Notes on Lewin Ch2: Self-adjointness

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Let be a separable Hilbert space over .

Graphs

  • Definition: the graph of an operator is the vector subpsace of given by

  • Lemma: is the graph of an operator iff

    1. (vector space) is a vector subspace of
    2. (linearity at zero) implies
    3. (dense projection to ) is dense in .

Spectrum

  • Definition: Let be an operator. the resolvent set is

    and the spectrum .

    • can belong to (1) if is not injective [if then is an eigenvalue of ], (2) if is injective but not surjective (so the inverse cannot be defined), or (3) if it is invertible but not boundedly.

      • right shift on is injective but not surjective - there is a left inverse but no right inverse. Said differently - there is an inverse but only on a non-dense subspace of .
      • for an example of the third kind: Consider with operator . Generically you can take with an unbounded operator with . The inverse is of course which is densely defined but is not bounded. We can consider also on for an operator that is unbounded with unbounded inverse.

  • Let be a d.d. operator with . The ball around with radius

    is contained in , and in this ball we have the norm convergent power series for the resolvent,

  • Remark: if is bounded.
  • Lemma:

Closure

  • We say is closed if is closed, i.e. if with and then and .
  • If is not closed then .
  • generically will always satisfy (i),(ii) of the above lemma characterising graphs. But it is (ii) that may fail, i.e. there can be some nonzero with , (giving the absurdity ). Evaluation on is like this, since convergence in of highly oscillatory functions to 0 can break evaluation (the next exercise).
  • We say that is closable if has at least one closed extension. Then for a uniquely defined operator , called the closure of .

  • For , on the domain of the closure are and respectively.

    • proof idea: the operator is clearly closed on and resp. To show that the closure of the operator is this closed operator, we can directly check what the closure of the graph is, which amounts to checking convergence of functions to Sobolev functions.

Adjoint

  • cute observation: the wanted equality can be written

    So we make the definition

    • clearly a vector space so (i) is OK. Suppose in . Then for all , As is dense this implies . So (ii) is also OK. Finally for (iii): recall . We want that But note iff (zeroing out the second inner product in the product space), and

      i.e. iff , which is precisely the condition needed for closability. It follows:

      • density of allows a definition of , but may not be dense. And it is dense iff is closeable. So we will henceforth only use closeable d.d. operators.

    • Once upon a time I learned that we should define

      this is a reformulation of the graph definition above, which you can see by projecting to the first coordinate.

  • If is closeable then is closed. Also, and .

  • If is a closeable operator, then .

    • Proof: Observe that

      Then note that iff for all ,

      i.e. iff .

  • Exercise: If then also. [this is an application of Riesz representation. The start of proof: Let be fixed, and consider …]

Symmetry

  • is symmetric if for all . Equivalently .
  • Exercise: show that if is symmetric then its closure is as well. [This is trivial as by minimality]

  • Theorem (Spectrum of symmetric operators): Let be a symmetric operator. Then its spectrum is nonempty, and is either:

    1. all of ,
    2. closed upper half plane ,
    3. closed lower half plane ,
    4. some nonempty subset of .

    Furthermore, any eigenvalues must be real, and if is surjective for , then , and is contained in the half plane that does not contain .

    • Proof uses the important identity for symmetric operators: for ,

      Hence

      1. if then implies . So all eigenvalues must be real.
      2. if and is invertible then it is automatically bounded by .

      So the only way to be in the spectrum when is for to be not surjective.

      • Proof: Suppose that there is a point in . Assume for a contradiction that is not empty. Let on the boundary of this set with , and let approach . By the above it follows that the ball of radius around is contained in . But eventually and , so in fact is not on the boundary which is absurd. It follows that if has a single point, it is the entire half plane. [If a half plane is excluded then the real line is a subset of since it is closed.]

      • For the fact that the spectrum is always non-empty: we know this already for bounded operators. Suppose not for an unbounded . Then so exists and is bounded and symmetric. We know that implies . But it turns out that the only bounde symmetric operator with zero spectrum is the trivial operator, which follows form the next exercise:

        • Exercise 2.24 (Bounded Symmetric Operators)

          Let be a bounded and symmetric operator on . Following [Bre11, Prop. 6.9], we will show that

          both belong to the spectrum of . We recall that is always real since is symmetric. Show that for all we have

          and deduce that

          Using a minimizing sequence for the minimization problem , deduce that . Similarly, we have (we will see later in Theorem 2.33 that ). Deduce from the fact that that if then . [proof is just Cauchy-Schwarz and reading the outline.]

Self-adjointness

An operator is self-adjoint if , i.e. if is symmetric and . is essentially self-adjoint if is self-adjoint.

