Notes on Lewin Ch2: Self-adjointness

Let $ℌ$ be a separable Hilbert space over $ℂ$.

Graphs

• Definition: the graph of an operator $(𝐴,𝐷(𝐴))$ is the vector subpsace of $ℌ\times ℌ$ given by

• Lemma: $𝐺\subset ℌ\times ℌ$ is the graph of an operator $(𝐴,𝐷(𝐴))$ iff

  1. (vector space) $𝐺$ is a vector subspace of $ℌ\times ℌ$
  2. (linearity at zero) $(0,𝑦)\in 𝐺$ implies $𝑦=0$
  3. (dense projection to $ℌ$) $𝐷≔\{𝑥\in ℌ:\exists 𝑦\in ℌ\text{such that}(𝑥,𝑦)\in 𝐺\}$ is dense in $ℌ$.

Spectrum

• Definition: Let $(𝐴,𝐷(𝐴))$ be an operator. the resolvent set $𝜌(𝐴)\subset ℂ$ is

  $$𝜌(𝐴)≔\{$$
  $𝑧\in ℂ$ such that $𝐴-𝑧:𝐷(𝐴)\to ℌ$ is invertible with bounded inverse
  }

  and the spectrum $𝜎(𝐴)≔ℂ\backslash 𝜌(𝐴)$.

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

    • right shift on ${\ell }^2(\mathbb{N})$ 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 $ℌ$.
    • for an example of the third kind: Consider $ℌ={𝐿}^2(0,1)$ with operator $𝐴𝑓≔𝑡𝑓$. Generically you can take $𝐴={(𝐵-𝑧)}^{-1}$ with $𝐵$ an unbounded operator with $𝑧\in 𝜌(𝐵)$. The inverse is of course ${𝐴}^{-1}=𝐵-𝑧$ which is densely defined but is not bounded. We can consider also $𝐴\oplus {𝐴}^{-1}$ on $ℌ\oplus ℌ$ for an operator that is unbounded with unbounded inverse.

• Let $(𝐴,𝐷(𝐴))$ be a d.d. operator with $𝑧\in 𝜌(𝐴)$. The ball around $𝑧$ with radius

  $$\frac{1}{\Vert {(𝐴-𝑧)}^{-1}\Vert }$$

  is contained in $𝜌(𝐴)$, and in this ball we have the norm convergent power series for the resolvent,

  $${(𝐴-{𝑧}^{\prime })}^{-1}={(𝐴-𝑧)}^{-1}\sum _{𝑛\ge 0}{({𝑧}^{\prime }-𝑧)}^{𝑛}{(𝐴-𝑧)}^{-𝑛}$$
• Remark: $𝜎(𝐴)\ne \varnothing$ if $𝐴$ is bounded.
• Lemma: $𝜎({(𝐴-𝑧)}^{-1})=\left\{\frac{1}{𝜆-𝑧}:𝜆\in 𝜎(𝐴)\right\}\cup \{\begin{array}{c}\varnothing \text{if}𝐴\text{is bounded}\\ \{0\}\text{otherwise}\end{array}$

Closure

• We say $𝐴$ is closed if $𝐺(𝐴)$ is closed, i.e. if ${𝑥}_{𝑛}\in 𝐷(𝐴)$ with ${𝑥}_{𝑛}\stackrel{ℌ}{\to }𝑥$ and ${𝑦}_{𝑛}\stackrel{ℌ}{\to }𝑥$ then $𝑥\in 𝐷(𝐴)$ and $𝐴𝑥=𝑦$.
• If $(𝐴,𝐷(𝐴))$ is not closed then $𝜎(𝐴)=ℂ$.
• generically $\overline{𝐺(𝐴)}$ 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 $𝑦$ with $(0,𝑦)\in 𝐺$, (giving the absurdity $𝐴0=𝑦\ne 0$). Evaluation on ${𝐿}^2\to {𝐿}^2$ is like this, since convergence in ${𝐿}^2$ of highly oscillatory functions to 0 can break evaluation (the next exercise).
• We say that $𝐴$ is closable if $𝐴$ has at least one closed extension. Then $\overline{𝐺(𝐴)}=𝐺(\overline{𝐴})$ for a uniquely defined operator $(\overline{𝐴},𝐷(\overline{𝐴}))$, called the closure of $𝐴$.

