Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $A: D(A) \subset H \to H$ be an unbounded linear operator that is densely defined. One says that
- $A$ is symmetric IFF $\langle Au, v \rangle = \langle u, A v \rangle$ for all $u,v\in D(A)$.
- $A$ is self-adjoint IFF $D(A) =D(A^*)$ and $A^*=A$.
I would like to verify a claim in Remark 7 at page 193 of Brezis' Functional Analysis, i.e.,
$A$ is symmetric IFF $A \subset A^*$, i.e., $D(A) \subset D(A^*)$ and $A^*=A$ on $D(A)$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
- $\implies$
Let $A$ be symmetric. We have $$ D(A^*) := \{v \in H : \exists c >0 \text{ s.t. } |\langle v, Au \rangle | \le c |u| \quad \forall u \in D(A)\}. $$
First, we prove $D(A) \subset D(A^*)$. Let $v \in D(A)$. By symmetry of $A$ and Cauchy-Schwarz inequality, $$ |\langle v, Au \rangle | = |\langle Av, u \rangle | \le |Au| \cdot |v| \quad \forall u\in D(A). $$
Then $v \in D(A^*)$.
Second, we prove $A^*=A$ on $D(A)$. We have $$ \begin{align*} \langle Au, v \rangle &= \langle u, Av \rangle \quad \forall u,v\in D(A), \\ \langle Au, v \rangle &= \langle u, A^* v \rangle \quad \forall u\in D(A), v\in D(A^*). \\ \end{align*} $$
Because $D(A) \subset D(A^*)$, $$ \langle u, Av \rangle = \langle u, A^* v \rangle \quad \forall u,v\in D(A). $$
Let $v \in D(A)$. Then $$ \langle u, Av \rangle = \langle u, A^* v \rangle \quad \forall u\in D(A). $$
Because $D(A)$ is dense in $H$, we get $Av = A^*v$.
- $\impliedby$
By definition of $A^*$, $$ \langle Au, v \rangle = \langle u, A^* v \rangle \quad \forall u\in D(A), v\in D(A^*). $$
Let $A \subset A^*$. It follows from $D(A) \subset D(A^*)$ that $$ \langle Au, v \rangle = \langle u, A^* v \rangle \quad \forall u\in D(A), v\in D(A). $$
It follows from $A^*=A$ on $D(A)$ that $$ \langle Au, v \rangle = \langle u, A v \rangle \quad \forall u\in D(A), v\in D(A). $$