Let $H=(H,(\cdot, \cdot))$ be a complex Hilbert space and $A:D(A)\subset H \longrightarrow H$ be a linear operator on $H$. I know there is a functional $ L: D(A) \longrightarrow \mathbb{C}$, associated with $ A $ given by $$L(u)=(A(u),u),\; \forall \; u \in D(A),$$ where $L$ is called the quadratic form associated to $A$.
Is the converse true?
That is, given a functional $\mathcal{L}:D(\mathcal{L})\subset H \longrightarrow \mathbb{C}$, then is there a linear operator $B: D(B) \subset H \longrightarrow H$ of $H$ associated with $\mathcal{L}$?
The answer is no.
For example, let $f$ be a functional that assigns $1$ to all of the vectors. Then we would like to have a linear operator $A$ with the property that $$1=(Ax|x) \, \forall x \in H$$ which is absurd.