Let
- $d\in\mathbb N$
- $\Lambda\subseteq\mathbb R^d$ be open
- $V$ be a closed subspace of $H_0^1(\Lambda,\mathbb R^d)$ and $$\mathfrak a(u,v):=\sum_{i=1}^d\langle\nabla u_i,\nabla v_i\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in V$$
It's easy to see that $$A_0u:=\mathfrak a(\;\cdot\;,v)\;\;\;\text{for }u\in V$$ is a bounded linear operator from $V$ to $V'$. On the other hand, $V$ equipped with the restriction of the inner product inherited from $H_0^1(\Lambda,\mathbb R^d)$ is a Hilbert space and hence $A_0$ can be considered as being a bounded linear operator $\tilde A_0$ from $V$ to $V$ as well.
The question is: I've read that $A_0$ is self-adjoint, but what's meant by that?
The notion of being self-adjoint that I know is only defined for bounded linear operators from a Hilbert space $H$ to $H$. So, one option would be that they mean that $\tilde A_0$ is self-adjoint.
But there is a notion of being symmetric, which is defined for operators from a space $U$ to a Hilbert space $H$ where $U$ is a dense subspace of $H$. So, another option would be that they mean that $A$ is symmetric.
And maybe the distinction is irrelevant and both interpretations are essentially the same.