DIARY OF THE TUTORIAL SESSIONS


T24. To conclude on the self-adjoint extensions of symmetric operators that are bounded below:
(i) There's the Friedrichs extension $ A^F$, with the same bound below as the original operator $ A$ and domain in the form domain of $ A$.
(ii) If $ A$ has finite deficiency indices, then any self-adjoint extension is bounded below.
(iii) In case of infinite deficiency indices, there might be extensions unbounded below.
(iv) For sure, unless $ A$ is already essentially self-adjoint, there will be other extensions, besides the Friedrichs extension, which are bounded below.
Practise with this phenomena in E26, E27.
Final remarks: the (von Neumann's) theory of self-adjoint extensions is so elegant and complete that it appears to close the subject, but this is definitely not the case for extensions of $ A$'s that are bounded below. In partic. (see also the historical note in T23), it's not clear which unitary labels the Friedrichs extension. There is a much deeper and most natural analysis for this case, the Birman-Krein-Vishik theory. Those who are interested may look up the expository paper by Alonso and Simon.


T23. The Friedrichs extension (recap). It's that self-adjoint extension $ A^F$ of a semi-bounded symmetric operator $ A$ (with bound, say, $ M$) which is uniquely identified by the quadratic form $ q_A$ associated with $ A$ (see T21, (ii)).
$ A^F$has these two nice features that make it a "distinguished" extension:
(i) the lower bound is preserved, that is, $ \inf(\sigma(A^F))=\inf(\sigma(A))=M$, or in other words $ \displaystyle\inf_{\varphi\in\mathcal{D}(A^F),\Vert\varphi\Vert=1}\langle\varphi,A^F\varphi\rangle=$ $ \displaystyle\inf_{\psi\in\mathcal{D}(A),\Vert\psi\Vert=1}\langle\psi,A\psi\rangle$,
(ii) it is the only self-adjoint extension of $ A$ whose domain $ \mathcal{D}(A^F)$ is contained in the form domain $ \mathcal{D}(q_A)$ of $ A$.
All other self-adjoint extensions have domain not included in $ \mathcal{D}(q_A)\supset\mathcal{D}(A^F)\supset\mathcal{D}(A)$ and a bound below that is equal or lower, possibly $ -\infty$. Practise with this phenomena in E23, E24, E25.
(Historical note: In his original paper on self-adjoint extensions von Neumann proved that a semi-bounded operator has semi-bounded extensions whose lower bound is arbitrarily close to the bound of the original operator, and conjectured that extensions exist with the same lower bound. This fact was proved by Friedrichs and Stone.)


T22. Recap on closed semi-bounded quadratic forms, i.e., semi-bounded forms $ q$ (say, with $ M:=\!\displaystyle\inf_{\Vert\psi\Vert=1} q(\psi,\psi)>-\infty$) whose domain $ \mathcal{D}(q)$ is complete under the scalar product $ \langle\varphi,\psi\rangle_{q}:=q(\varphi,\psi)+(1\!-\!M)\langle\varphi,\psi\rangle$.
(i) Closure of a form, examples of non-closed / non-closable forms. The form $ q_A$associated to a symmetric operator $ A$ bounded below (see T21) is closed and bounded below.
(ii) A closed semi-bounded quadratic form $ q$ identifies a unique self-adjoint operator $ A$ such that $ q(\varphi,\psi)=\langle\varphi,A\psi\rangle$ $ \forall\varphi,\psi\in\mathcal{D}(A)\subset\mathcal{D}(q)$. $ A$ has domain $ \mathcal{D}(A)=\{\xi\in\mathcal{D}(q)\,\vert\,\exists\psi_\xi\in{\cal H}\textr... ...h }q(\cdot,\xi)+(1\!-\!M)\langle\cdot,\xi\rangle=\langle\cdot,\psi_\xi\rangle\}$ and acts as $ A\xi=\psi_\xi-(1\!-\!M)\xi$. The fact that $ \mathcal{D}(A)$ is dense in $ {\cal H}$ and $ \psi_\xi$ is uniquely determined follows from Riesz' lemma (and from the closedness and symmetry of $ q$). The fact that $ A$ is symmetric follows from the symmetry of $ q$. The fact that $ A=A^*$ follows from the self-adjointness of the bounded linear operator $ (A+1-M)^{-1}$, which is easier to check and follows again from Riesz' lemma.
(iii) Thus: there always exist closed extensions of a symmetric operator $ A$ (for instance, $ A^{**}$) although it is possible that none of these is self-adjoint, while semi-bounded forms need not have any closed extensions, but when they have and are semi-bounded, they are the quadratic forms associated with self-adjoint operators.
(iv) A closed semi-bounded quadratic form $ q$ is bounded iff the associated self-adjoint operator $ A$ is, in which case $ \displaystyle\sup_{\Vert\psi\Vert=1}\vert q(\psi,\psi)\vert=\Vert A\Vert$ (it follows by polarisation). This also proves that $ \Vert A\Vert=\displaystyle\sup_{\Vert\psi\Vert=1}\vert\langle \psi,A\psi\rangle\vert$ for any symmetric operator $ A$, thus extending the same formula discussed in T6 to the unbounded case. (Note that this formula is wrong if $ A$ is not symmetric!)