• For $-𝑖{𝜕}_{𝑗}$, $\mathrm{\Delta }$ on ${𝐶}_{𝑐}^{\infty }({ℝ}^{𝑑})$ the domain of the closure are ${𝐻}^1$ and ${𝐻}^2$ respectively.

  • proof idea: the operator is clearly closed on ${𝐻}^1$ and ${𝐻}^2$ 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 ${𝐶}_{𝑐}^{\infty }$ functions to Sobolev functions.

Adjoint

• cute observation: the wanted equality $⟨𝑣\text{,}𝐴𝑢⟩=⟨{𝐴}^{\ast }𝑣\text{,}𝑢⟩$ can be written

  $${⟨(𝑣,{𝐴}^{\ast }𝑣)\text{,}(𝐴𝑢,-𝑢)⟩}_{ℌ\times ℌ}=0$$

  So we make the definition

  $$𝐺({𝐴}^{\ast })≔{\{(𝐴𝑢,-𝑢):𝑢\in 𝐷(𝐴)\}}^{⟂}$$

  • clearly a vector space so (i) is OK. Suppose $(0,𝑦)$ in $𝐺({𝐴}^{\ast })$. Then for all $𝑢\in 𝐷(𝐴)$, $0={⟨(0,𝑦)\text{,}(𝐴𝑢,-𝑢)⟩}_{ℌ\times ℌ}=⟨𝑦\text{,}-𝑢⟩.$ As $𝐷(𝐴)$ is dense this implies $𝑢=0$. So (ii) is also OK. Finally for (iii): recall $𝐷≔\{𝑥\in ℌ:\exists 𝑦\in ℌ\text{such that}(𝑥,𝑦)\in 𝐺\}$. We want that ${𝐷}^{⟂}=\{0\}.$ But note $𝑣\in {𝐷}^{⟂}$ iff $(𝑣,0)\in {𝐺}^{⟂}$ (zeroing out the second inner product in the product space), and

    $$(𝑣,0)\in {𝐺}^{⟂}={\{(𝐴𝑢,-𝑢):𝑢\in 𝐷(𝐴)\}}^{⟂⟂}=\overline{\{(𝐴𝑢,-𝑢):𝑢\in 𝐷(𝐴)\}}$$

    i.e. iff $(0,𝑣)\in \overline{𝐺(𝐴)}$, which is precisely the condition needed for closability. It follows:

    • density of $𝐷(𝐴)$ allows a definition of ${𝐴}^{\ast }$, but $𝐷({𝐴}^{\ast })$ may not be dense. And it is dense iff $𝐴$ is closeable. So we will henceforth only use closeable d.d. operators.

  • Once upon a time I learned that we should define

    $$𝐷({𝐴}^{\ast })=\{𝑣\in ℌ:\exists 𝑤\text{such that for all}𝑥\in 𝐷(𝐴),⟨𝑣\text{,}𝐴𝑥⟩=⟨𝑤\text{,}𝑥⟩\}$$

    this is a reformulation of the graph definition above, which you can see by projecting $𝐺({𝐴}^{\ast })$ to the first coordinate.

• If $𝐴$ is closeable then ${𝐴}^{\ast }$ is closed. Also, $\overline{𝐴}={𝐴}^{\ast \ast }$ and ${𝐴}^{\ast }={\overline{𝐴}}^{\ast }$.

• If $(𝐴,𝐷(𝐴))$ is a closeable operator, then $ker({𝐴}^{\ast }-\overline{𝑧})={ran(𝐴-𝑧)}^{⟂}={ran(\overline{𝐴}-𝑧)}^{⟂}$.

  • Proof: Observe that

    $$𝑣\in ker({𝐴}^{\ast }-\overline{𝑧})\text{iff}{𝐴}^{\ast }𝑣=\overline{𝑧}𝑣.$$

    Then note that $(𝑣,𝑧𝑣)\in 𝐺({𝐴}^{\ast })$ iff for all $𝑢\in 𝐷(𝐴)$, $log(𝑥)$

    $$⟨𝑣\text{,}𝐴𝑢⟩=⟨\overline{𝑧}𝑣\text{,}𝑢⟩=⟨𝑣\text{,}𝑧𝑢⟩,$$

    i.e. iff $𝑣\in {ran(𝐴-𝑧)}^{⟂}$.

