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🧐Functional Analysis Unit 11 Review

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11.3 Adjoints of unbounded operators

11.3 Adjoints of unbounded operators

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🧐Functional Analysis
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Unbounded operators in Hilbert spaces are crucial in functional analysis. They extend beyond bounded operators, allowing for a wider range of applications in physics and mathematics. Understanding their adjoints is key to grasping their behavior and properties.

Adjoints of unbounded operators help analyze operator properties like symmetry and self-adjointness. These concepts are fundamental in quantum mechanics and differential equations, where unbounded operators often represent physical observables or differential operators.

Adjoints of Unbounded Operators

Definition of unbounded operator adjoint

  • Considers a densely defined unbounded operator A:D(A)HHA: D(A) \subset H \to H on a Hilbert space HH
  • Defines the adjoint AA^* of AA as follows:
    • Domain D(A)D(A^*) consists of all yHy \in H for which there exists zHz \in H satisfying Ax,y=x,z\langle Ax, y \rangle = \langle x, z \rangle for all xD(A)x \in D(A)
    • For each yD(A)y \in D(A^*), AyA^*y is the unique element zHz \in H fulfilling the condition Ax,y=x,z\langle Ax, y \rangle = \langle x, z \rangle for all xD(A)x \in D(A)
Definition of unbounded operator adjoint, Hilbert space - Wikipedia

Properties of unbounded operator adjoints

  • Proves that if AA is densely defined, then its adjoint AA^* is a closed operator
    • Considers a sequence (yn)D(A)(y_n) \subset D(A^*) converging to yy with AynzA^*y_n \to z
    • Shows that for any xD(A)x \in D(A), x,z=limnx,Ayn=limnAx,yn=Ax,y\langle x, z \rangle = \lim_{n \to \infty} \langle x, A^*y_n \rangle = \lim_{n \to \infty} \langle Ax, y_n \rangle = \langle Ax, y \rangle
    • Concludes that yD(A)y \in D(A^*) and Ay=zA^*y = z, establishing the closedness of AA^*
  • Demonstrates that if AA is densely defined and ABA \subset B, then BAB^* \subset A^*
    • Takes yD(B)y \in D(B^*) and shows that for any xD(A)D(B)x \in D(A) \subset D(B), Ax,y=Bx,y=x,By\langle Ax, y \rangle = \langle Bx, y \rangle = \langle x, B^*y \rangle
    • Deduces that yD(A)y \in D(A^*) and Ay=ByA^*y = B^*y, confirming BAB^* \subset A^*
  • Establishes that if AA is densely defined, then (A)=A(A^*)^* = \overline{A}, the closure of AA
    • Proves A(A)A \subset (A^*)^* by showing that for xD(A)x \in D(A) and yD(A)y \in D(A^*), (A)x,y=x,Ay=Ax,y\langle (A^*)^*x, y \rangle = \langle x, A^*y \rangle = \langle Ax, y \rangle
    • Utilizes the closedness of (A)(A^*)^* as the adjoint of the densely defined operator AA^*
    • Argues that A(A)\overline{A} \subset (A^*)^* since A(A)A \subset (A^*)^* and (A)(A^*)^* is closed, and the reverse inclusion follows from the previous property
Definition of unbounded operator adjoint, Adjoint functors - Wikipedia

Computation of common unbounded adjoints

  • Considers the multiplication operator Mf:D(Mf)L2(Ω)L2(Ω)M_f: D(M_f) \subset L^2(\Omega) \to L^2(\Omega) defined by Mfφ=fφM_f\varphi = f\varphi, where fL(Ω)f \in L^\infty(\Omega) and D(Mf)={φL2(Ω):fφL2(Ω)}D(M_f) = \{\varphi \in L^2(\Omega): f\varphi \in L^2(\Omega)\}
    • Computes the adjoint as Mf=MfM_f^* = M_{\overline{f}} with D(Mf)=D(Mf)D(M_f^*) = D(M_f)
  • Examines the differential operator A:D(A)L2(0,1)L2(0,1)A: D(A) \subset L^2(0,1) \to L^2(0,1) defined by Aφ=iφA\varphi = -i\varphi' with D(A)={φH1(0,1):φ(0)=φ(1)=0}D(A) = \{\varphi \in H^1(0,1): \varphi(0) = \varphi(1) = 0\}
    • Determines the adjoint as A=iddxA^* = i\frac{d}{dx} with D(A)=H1(0,1)D(A^*) = H^1(0,1)

Unbounded operators vs their adjoints

  • Shows that if AA is symmetric (i.e., Ax,y=x,Ay\langle Ax, y \rangle = \langle x, Ay \rangle for all x,yD(A)x, y \in D(A)), then AAA \subset A^*
    • Demonstrates that for any x,yD(A)x, y \in D(A), Ax,y=x,Ay\langle Ax, y \rangle = \langle x, Ay \rangle implies yD(A)y \in D(A^*) and Ay=AyA^*y = Ay
  • Proves that if AA is self-adjoint (i.e., A=AA = A^*), then AA is closed
    • Argues that since A=AA = A^* and AA^* is always closed, AA must be closed
  • Establishes that if AA is positive (i.e., Ax,x0\langle Ax, x \rangle \geq 0 for all xD(A)x \in D(A)), then AA^* is also positive
    • Considers yD(A)y \in D(A^*) and a sequence (xn)D(A)(x_n) \subset D(A) such that xnyx_n \to y and AxnAyAx_n \to A^*y
    • Shows that Ay,y=limnAxn,xn0\langle A^*y, y \rangle = \lim_{n \to \infty} \langle Ax_n, x_n \rangle \geq 0, confirming the positivity of AA^*
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