T21. Recap on quadratic forms on a Hilbert space $ {\cal H}$. (i) Form domain, symmetric forms, forms bounded below (they are automatically symmetric if $ {\cal H}$ is complex). (ii) The form associated with a symmetric operator $ A$ bounded below (say, $ A\geqslant M\mathbbm{1}$, where $ M=\displaystyle\inf_{\Vert\psi\Vert=1} \langle\psi,A\psi\rangle>-\infty$): it's the quadratic form $ q_A$ with domain $ \mathcal{D}(q_A)$ given by the completion of $ \mathcal{D}(A)$ (the domain of $ A$) with respect to the scalar product $ \langle\varphi,\psi\rangle_{\!A}:=\langle\varphi,(A+(1\!-\!M)\mathbbm{1})\psi\rangle$ and defined by $ q_A(\varphi,\psi):=\langle\varphi,\psi\rangle_{\!A}-(1\!-\!M)\langle\varphi,\psi\rangle$ $ \forall\varphi,\psi\in\mathcal{D}(q_A)$. One has $ \mathcal{D}(A)\subset\mathcal{D}(q_A)\subset{\cal H}$ and $ \langle\varphi,A\psi\rangle=q_A(\varphi,\psi)$ $ \forall\varphi,\psi\in\mathcal{D}(A)$. $ \mathcal{D}(q_A)$, often denoted also by $ \mathcal{Q}(A)$, is called the form domain of $ A$. Important: the form $ q_A$is closed (see T22) and bounded below.  Practise with E22, E23.


T20.  A useful criterion for a symmetric operator $ A$ to have self-adjoint extensions (von Neumann): if there exists a conjugation $ C$ such that $ C\mathcal{D}(A)\subset\mathcal{D}$ and $ AC=CA$, then $ A$ has equal deficiency indices and therefore has self-adjoint extensions. Recall that a conjugation $ C:{\cal H}\to{\cal H}$ is an antilinear map that preserves norm and such that $ C^2=\mathbbm{1}$. To apply this criterion to Schrödinger operators $ -\Delta+V$ ($ V$ real-valued potential) just choose the complex conjugation.


T19. A message from class: different self-adjoint extensions correspond to different physics. See it in the easy example of the motion by translations on $ [0,1]$: the generator is a self-adjoint extension of $ -\mathrm{i}\frac{\rm d}{\rm d x}$ (see E17, E18), unitarity still leaves the freedom of choosing the phase of the wave packet as it comes in at zero, by the superposition principle all functions must change by the same phase when they come back in. Thus, the different translations are just given by specifying $ \alpha$, $ \vert\alpha\vert=1$, and by requiring that all reasonable wave packets $ \psi_y(x)\equiv\varphi(\cdot+y)$ satisfy $ \psi_y(1)=\alpha\psi_y(0)$ for all ``times'' $ y$.


