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I am wondering if it is possible to say that if a symmetric differential operator is densely defined then the operator is self-adjoint indeed?

More Precisely, Let $A:D(A)(\subset H)\to H$ a densely defined operator on the Hilber space $H$ and we have $<Ax,y>=<Ay,x>$; is $A$ a self-adjoint operator?

What more condition does it need to become self adjoint?

Please help me with this, I think in an infinite dimensional spaces every assertion would has some subtleties as in this case?

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no of course not. en effet, you must require $A=A^\ast$, you are missing $D(A)=D(A^\ast)$.

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Exactly, you are right, it requires $D(A)=D(A^*)$; but I emphasized on densely defined operator.

In fact, is it possible to say that when an operator is densely defined on a suitable Hilbert space and also symmetric then $D(A)=D(A^*)$?

what more condition does it need to give $D(A)=D(A^*)$?

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If $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is symmetric on its domain, then $A$ is selfadjoint iff $A\pm iI$ are surjective. If these operators are surjective, then the domain is automatically dense, which saves some checking.

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  • $\begingroup$ Thank you, but, Why does the surjectivity of $A\pm iI$ give the denseness of domain? $\endgroup$ – Sara Winslet Oct 12 '14 at 9:01
  • $\begingroup$ Suppose $A-iI$ is surjective and $A$ is symmetric. Suppose $y \perp \mathcal{D}(A)$. Then $y=(A-iI)x$ for some $x\in\mathcal{D}(A)$, and $y\perp x$, which gives $((A-iI)x,x)=0$. However, the imaginary part of $((A-iI)x,x)$ is $-\|x\|^{2}$, which must be $0$. So $x=0$, which implies $y=(A-iI)x=0$. Hence $\mathcal{D}(A)^{\perp}=\{0\}$. $\endgroup$ – DisintegratingByParts Oct 12 '14 at 11:30

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