  • Interpretation. For any symmetric operator we always have , i.e. we have the inclusion of graphs

    In finite dimension these two sets coincide (as they are both -dimensional subspaces of ), but in infinite dimension the second can be strictly larger. Self-adjointness is the condition that they coincide,

    An operator is self-adjoint iff it satisfies for any

    the left hand side is a weak formulation of the equation . So a self-adjoint operator is one for which a weak solution is in fact strong.

  • All self-adjoint extensions (note: plural) are between and .
  • Exercise: Let be a symmetric operator. Show that is self-adjoint iff is symmetric. [ If is self-adjoint, then is symmetric. Conversely, if is symmetric then so is self-adjoint.]
  • If and is symmetric then is self-adjoint and bounded. [This is because . Thus is a closed operator defined on the full Banach space - by the closed graph theorem it is bounded.]

Characterisation and Weyl Sequences

  • Theorem (Characterisation) Let be a symmetric operator. Then the following are equivalent:

    1. is self-adjoint.
    2. .
    3. there exists such that and are both surjective from to .

    NB 2 is a priori much stronger than 3 as it implies that 3 holds for all .

  • Exercise: Let be a symmetric operator. Show that is essentially self-adjoint iff there exists such that and have dense range in .

  • Theorem. Let be self-adjoint. Then TFAE:

    1. .
    2. = 0.
    3. there exists a sequence with and .

    A sequence as in (3) is called a Weyl sequence for and .

    Proof: is trivial. For : if is not invertible we are done. If is invertible, then (2) implies it cannot be bounded. Indeed, for any with we would have

    which contradicts (2), so (1) holds. Conversely if the infimum is some then we would have , so injective i.e. , and hence has dense range. If then this inequality implies is Cauchy, so the range is closed. Hence is surjective, so .

  • Theorem. Let be self-adjoint. We have

    In particular, the spectrum is bounded below iff for all .

Diagonalization

  • Theorem Let be a symmetric operator on a domain , such that there exists a Hilbert basis of composed of elements of , which are all eigenvectors: with . Then, the closure of is the operator

The latter is a self-adjoint operator whose spectrum is

In other words, is essentially self-adjoint.

Momentum and Laplacian on

  • with and are self-adjoint. Moreover, and , and they have no eigenvalues.

    • Only proof for is given. Symmetry (integration by parts) is easy. Then to check self-adjointness we can check the surjectivity of for , which is convenient as this gives us a japanese bracket on the Fourier side to work with. The claim on the spectrum comes from

Momentum on

  • Let , and . Then

    in particular/addition

    • is symmetric but not self-adjoint, and its closure is .
    • , and is not symmetric.

    • is closed and symmetric but not self-adjoint.

      • A symmetric operator has . So is always a closed extension of such an operator. But this example shows that the closure of a symmetric operator need not be self-adjoint, and that the adjoint of a symmetric operator need not be symmetric. When we try to compute the adjoint we see the issues: the domain has all functions in satisfying the adjoint equation, and the vanishing of gives freedom to have a nonzero value at the boundary for , and breaks symmetry.

    • We have . The spectrum of consists only of eigenvalues, while the other spectra have none.

      • this is because all solutions to the eigenvalue equation for are , and also separately check that .

  • The strict symmetric extensions of are defined by on the domain

    These operators are all self-adjoint, and their spectrum is , consisting only of eigenvalues.

    • the condition is known as the Born-von Karman boundary condition.
    • In the proof it is shown that is codimension 1 in .

Momentum on

Making the similar definitions of , and mutatis mutandis we have Equation 22, but in this case has no self-adjoint extensions, with , without eigenvalues.

Exercises

  1. ( and Fourier Series) Skipping for now as it seems routine
  2. (Beware of Commutators) The issue here is that wont necessarily be defined as could take you out of .

  3. Exercise 2.48 (Deficiency Indices) Let be a closed symmetric operator.

    1. Show that for all . Deduce that the operator is an isometry from into .

      • This follows from the nice identity from before that

        with and . (GPT tells me the formula for has the inverse defined only as a map on , not necessarily a proper resolvent.)

    2. What can we conclude in finite dimension?

      • Peeking ahead at the later parts of the exercise, I think we are supposed to see that (using the finiteness of dimension to conclude from the rank).

    3. Let be a self-adjoint extension of . Show that is an isometry from into , which is an extension of , that is, such that for all .

      • This is easy because we already know that and that is surjective.

    4. Then show that the image of by contains .

      • We want to show that . Let ; we want to show that there is in such that . Clearly such would be defined by . Lets check that . For :

    5. Deduce that a symmetric operator can only have self-adjoint extensions if (which can be finite or infinite). [Follows as is an isometry]

    6. Reinterpret the result of Theorem 2.40 in this perspective.

      • (This theorem concerns the lack of self adjoint extensions for the momentum operator on the half line.) The result on the spectrum is enough to show that there can be no self adjoint extension.

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