• Exercise: If $𝐴\in 𝐵(ℌ)$ then $𝐷({𝐴}^{\ast })=ℌ$ also. [this is an application of Riesz representation. The start of proof: Let $𝑣$ be fixed, and consider ${𝑓}_{𝑣}(𝑥)=⟨𝑣\text{,}𝐴𝑥⟩$…]

Symmetry

• $(𝐴,𝐷(𝐴))$ is symmetric if $⟨𝑣\text{,}𝐴𝑤⟩=⟨𝐴𝑣\text{,}𝑤⟩$ for all $𝑣,𝑤\in 𝐷(𝐴)$. Equivalently $𝐺(𝐴)\subset 𝐺({𝐴}^{\ast })$.
• Exercise: show that if $𝐴$ is symmetric then its closure is as well. [This is trivial as $𝐺(𝐴)\subset 𝐺(\overline{𝐴})\subset 𝐺({𝐴}^{\ast })$ by minimality]

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

  1. all of $ℂ$,
  2. closed upper half plane $\{\Im 𝑧\ge 0\}$,
  3. closed lower half plane $\{\Im 𝑧\le 0\}$,
  4. some nonempty subset of $ℝ$.

  Furthermore, any eigenvalues must be real, and if $𝐴-𝑧$ is surjective for $𝑧\notin ℝ$, then $𝑧\in 𝜌(𝐴)$, and $𝜎(𝐴)$ is contained in the half plane that does not contain $𝑧$.

  • Proof uses the important identity for symmetric operators: for $𝑎,𝑏\in ℝ$,

    $${\Vert (𝐴-𝑎-𝑖𝑏)𝑢\Vert }^2={\Vert (𝐴-𝑎)𝑢\Vert }^2+{𝑏}^2{\Vert 𝑢\Vert }^2>{𝑏}^2{\Vert 𝑢\Vert }^2.$$

    Hence

    1. if $𝑏\ne 0$ then $(𝐴-𝑎-𝑖𝑏)𝑢=0$ implies $𝑢=0$. So all eigenvalues must be real.
    2. if $𝑏\ne 0$ and $𝐴-𝑎-𝑖𝑏$ is invertible then it is automatically bounded by $\frac{1}{𝑏}$.

    So the only way to be in the spectrum when $𝑏\ne 0$ is for $𝐴-𝑎-𝑖𝑏$ to be not surjective.

    • Proof: Suppose that there is a point in $𝜌(𝐴)\cap \{\Im 𝑧>0\}$. Assume for a contradiction that $𝜎(𝐴)\cap \{\Im 𝑧>0\}$ is not empty. Let $𝑧=𝑎+𝑖𝑏$ on the boundary of this set with $𝑏>0$, and let ${𝑧}_{𝑛}={𝑎}_{𝑛}+𝑖{𝑏}_{𝑛}\in 𝜌(𝐴)$ approach $𝑧$. By the above it follows that the ball of radius $|{𝑏}_{𝑛}|$ around ${𝑧}_{𝑛}$ is contained in $𝜌(𝐴)$. But eventually $|{𝑏}_{𝑛}|>\frac{𝑏}{2}$ and $|𝑧-{𝑧}_{𝑛}|<\frac{𝑏}{2}$, so in fact $𝑧$ is not on the boundary which is absurd. It follows that if $𝜌(𝐴)\cap \{\Im 𝑧>0\}$ has a single point, it is the entire half plane. [If a half plane is excluded then the real line is a subset of $𝜎(𝐴)$ 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 $𝐴$. Then $0\in 𝜌(𝐴)$ so ${𝐴}^{-1}$ exists and is bounded and symmetric. We know that $𝜎(𝐴)=\{\}$ implies $𝜎({𝐴}^{-1})=\{0\}$. 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 $𝐵$ be a bounded and symmetric operator on $𝐷(𝐵)=𝐻$. Following [Bre11, Prop. 6.9], we will show that

