Let
- $H$ be a $\mathbb R$-Hilbert space
- $D(\mathfrak a)$ be a dense subspace of $H$
- $\mathfrak a:D(\mathfrak a)\times D(\mathfrak a)\to\mathbb R$ be bounded, i.e. $\exists c\ge 0$ with $$\left|\mathfrak a(u,v)\right|\le c\left\|u\right\|_{\mathfrak a}\left\|v\right\|_{\mathfrak a}\;\;\;\text{for all }u,v\in D(\mathfrak a)$$ where $$\left\|u\right\|_{\mathfrak a}:=\sqrt{\mathfrak a(u,u)+\left\|u\right\|_H^2}\;\;\;\text{for all }u\in D(\mathfrak a)$$
Now, let $$D(A):=\left\{u\in D(\mathfrak a)\mid\exists v\in H:\mathfrak a(\;\cdot\;,u)=\left.\langle\;\cdot\;,v\rangle_H\right|_{D(\mathfrak a)}\right\}\;.\tag 1$$ It's easy to see that $$u\in D(A)\Leftrightarrow u\in D(\mathfrak a)\text{ and }\mathfrak a(\;\cdot\;,u)\text{ is }H\text{-continuous}\;.\tag 2$$
I've seen that many people write $$D(A)=\left\{u\in D(\mathfrak a):\mathfrak a(\;\cdot\;,u)\in H\right\}\;.\tag 3$$ So, my question is: Why are $(1)$ and $(2)$ equivalent?
I guess $(2)$ has to be interpreted in a suitable sense. Clearly, if $u\in D(A)$, we know that $u\in D(\mathfrak a)$ such that $\mathfrak a(\;\cdot\;,u)$ is $H$-continuous. Since $D(\mathfrak a)$ is dense in $H$, there is a unique $F\in H'$ with $$\mathfrak a(\;\cdot\;,u)=\left.F\right|_{D(\mathfrak a)}$$ by the bounded linear transformation theorem. Thus, there is a unique extension of $\mathfrak a(\;\cdot\;,u)$ to $H$. Since $H'\cong H$ we could justify the notation $\mathfrak a(\;\cdot\;,u)\in H$.
For the other direction, $\mathfrak a(\;\cdot\;,u)\in H$ would imply the existence of a unique extension $F\in H'$ of $\mathfrak a(\;\cdot\;,u)$ to $H$. Since $H'\cong H$, there is a unique $v\in H$ with $$F=\langle\;\cdot\;,v\rangle_H$$ and hence $$\mathfrak a(\;\cdot\;,u)=\left.\langle\;\cdot\;,v\rangle_H\right|_{D(\mathfrak a)}\;.$$
Is this the way we need to understand $(3)$?
If you would ask me, it is meant in the sense you mentioned. Since a Hilbert space is isometric isomorphic to its dual space due to the Riesz representation theorem you can identify the functionals in $H'$ with their riesz representers in $H$. I did that frequently in my bachelor thesis. It is not the most formal way but you can give shorter definitions and assumptions by using this identification.