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๐ŸงFunctional Analysis Unit 11 Review

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11.3 Adjoints of unbounded operators

๐ŸงFunctional Analysis
Unit 11 Review

11.3 Adjoints of unbounded operators

Written by the Fiveable Content Team โ€ข Last updated September 2025
Written by the Fiveable Content Team โ€ข Last updated September 2025
๐ŸงFunctional Analysis
Unit & Topic Study Guides

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 $A: D(A) \subset H \to H$ on a Hilbert space $H$
  • Defines the adjoint $A^$ of $A$ as follows:
    • Domain $D(A^)$ consists of all $y \in H$ for which there exists $z \in H$ satisfying $\langle Ax, y \rangle = \langle x, z \rangle$ for all $x \in D(A)$
    • For each $y \in D(A^*)$, $A^*y$ is the unique element $z \in H$ fulfilling the condition $\langle Ax, y \rangle = \langle x, z \rangle$ for all $x \in D(A)$

Properties of unbounded operator adjoints

  • Proves that if $A$ is densely defined, then its adjoint $A^$ is a closed operator
    • Considers a sequence $(y_n) \subset D(A^*)$ converging to $y$ with $A^*y_n \to z$
    • Shows that for any $x \in D(A)$, $\langle x, z \rangle = \lim_{n \to \infty} \langle x, A^y_n \rangle = \lim_{n \to \infty} \langle Ax, y_n \rangle = \langle Ax, y \rangle$
    • Concludes that $y \in D(A^*)$ and $A^*y = z$, establishing the closedness of $A^$
  • Demonstrates that if $A$ is densely defined and $A \subset B$, then $B^* \subset A^*$
    • Takes $y \in D(B^*)$ and shows that for any $x \in D(A) \subset D(B)$, $\langle Ax, y \rangle = \langle Bx, y \rangle = \langle x, B^*y \rangle$
    • Deduces that $y \in D(A^*)$ and $A^*y = B^y$, confirming $B^ \subset A^$
  • Establishes that if $A$ is densely defined, then $(A^)^ = \overline{A}$, the closure of $A$
    • Proves $A \subset (A^)^$ by showing that for $x \in D(A)$ and $y \in D(A^)$, $\langle (A^)^*x, y \rangle = \langle x, A^*y \rangle = \langle Ax, y \rangle$
    • Utilizes the closedness of $(A^)^$ as the adjoint of the densely defined operator $A^$
    • Argues that $\overline{A} \subset (A^)^$ since $A \subset (A^)^$ and $(A^)^$ is closed, and the reverse inclusion follows from the previous property

Computation of common unbounded adjoints

  • Considers the multiplication operator $M_f: D(M_f) \subset L^2(\Omega) \to L^2(\Omega)$ defined by $M_f\varphi = f\varphi$, where $f \in L^\infty(\Omega)$ and $D(M_f) = {\varphi \in L^2(\Omega): f\varphi \in L^2(\Omega)}$
    • Computes the adjoint as $M_f^* = M_{\overline{f}}$ with $D(M_f^*) = D(M_f)$
  • Examines the differential operator $A: D(A) \subset L^2(0,1) \to L^2(0,1)$ defined by $A\varphi = -i\varphi'$ with $D(A) = {\varphi \in H^1(0,1): \varphi(0) = \varphi(1) = 0}$
    • Determines the adjoint as $A^* = i\frac{d}{dx}$ with $D(A^*) = H^1(0,1)$

Unbounded operators vs their adjoints

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