T18.  Recap on symmetric/self-adjoint extensions of a closed symmetric operator $ A$.
(i) The closed symmetric extensions $ B$ of $ A$are restrictions of $ A^*$: $ A\subset B\subset B^*\subset A^*$. Thus, one needs to control the structure of the domain of $ A^*$. One has $ \mathcal{D}(A^*)=\mathcal{D}(A)\oplus_A{\cal H}_+\oplus_A{\cal H}_-$ where $ \mathcal{H}_\pm:=\mathrm{Ker}(A^*\mp\mathrm{i})$ are the deficiency spaces of $ A$ and the direct sum is with respect to the scalar product $ \langle \varphi,\psi\rangle_A:=\langle A^*\varphi,A^*\psi\rangle+\langle \varphi,\psi\rangle$ $ \forall\varphi,\psi\in\mathcal{D}(A^*)$(in partic., $ \mathcal{D}(A)$, $ {\cal H}_+$, and $ {\cal H}_-$ are all closed w.r.t. this scalar product).
(ii) The closed symmetric extensions are in one-to-one correspondence with partial isometries $ U:\mathcal{H}_+\to\mathcal{H}_-$. (A partial isometry is a bounded linear map $ U:{\cal H}\to{\cal H}$ such that $ (\mathrm{Ker}U)^\perp\xrightarrow[]{U}\mathrm{Ran}U$ is norm-preserving; this implies that $ U^*$ too is a partial isometry, since $ \mathrm{Ran}U\xrightarrow[]{U^*}(\mathrm{Ker}U)^\perp$ is norm-preserving; in partic., if $ \mathrm{Ker}U=\{0\}$ then $ U$ is an isometry in the usual sense.)
(iii) If $ U$ is such an isometry, with $ (\mathrm{Ker}U)^\perp\subset{\cal H}_+$, then the corresponding closed symmetric extension $ A_U$ has domain $ \mathcal{D}(A_U)=\{\psi+\psi_+ +U\psi_+\vert\psi\in\mathcal{D}(A),\psi_+\in(\mathrm{Ker}U)^\perp \}$ and action  $ A_U(\psi+\psi_+ +U\psi_+)=$ $ A\psi+\mathrm{i}\psi_+-\mathrm{i}U\psi_+$ .
(iv) If $ \dim (\mathrm{Ker}U)^\perp<\infty$ then the deficiency indices of $ A_U$ are $ n_\pm(A_U)=n_\pm(A)-\dim (\mathrm{Ker}U)^\perp$. (Recall: $ n_\pm(A):=\dim\mathcal{H}_\pm$, the deficiency indices of $ A$.)
(v) Thus, $ A=A^*$ iff $ n_\pm(A)=0$.
(vi) $ A$ has self-adjoint extensions iff $ n_+(A)=n_-(A)$. In this case each extension is labelled by a unitary map $ U:\mathcal{H}_+\to\mathcal{H}_-$.
(vii) If either $ n_+(A)=0\neq n_-(A)$ or $ n_-(A)=0\neq n_+(A)$, then $ A$ has no non-trivial symmetric extensions ($ A$ is "maximally symmetric").  Practise with E18, E19, E20, E21.


T17. Review from class: the step-by-step determination of $ A_0^*$ and $ \overline{A_0}$, where $ A_0=-\mathrm{i}\frac{{\rm d}}{{\rm d}x}$, $ \mathcal{D}(A_0)=\{f\vert f\in C^1([0,1])\,,f(0)=f(1)=0\}$. General comment: symmetry is easy to check (in case of differential operators it usually boils down to integration by parts), computing the adjoint is a non-trivial job even in simple situations. However, self-adjointness is a much stronger property than symmetry, which justifies the additional effort needed to prove it.


T16. Recap: symmetry (hermiticity) vs self-adjointness.
(i) $ T$ symmetric means $ T\subset \overline{T}=T^{**}\subset T^*$
(ii) $ T$ closed and symmetric means $ T=\overline{T}=T^{**}\subset T^*$
(iii) $ T$ self-adjoint means $ T=\overline{T}=T^{**}=T^*$
(iv) $ T$ essentially self-adjoint means $ T\subset \overline{T}=T^{**}=T^*$ (and its unique self-adjoint extension is $ \overline{T}=T^{**}=T^*$).
(v) The basic criterion for the self-adjointess of a symmetric operator $ T$ on a Hilbert space $ {\cal H}$: $ T$ is self-adjoint iff $ T$ is closed and $ \textrm{Ker}(T^*\pm\mathrm{i})=\{0\}$, equivalently, iff $ \textrm{Ran}(T\pm \mathrm{i})={\cal H}$.
(vi) Criterion for the essential self-adjointess of a symmetric operator $ T$ on a Hilbert space $ {\cal H}$: $ T$is essentially self-adjoint iff $ \textrm{Ker}(T^*\pm\mathrm{i})=\{0\}$, equivalently, iff $ \textrm{Ran}(T\pm \mathrm{i})$ are dense in $ {\cal H}$.


T15. New phenomena involving unbounded operators on a Hilbert space with respect to the bounded case:
(i) If $ T$ is unbounded, $ T^*$may not be densely defined (example: $ T:C^\infty_0\to L^2(\mathbb{R})$, $ (Tf)(x)=(\int_{\mathbb{R}} (1+\vert t\vert)^{-1/2} f(t){\rm d}t)e^{-x^2}$; then $ \mathcal{D}(T^*)=\{e^{-x^2}\}^\perp$ and $ T^*=\mathbb{O}$).
(ii) If unbounded, $ T$ may not be closed (the closure of $ T$, if it exists, is the unique operator $ \overline{T}$ satisfying $ \Gamma(\overline{T})=\overline{\Gamma(T)}$ and is the simplest and smallest extension of $ T$).
(iii) If densely defined, $ T$ is closable iff $ T^*$ is densely defined, in which case $ \overline{T}=T^{**}$.


