proof of basic criterion for self-adjointness

  1. 1.

    (12) If A is self-adjoint and Ax=ix, then


    so x=0. Similarly we prove that Ax=-ix implies x=0. That A is closed follows from the fact that the adjoint of an operator is always closed.

  2. 2.

    (23) If 2 holds, then {0}=Ker(A*±i)*=Ker(Ai)*=Ran(Ai), so that RanAi is dense in . Also, since A is symmetric, for xD(A),


    because (Ax,ix)=(x,iA*x)=(x,iAx)=-(ix,Ax). Hence x(A+i)x, so that given a sequence xnD(A) such that (A+i)xny, we have that {(A+i)xn} is a Cauchy sequenceMathworldPlanetmathPlanetmath and thus {xn} itself is a Cauchy sequence. Hence {xn} converges to some x and since A is closed it follows that xD(A) and (A+i)x=y. This proves that yRan(A+i), so that Ran(A+i) is closed (and similarly, Ran(A-i) is closed. Thus Ran(A±i)=.

  3. 3.

    (31) Suppose 3. If yD(A*), then there is xD(A) such that (A+i)x=(A*-i)y. Since A is symmetric, (A+i)x=(A*+i)x=(A-i)*x, so that (A*-i)(x-y)=0. But since Ker(A*-i)=Ran(A+i)={0}, it follows that x=y, so that yD(A). Hence D(A)=D(A*), and therefore A is self-adjoint.

Title proof of basic criterion for self-adjointness
Canonical name ProofOfBasicCriterionForSelfadjointness
Date of creation 2013-03-22 14:53:05
Last modified on 2013-03-22 14:53:05
Owner Koro (127)
Last modified by Koro (127)
Numerical id 5
Author Koro (127)
Entry type Proof
Classification msc 47B25