Is every symmetric bilinear form on a Hilbert space a weighted inner product?

Question

Is every symmetric bilinear form on a Hilbert space a weighted inner product? i.e. can I write that $b(u,v) = (wu,v)_H$ for all $u, v \in H$? I am not sure about this. Maybe something to do with Riesz theorem..

Answer 1

Since you're interested in bilinear forms, I'll assume you're working with a real Hilbert space $H$; in the complex case, replace "bilinear" with "sesquilinear."

1. First, suppose that you have a bounded bilinear form $b$, so that there exists some constant $C \geq 0$ such that $b(u,v) \leq C\|u\|\|v\|$ for all $u$, $v \in H$. Then for all $u \in H$, $b(u,\cdot)$ defines a bounded linear functional on $H$ (why?), so that by the Riesz theorem, $b(u,\cdot) = (w(u),\cdot)_H$ for a unique $w(u) \in H$; check (how?) that $u \mapsto w(u)$ defines a bounded linear operator on $H$.
2. Now, suppose in addition that $b$ is symmetric. Check that $w$ is therefore self-adjoint. In fact, $b$ is symmetric if and only if $w$ is self-adjoint (why?).
3. Finally, suppose in addition that $b$ is positive-definite, i.e., $b(u,u) > 0$ for all $u \neq 0$. Check, then, that $w$ is therefore positive-definite, so that $b$ is indeed a weighted inner product. In fact, $b$ is positive-definite (and thus a weighted inner product) if and only if $w$ is positive definite (why?).

Throughout all this, all you really need is to remember what the Riesz theorem says, and what it means for an operator to be self-adjoint and positive-definite. In particular, observe the role (indeed, necessity and sufficiency) of the hypotheses of boundedness, symmetry (which you already had), and positive-definiteness.