1 Answer
The Riesz Representation Theorem holds for any bounded linear functional on a Hilbert space. There is no mention of coercivity in the theorem.
Besides the fact that a functional can never be coercive, as it always has a nontrivial kernel.
Edit: attempt at clarification.
These are the three relevant results to the discussion:
Riesz Representation Theorem: given $f\in H^*$, there exists $v\in H$ such that $f(x)=\langle x,v\rangle$ for all $x\in H$.
Corollary of the RRT: If $b$ is bilinear and bounded on $H$ and $u\in H$, then there exists $v\in H$ such that $b(u,x)=\langle x,v\rangle$ for all $x\in H$.
Lax-Milgram: given $b$ is bilinear, bounded, and coercive, and $f\in H^*$, then there exists $v\in H$ such that $f(x)=b(x,v)$ for all $x\in H$.
So, the RRT is a particular case of LM, where $b$ is taken to be the scalar product.
On the other hand, the corollary---which seems to be the result quoted in the question---does not attempt to represent every functional using $b$, but rather represent $b$ when one coordinate is fixed.