T14. Unboundedness is "unavoidable". Position and momentum are unbounded observables. The energy of the harmonic oscillator is an unbounded observable. The Hellinger-Toepliz theorem (follows from Closed Graph).


T13. Recap: the functional calculus for bounded Borel functions of bounded self-adjoint operators on a Hilbert space. The key point is to establish the identity $ \Vert P(A)\Vert=\Vert P\Vert _{\textrm{sup}}$ for polynomials $ P$ (the rest is just Weiestrass + B.L.T. + bla bla), and it is here that the self-adjointness of $ A$ is crucially needed.


T12. The Weyl's criterion. Let $ A$ be a bounded, self-adjoint operator on a Hilbert space. $ \lambda\in\sigma(A)$ iff there exists a sequence $ \Vert\psi_n\}_{n=1}^\infty$ of normalised vectors $ (\Vert\psi_n\Vert=1)$ such that $ \Vert A\psi_n-\lambda\psi_n\Vert\to 0$ as $ n\to\infty$. See also exercise 3 in FA2, SoSe2009.


T11. Message from class: the spectral theory of compact operators on a Hilbert space has many similarities with the spectral theory on finite-dimensional spaces (for example, with the possible exception of zero, each eigenvalue of a compact operator has finite multiplicity). However, there are compact operators with no eigenvalues. For example, the Volterra operator [E5]. Or the operator $ T:\ell^2\to\ell^2$, $ T(x_1,x_2,x_3,\dots)=(0,x_1,x_2/2,x_3/3,\dots)$: $ T$ is compact and $ 0\in\sigma(T)$ since $ T^{-1}$ is not bounded, however $ T$ has no eigenvalues. On the other hand, every non-zero compact and self-adjoint operator $ T$ on a Hilbert space has a non-zero eigenvalue, since either $ -\Vert T\Vert$ or $ \Vert T\Vert$ is an eigenvalue.


T10.  Review of useful identities and estimates involving resolvents:
(i) $ R_\lambda(T)-R_\mu(T)=(\lambda-\mu)R_\lambda(T)R_\mu(T)$
(ii) $ R_\lambda(T)-R\lambda(S)=R_\lambda(T)(S-T)R_\lambda(S)$ $ =R_\lambda(S)(S-T)R_\lambda(T)$
(iii) $ R_\lambda(T)-R_\lambda(S)=(\mathbbm{1}-(\lambda-\mu)R_\lambda(T)(R_\mu(T)-R_\mu(S))(\mathbbm{1}-(\lambda-\mu)R_\lambda(S))$
(iv) $ R_\lambda(T)=\sum_{n=0}^\infty(\lambda-\lambda_0)^n R_{\lambda_0}(T)^{n+1}$ for $ \vert\lambda-\lambda_0\vert<\Vert R_{\lambda_0}(T)\Vert^{-1}$
(v) $ R_\lambda(T)=\sum_{n=0}^\infty\lambda^{-1-n}T^n$ for $ \vert\lambda\vert>\Vert T\Vert$ (the Neumann series)
(vi) $ \Vert R_\lambda(T)\Vert\geqslant(\textrm{dist}(\lambda,\sigma(T)))^{-1}$
(vii) if $ T$ is self-adjoint then $ \Vert R_\lambda(T)\Vert\leqslant\vert\mathfrak{Im}(\lambda)\vert^{-1}$


T9. Misc on orthogonal projections on a Hilbert space. Definition: $ P=P^2=P^*$. $ \textrm{Ran}P$is closed. $ P$ acts as the identity on its image and as the zero operator on the orthogonal to its image. The Projection Theorem establishes a 1:1 correspondence between orthogonal projections and closed subspaces. Also: $ \Vert P\Vert=1$, $ P$ is normal, $ P\geqslant\mathbb{O} $. Spectrum: $ \sigma(P)=\sigma_\textrm{point}(P)=\{0,1\}$. Resolvent: $ (\lambda-P)^{-1}=\frac{1}{\lambda-1}P+\frac{1}{\lambda}(\mathbbm{1}-P)$.  To practise further with orthogonal projections:
(i) Exercise 24, exercise 44, problem 18 in FA, SoSe2010.