        $$𝑚≔\underset{𝑣\in 𝐻,\Vert 𝑣\Vert =1}{inf}⟨𝑣,𝐵𝑣⟩,𝑀≔\underset{𝑣\in 𝐻,\Vert 𝑣\Vert =1}{sup}⟨𝑣,𝐵𝑣⟩$$

        both belong to the spectrum of $𝐵$. We recall that $⟨𝑣,𝐵𝑣⟩$ is always real since $𝐵$ is symmetric. Show that for all $𝑣,𝑤\in 𝐻$ we have

        $$|⟨𝑤\text{,}(𝐵-𝑚)𝑣⟩|\le ⟨𝑣,(𝐵-𝑚)𝑣{⟩}^{\frac{1}{2}}⟨𝑤,(𝐵-𝑚)𝑤{⟩}^{\frac{1}{2}}$$

        and deduce that

        $$\Vert (𝐵-𝑚)𝑣\Vert \le ⟨𝑣,(𝐵-𝑚)𝑣{⟩}^{\frac{1}{2}}{\Vert 𝐵-𝑚\Vert }^{\frac{1}{2}}.$$

        Using a minimizing sequence for the minimization problem $𝑚$, deduce that $𝑚\in 𝜎(𝐵)$. Similarly, we have $𝑀\in 𝜎(𝐵)$ (we will see later in Theorem 2.33 that $𝜎(𝐵)\subseteq [𝑚,𝑀]$). Deduce from the fact that $𝑚,𝑀\in 𝜎(𝐵)$ that if $𝜎(𝐵)=\{0\}$ then $𝐵\equiv 0$. [proof is just Cauchy-Schwarz and reading the outline.]

Self-adjointness

An operator is self-adjoint if $𝐴={𝐴}^{\ast }$, i.e. if $𝐴$ is symmetric and $𝐷(𝐴)=𝐷({𝐴}^{\ast })$. $𝐴$ is essentially self-adjoint if $\overline{𝐴}$ is self-adjoint.

• Interpretation. For any symmetric operator we always have $𝐴\subset {𝐴}^{\ast }$, i.e. we have the inclusion of graphs

  $$\{(𝑣,𝐴𝑣):𝑣\in 𝐷(𝐴)\}\subset \{(𝑣,{𝐴}^{\ast }𝑣):𝑣\in 𝐷({𝐴}^{\ast })\}.$$

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

  $$\{(𝑣,𝐴𝑣):𝑣\in 𝐷(𝐴)\}=\{(𝑣,{𝐴}^{\ast }𝑣):𝑣\in 𝐷({𝐴}^{\ast })\}.$$

  An operator is self-adjoint iff it satisfies for any $𝑣,𝑤\in ℌ$

  $$⟨𝑣\text{,}𝐴𝑧⟩=⟨𝑤\text{,}𝑧⟩\text{for all}𝑧\in 𝐷(𝐴)\text{implies}𝑣\in 𝐷(𝐴)\text{and}𝐴𝑣=𝑤$$

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

• All self-adjoint extensions (note: plural) are between $\overline{𝐴}$ and ${𝐴}^{\ast }$.
• Exercise: Let $(𝐴,𝐷(𝐴))$ be a symmetric operator. Show that $\overline{𝐴}$ is self-adjoint iff ${𝐴}^{\ast }$ is symmetric. [ If $\overline{𝐴}$ is self-adjoint, then ${𝐴}^{\ast }={(\overline{𝐴})}^{\ast }=\overline{𝐴}$ is symmetric. Conversely, if ${𝐴}^{\ast }$ is symmetric then $\overline{𝐴}\subset {𝐴}^{\ast }\subset {𝐴}^{\ast \ast }=\overline{𝐴}$ so $\overline{𝐴}={𝐴}^{\ast }$ is self-adjoint.]
• If $𝐷(𝐴)=ℌ$ and $𝐴$ is symmetric then $𝐴$ is self-adjoint and bounded. [This is because $𝐷({𝐴}^{\ast })\supset 𝐷(𝐴)$. Thus $𝐴$ 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 $(𝐴,𝐷(𝐴))$ be a symmetric operator. Then the following are equivalent:

  1. $𝐴$ is self-adjoint.
  2. $𝜎(𝐴)\subset ℝ$.
  3. there exists $𝜆\in ℂ$ such that $𝐴-𝜆$ and $𝐴-\overline{𝜆}$ are both surjective from $𝐷(𝐴)$ to $ℌ$.

