Is $AA^*$ and $A^*A$ self-adjoint?

if I have a densely defined closed linear operator $A$ and $A^* = -A$(same domain also closed). Is this sufficient that $AA^*$ and $A^*A$ are proper self-adjoint operators, assuming that we can also define both of them on a dense domain of our underlying Hilbert space. If this is true( I guess so), could anybody state the actual theorem that shows this or better: explain why the fact that the operators are closed and densely defined are important here?

• Doesn't $\langle ABx,y\rangle=\langle Bx,A^*y\rangle=\langle x,B^*A^*y\rangle \implies (AB)^*=B^*A^*$? – blue Aug 2 '14 at 9:53
• well, I thought that an operator $T$ is self-adjoint if the operator is closed, symmetric, densely-defined and $T-i$,$T+i$ are onto. I think your argument only shows symmetry. – user167575 Aug 2 '14 at 10:41
• Since $A^*=-A$, $H=iA$ is self-adjoint and so is $H^2$. Now $AA^*$ and ${A^*}A$ are defined and coincide on the domain of $H^2$. – Urgje Aug 2 '14 at 12:10
• @Urgje okay, you did this by using what $A^*$ actually is, in this example, thank you. Is the general case also true, if yes, could I encourage you to post this as an answer? I think this will be also more valuable to other people seeing this thread who may be more interested in the general case :-) – user167575 Aug 2 '14 at 12:12

Lemma: Let $B : \mathcal{D}(B)\subseteq X\rightarrow X$ be a symmetric linear operator on a Hilbert space $X$ for which $(Bx,x) \ge 0$. If $I+B$ is surjective, then $B$ is closed, densely-defined and selfadjoint.
Proof: Let $B$ as stated. Suppose that $B+I$ is surjective. To show that $B$ must be densely-defined, suppose that $y \perp \mathcal{D}(B)$ and write $y=(I+B)x$ for some $x\in \mathcal{D})$. Then $y \perp x$ gives $$0 = (y,x)=((I+B)x,x) \ge \|x\|^{2} \implies x = 0 \implies y = (I+B)x=0.$$ So the domain of $B$ must be dense. $B$ is closed because $(I+B)^{-1}$ is bounded and defined on all of $X$, owing to the fact that $$\|x\|^{2} \le ((I+B)x,x) \le \|(I+B)x\|\|x\|,\;\;\; x \in \mathcal{D}(B).$$ Therefore $B^{\star}$ is also closed and densely-defined. Because $B$ is symmetric, then $B\preceq B^{\star}$ (meaning that $B^{\star}$ extends $B$.) To see that $B=B^{\star}$, suppose that $y \in \mathcal{D}(B^{\star})$. The proof is finished by showing that $y \in \mathcal{D}(B)$: to this end, note that there exists $z \in \mathcal{D}(B)$ such that $(I+B^{\star})y=(I+B)z$. So, $$((I+B)x,y)=(x,(I+B^{\star})y)=(x,(I+B)z)=((I+B)x,z),\;\;\; x \in \mathcal{D}(B).$$ Because $(I+B)$ is surjective, then $z=y \in \mathcal{D}(B)$. So $B=B^{\star}$. $\Box$
Theorem [von Neumann]: Let $X$ be a Hilbert space, and let $A : \mathcal{D}(A)\subseteq X\rightarrow X$ be a closed densely-defined linear operator. Define $A^{\star}A$ on the domain $$\mathcal{D}(A^{\star}A)=\{ x \in \mathcal{D}(A) : Ax \in \mathcal{D}(A^{\star})\}.$$ Then $A^{\star}A$ is a closed densely-defined selfadjoint linear operator. Because $A^{\star}$ is also closed and densely-defined, then $AA^{\star}$ is also a closed densely-defined selfadjoint linear operator.
Proof: Let $J : X\times X\rightarrow X\times X$ be defined as $J\langle x,y\rangle=\langle-y,x\rangle$. Let $\mathcal{G}(A)=\{ \langle x,Ax\rangle : x\in\mathcal{D}(A)\}\subset X\times X$ be the graph of $A$. Then it is well-known that $\mathcal{G}(A^{\star})=J(\mathcal{G}(A)^{\perp})=(J\mathcal{G}(A))^{\perp}$, where the orthogonal complement is taken in $X\times X$. Because $X\times X$ is a Hilbert space, then $X\times X = \mathcal{M}\oplus\mathcal{M}^{\perp}$ for any closed linear subspace $\mathcal{M}$ of $X\times X$. Therefore, $$X\times X = \mathcal{G}(A)\oplus J\mathcal{G}(A^{\star}).$$ Hence, for every $a \in X$, there are unique $x \in \mathcal{D}(A)$, $y\in\mathcal{D}(A^{\star})$ such that $$\langle a,0\rangle = \langle x,Ax\rangle + \langle -A^{\star}y,y\rangle, \\ a = x - A^{\star}y,\;\; y=-Ax.$$ So $x \in \mathcal{D}(A^{\star}A)$ and $(I+A^{\star}A)x=a$. That is, $I+A^{\star}A$ is surjective on its natural domain. Now the previous lemma applies to $B=A^{\star}A$. $\;\;\Box$