(ii) Given two orthogonal projections $ P_1$, $ P_2$, one has $ P_1\leqslant P_2$ iff $ \textrm{Ran}P_1\subseteq\textrm{Ran}P_2$, in which case they commute and $ P_2-P_1$ is also an orthogonal projection.
(iii) Given two orthogonal projections $ P_1$,$ P_2$,$ \displaystyle\lim_{n\to\infty}(P_1 P_2)^n$ exists and is equal to the orthogonal projection onto $ \textrm{Ran}P_1\cap\textrm{Ran}P_2$. (Hint: use the monotonicity of $ (P_1P_2P_1)^n$.)


T8. Hilbert adjoint vs Banach adjoint. Comparison between the two definitions given in class. Examples (make sure you can work them out again!):
(i) the adjoint of the $ \ell^p\to\ell^p$right shift (see [E2.(v)]) is the $ \ell^{p'}\to\ell^{p'}$left shift
(ii) the adjoint of the $ \mathcal{D}\to\mathcal{D}$ Fourier transform (i.e., on test functions), is the $ \mathcal{D}'\to\mathcal{D}'$Fourier transform (i.e., on distributions)
(iii) in [E5] we determined the adjoint of the Volterra integral operator
(iv) a multiplication operator with a bounded measurable funct. is selfadjoint in $ L^2$
(v) the adjoint of the Hilbert-Schmidt integral operator [E1.(iv)] acts as$ (T^*f)(x):=\displaystyle\int_M \overline{K(y,x)}f(y)\mathrm{d}y$

T7. Misc on bounded linear operators on a Hilbert space. One has $ \textrm{Ker} A^*=(\textrm{Ran}A)^\perp$, whence (Projection Theorem)$ {\cal H}=\overline{\textrm{Ran}A}\oplus\textrm{Ker} A^* $. If $ A$ is normal, $ \textrm{Ker} A$$ =\textrm{Ker} A^*$$ =(\textrm{Ran}A)^\perp$$ =(\textrm{Ran}A^*)^\perp$. Recall the unique decomposition $ A=B+iC$ with $ B=B^*$ and $ C=C^*$. In partic., $ A$ is normal iff $ B$ and $ C$ commute, $ A$ is unitary iff it is normal and $ B^2+C^2=\mathbbm{1}$.


T6. Recall that a bounded linear operator $ A$ on a Hilbert space is known when one knows all its matrix elements $ \langle\varphi,A\psi\rangle$, and $ \Vert A\Vert=\displaystyle\sup_{\Vert\varphi\Vert=\Vert\psi\Vert=1}\vert\langle\varphi,A\psi\rangle\vert$. Via polarisation $ A$ can be reconstructed by its expectations $ \langle\psi,A\psi\rangle$. In partic., $ A=A^*$ is equivalent to $ \langle\psi,A\psi\rangle=\langle A\psi,\psi\rangle$ $ \forall\psi\in{\cal H}$, and via polarisation+parallelogram law it follows that the norm of a self-adjoint operator is given by (very useful!)$ \Vert A\Vert=\displaystyle\sup_{\Vert\psi\Vert=1}\vert\langle \psi,A\psi\rangle\vert$.


T5. The subadditivity argument needed in the proof (given in class) of the spectral radius formula. Note that the proof given for self-adjoint operators extends to normal operators.


T4. Picture, in the complex plane, of the resolvent set and of the spectrum of a bounded linear operator. The generic case (make sure you can draw it!). For self-adjoint operators the spectrum is on the real line, for unitary operators it is on the unit circle.


T3. A chain of Banach-space theorems (mentioned in class, used ubiquitously): Uniform Boundedness (aka Banach-Steinhaus), Open Mapping, Inverse Mapping, Closed Graph. The underlying Banach structure is crucial: see, e.g., the provocation in problem 42 in FA, SoSe2010. To play a bit with these theorems, look up exercise 44, problem 37, problem 42, and problem 4 (final test) in FA, SoSe2010.


T2. The Riesz Lemma (recap). The proof is based on the Projection Theorem. Corollary: a bounded sesquilinear form identifies uniquely a bounded linear operator: $ q(\varphi,\psi)=\langle\varphi,A\psi\rangle$. Note that the orthogonal complement to the kernel of a bounded linear functional is at most one-dimensional.


T1. The Projection Theorem $ \mathcal{H}=\overline{\mathcal{M}}\oplus\mathcal{M}^\perp$. We gave the proof in Functional Analysis (see e.g. exercise 3 in FA, SoSe2010). Recall that in infinite dimensions the proof exploits the fact that $ \overline{\mathcal{M}}$ is closed (of course!) and the parallelogram law (this is a Hilbert-space-only feature! see, e.g., exercise 20 in FA, SoSe2010).