  NB 2 is a priori much stronger than 3 as it implies that 3 holds for all $𝜆\in ℂ\backslash ℝ$.

• Exercise: Let $(𝐴,𝐷(𝐴))$ be a symmetric operator. Show that $𝐴$ is essentially self-adjoint iff there exists $𝜆\in ℂ$ such that $𝐴-𝜆$ and $𝐴-\overline{𝜆}$ have dense range in $ℌ$.

• Theorem. Let $(𝐴,𝐷(𝐴))$ be self-adjoint. Then TFAE:

  1. $𝜆\in 𝜎(𝐴)$.
  2. ${inf}_{\begin{array}{c}𝑣\in 𝐷(𝐴)\\ \Vert 𝑣\Vert =1\end{array}}\Vert (𝐴-𝜆)𝑣\Vert$ = 0.
  3. there exists a sequence ${𝑣}_{𝑛}\in 𝐷(𝐴)$ with $\Vert {𝑣}_{𝑛}\Vert =1$ and $\Vert (𝐴-𝜆){𝑣}_{𝑛}\Vert \to 0$.

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

  Proof: $(2)⟺(3)$ is trivial. For $(2)⟹(1)$: if $𝐴-𝜆$ is not invertible we are done. If $𝐴-𝜆$ is invertible, then (2) implies it cannot be bounded. Indeed, for any $𝑤\in ℌ$ with $𝑣={(𝐴-𝜆)}^{-1}𝑤$ we would have

  $$\Vert {(𝐴-𝜆)}^{-1}𝑤\Vert \le 𝐶\Vert 𝑤\Vert \text{i.e.}\Vert 𝑣\Vert \le 𝐶\Vert (𝐴-𝜆)𝑣\Vert$$

  which contradicts (2), so (1) holds. Conversely if the infimum is some $𝜀>0$ then we would have $\Vert (𝐴-𝜆)𝑣\Vert \ge 𝜀\Vert 𝑣\Vert$, so injective i.e. $ker(𝐴-𝜆)=\{0\}$, and hence has dense range. If $(𝐴-𝜆){𝑣}_{𝑛}\to 𝑤$ then this inequality implies ${𝑣}_{𝑛}$ is Cauchy, so the range is closed. Hence $𝐴-𝜆$ is surjective, so $𝜆\notin 𝜎(𝐴)$.

• Theorem. Let $(𝐴,𝐷(𝐴))$ be self-adjoint. We have

  $$inf𝜎(𝐴)=\underset{\begin{array}{c}𝑣\in 𝐷(𝐴)\\ \Vert 𝑣\Vert =1\end{array}}{inf}⟨𝑣\text{,}𝐴𝑣⟩,sup𝜎(𝐴)=\underset{\begin{array}{c}𝑣\in 𝐷(𝐴)\\ \Vert 𝑣\Vert =1\end{array}}{sup}⟨𝑣\text{,}𝐴𝑣⟩.$$

  In particular, the spectrum is bounded below iff $⟨𝑣\text{,}𝐴𝑣⟩\gtrsim {\Vert 𝑣\Vert }^2$ for all $𝑣\in 𝐷(𝐴)$.

Diagonalization

• Theorem Let $𝐴$ be a symmetric operator on a domain $𝐷(𝐴)\subset 𝐻$, such that there exists a Hilbert basis ${({𝑒}_{𝑛})}_{𝑛\ge 1}$ of $𝐻$ composed of elements of $𝐷(𝐴)$, which are all eigenvectors: $𝐴{𝑒}_{𝑛}={𝜆}_{𝑛}{𝑒}_{𝑛}$ with ${𝜆}_{𝑛}\in ℝ$. Then, the closure of $𝐴$ is the operator

The latter is a self-adjoint operator whose spectrum is

In other words, $(𝐴,𝐷(𝐴))$ is essentially self-adjoint.

Momentum and Laplacian on ${ℝ}^{𝑑}$

• ${𝑃}_{𝑗}≔-𝑖{𝜕}_{𝑗},𝐴≔-\mathrm{\Delta }$ with $𝐷({𝑃}_{𝑗})≔\{𝑓\in {𝐿}^2:{𝜕}_{𝑗}𝑓\in {𝐿}^2\}\subset ℌ={𝐿}^2$ and $𝐷(𝐴)={𝐻}^2$ are self-adjoint. Moreover, $𝜎({𝑃}_{𝑗})=ℝ$ and $𝜎(𝐴)=\left[0,\infty \right)$, and they have no eigenvalues.

  • Only proof for $\mathrm{\Delta }$ is given. Symmetry (integration by parts) is easy. Then to check self-adjointness we can check the surjectivity of $𝐴-𝜆$ for $𝜆=-1=\overline{𝜆}$, 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 $(0,1)$

• Let $𝐷({𝑃}^{\text{min}})={𝐶}_{𝑐}^{\infty }$, $𝐷({𝑃}^{\text{max}})={𝐻}^1$ and $𝐷({𝑃}_0)={𝐻}_0^1(0,1)$. Then

  $${𝑃}^{\text{min}}\subset {({𝑃}^{\text{max}})}^{\ast }={𝑃}_0⊊{𝑃}_0^{\ast }={({𝑃}^{\text{min}})}^{\ast }={𝑃}^{\text{max}}$$

  in particular/addition

  • ${𝑃}^{\text{min}}$ is symmetric but not self-adjoint, and its closure is ${𝑃}_0$.
  • ${𝑃}^{\text{max}}={({𝑃}^{\text{min}})}^{\ast }={𝑃}_0^{\ast }$, and is not symmetric.

  • ${𝑃}_0$ is closed and symmetric but not self-adjoint.

    • A symmetric operator has $𝐴\subset {𝐴}^{\ast }$. So ${𝐴}^{\ast }$ 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 ${𝐿}^2$ satisfying the adjoint equation, and the vanishing of $𝑣\in 𝐷({𝑃}_0)$ gives freedom to have a nonzero value at the boundary for $𝑣\in 𝐷({𝑃}_0^{\ast })$, and breaks symmetry.

  • We have $𝜎({𝑃}_0)=𝜎({𝑃}^{\text{min}})=𝜎({𝑃}^{\text{max}})=ℂ$. The spectrum of ${𝑃}^{\text{max}}$ consists only of eigenvalues, while the other spectra have none.

    • this is because all solutions to the eigenvalue equation for ${𝑃}^{\text{max}}$ are ${𝑒}^{𝑖𝜆𝑥}$, and also separately check that $0\ne {𝑒}^{𝑖𝜆𝑥}\in {ran({𝑃}_0-𝜆)}^{⟂}$.

• The strict symmetric extensions of ${𝑃}_0$ are defined by ${𝑃}_{\text{per},𝜃}𝑓=-𝑖{𝑓}^{\prime }$ on the domain

  $$𝐷({𝑃}_{\text{per},𝜃})=\{𝑓\in {𝐻}^1:𝑓(1)={𝑒}^{𝑖𝜃}𝑓(0)\}.$$

  These operators are all self-adjoint, and their spectrum is $𝜎({𝑃}_{\text{per},𝜃})=\{2𝜋𝑛+𝜃:𝑛\in \mathbb{Z}\}$, consisting only of eigenvalues.

  • the condition $𝑓(1)={𝑒}^{𝑖𝜃}𝑓(0)$ is known as the Born-von Karman boundary condition.
  • In the proof it is shown that ${𝐻}_0^1$ is codimension 1 in ${𝐻}_{\text{per},𝜃}$.

Momentum on ${ℝ}_{+}$

Making the similar definitions of ${𝐻}^1,{𝐻}_0^1,{𝑃}^{\text{min}},{𝑃}^{\text{max}}$, and ${𝑃}_0$ mutatis mutandis we have Equation 22, but ${𝑃}_0$ in this case has no self-adjoint extensions, with $𝜎({𝑃}_0)={ℂ}_{-}≔\{𝑧\in ℂ:\Im (𝑧)<0\}$, without eigenvalues.

Exercises

1. (${𝐻}_{\text{per}}^1$ and Fourier Series) Skipping for now as it seems routine
2. (Beware of Commutators) The issue here is that ${𝑃}_{\text{per}}𝑋$ wont necessarily be defined as $𝑋$ could take you out of $𝐷({𝑃}_{\text{per}})$.

3. Exercise 2.48 (Deficiency Indices) Let $(𝐴,𝐷(𝐴))$ be a closed symmetric operator.

   1. Show that $\Vert (𝐴+𝑖)𝑣\Vert =\Vert (𝐴-𝑖)𝑣\Vert$ for all $𝑣\in 𝐷(𝐴)$. Deduce that the operator $𝑈=(𝐴+𝑖){(𝐴-𝑖)}^{-1}$ is an isometry from $ran(𝐴-𝑖)$ into $ran(𝐴+𝑖)$.

      • This follows from the nice identity from before that

        $$\Vert 𝐴-𝑎-𝑖𝑏\Vert {𝑢}^2={\Vert (𝐴-𝑎)𝑢\Vert }^2+{𝑏}^2{\Vert 𝑢\Vert }^2$$

        with $𝑎=0$ and $𝑏=\pm 1$. (GPT tells me the formula for $𝑈$ has the inverse defined only as a map on $ran(𝐴-𝑖)$, 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 $dimker(𝐴-𝑖)=dimker(𝐴+𝑖)$ (using the finiteness of dimension to conclude from the rank).

   3. Let $𝐵={𝐵}^{\ast }$ be a self-adjoint extension of $𝐴$. Show that $𝑉=(𝐵+𝑖){(𝐵-𝑖)}^{-1}$ is an isometry from $ℌ$ into $ℌ$, which is an extension of $𝑈$, that is, such that $𝑉𝑓=𝑈𝑓$ for all $𝑓\in 𝐷(𝐴)$.

      • This is easy because we already know that $𝑖\notin 𝜎(𝐵)$ and that $𝐵+𝑖$ is surjective.

   4. Then show that the image of $ran{(𝐴+𝑖)}^{⟂}=ker({𝐴}^{\ast }-𝑖)$ by $𝑉$ contains $ran{(𝐴-𝑖)}^{⟂}=ker({𝐴}^{\ast }+𝑖)$.

      • We want to show that $𝑉(ker({𝐴}^{\ast }-𝑖))\supset ker({𝐴}^{\ast }+𝑖)$. Let ${𝐴}^{\ast }𝑣=-𝑖𝑣$; we want to show that there is $𝑤$ in $ker({𝐴}^{\ast }-𝑖)$ such that $𝑉𝑤=𝑣$. Clearly such $𝑤$ would be defined by $𝑤={𝑉}^{-1}𝑣=(𝐵-𝑖){(𝐵+𝑖)}^{-1}𝑣$. Lets check that $𝑤\in ker({𝐴}^{\ast }+𝑖)={ran(𝐴-𝑖)}^{⟂}$. For $𝑢\in 𝐷(𝐴)\subset 𝐷(𝐵)$:
      $$\begin{array}{cc}⟨(𝐴-𝑖)𝑢\text{,}𝑤⟩& =⟨𝑢\text{,}(𝐵+𝑖)(𝐵-𝑖){(𝐵+𝑖)}^{-1}𝑣⟩\\ & =⟨𝑢\text{,}(𝐵-𝑖)(𝐵+𝑖){(𝐵+𝑖)}^{-1}𝑣⟩\\ & =⟨𝑢\text{,}(𝐵-𝑖)𝑣⟩\\ & =⟨(𝐵+𝑖)𝑢\text{,}𝑣⟩\\ & =⟨(𝐴+𝑖)𝑢\text{,}𝑣⟩=0.\end{array}$$

   5. Deduce that a symmetric operator can only have self-adjoint extensions if $dimker({𝐴}^{\ast }-𝑖)=dimker({𝐴}^{\ast }+𝑖)$ (which can be finite or infinite). [Follows as $𝑉$ 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 $𝜎(𝑃)={ℂ}_{-}$ is enough to show that there can be no self adjoint extension.