\documentclass{slides} \usepackage{amssymb, amsmath, amscd, color, epsfig} %\usepackage[matrix,arrow]{xy} \newcommand{\green}{\color[rgb]{0,0.4,0}} \newcommand{\purple}{\color[rgb]{0.4,0,0.4}} \newcommand{\red}{\color[rgb]{0.7,0,0}} \newcommand{\blue}{\color{blue}} \def\eqref#1{(\ref{#1})} \newcommand{\goth}{\mathfrak} \newcommand{\g}{{\frak g}} \newcommand{\arrow}{{\:\longrightarrow\:}} \newcommand{\Z}{{\Bbb Z}} \newcommand{\C}{{\Bbb C}} \newcommand{\R}{{\Bbb R}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\6}{\partial} \def\1{\sqrt{-1}\:} \newcommand{\restrict}[1]{{\left|_{{#1}}\right.}} \newcommand{\cntrct} % contraction with a vector field {\hspace{2pt}\raisebox{1pt}{\text{$\lrcorner$}}\hspace{2pt}} \def\Bbb#1{\mathbb #1} \newcommand{\calo}{{\cal O}} \newcommand{\cac}{{\cal C}} % Correcting TeX... %\let\oldtilde=\tilde %\renewcommand{\tilde}{\widetilde} \renewcommand{\bar}{\overline} \renewcommand{\phi}{\varphi} \renewcommand{\epsilon}{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant} % Operatornames \newcommand{\even}{{\rm even}} \newcommand{\ev}{{\rm even}} \newcommand{\odd}{{\rm odd}} \newcommand{\const}{{\sf const}} \newcommand{\fl}{{\rm fl}} \newcommand{\im}{\operatorname{im}} \newcommand{\End}{\operatorname{End}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Hol}{\operatorname{{\cal H}ol}} \newcommand{\Tot}{\operatorname{Tot}} \newcommand{\Id}{\operatorname{Id}} \newcommand{\id}{\operatorname{\text{\sf id}}} \newcommand{\symb}{\operatorname{\text{\sf symb}}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\Var}{\operatorname{Var}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Alt}{\operatorname{Alt}} \newcommand{\Iso}{\operatorname{Iso}} \newcommand{\Sec}{\operatorname{Sec}} \newcommand{\Can}{\operatorname{Can}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Spin}{\operatorname{Spin}} \newcommand{\codim}{\operatorname{codim}} \newcommand{\coim}{\operatorname{coim}} \newcommand{\coker}{\operatorname{coker}} \newcommand{\ind}{\operatorname{\sf ind}} \newcommand{\rk}{\operatorname{rk}} \newcommand{\Def}{\operatorname{Def}} \newcommand{\Lie}{\operatorname{Lie}} \newcommand{\Tw}{\operatorname{Tw}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Diff}{\operatorname{Diff}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\inbfpare}[1]{{% \mbox{\tt (}\hspace{-5pt}\mbox{\tt (} #1 % \mbox{\tt )}\hspace{-5pt}\mbox{\tt )}% }} \newcommand{\comment}[1]{{}} \def\blacksquare{\hbox{\vrule width 10pt height 10pt depth 0pt}} \def\endproof{\blacksquare} \def\shortdash{\mbox{\vrule width 4.5pt height 0.55ex depth -0.5ex}} \makeatletter %\@ifundefined{Bbb} % {\newcommand{\Bbb}[1]{{\mathbb #1}}}% %{}% {\edef\Bbb#1{{\Bbb #1}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Pagestyle % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\ps@verbit}{% \renewcommand{\@oddhead}{% \scriptsize {\it \small Hodge theory, lecture 3 \hfil \tiny M. Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{{\bf \green EXERCISE:\ }} \newcommand{\pstep}{{\bf \green Proof. Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Hodge theory \\[15mm] \small lecture 3: compact operators} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, January 31, 2018 } \end{center} \newpage {\bf \blue Hilbert spaces} \definition {\bf \blue Hilbert space} is a complete, infinite-dimensional Hermitian space which is second countable (that is, has a countable dense set). \definition {\bf \blue Orthonormal basis} in a Hilbert space $H$ is a set of pairwise orthogonal vectors $\{x_\alpha\}$ which satisfy $|x_\alpha|=1$, and such that $H$ is the closure of the subspace generated by the set $\{x_\alpha\}$. \theorem {\bf \red Any Hilbert space has a basis, and all such bases are countable.} \proof A basis is found using Zorn lemma. If it's not countable, open balls with centers in $x_\alpha$ and radius $\epsilon < 2^{-1/2}$ don't intersect, which means that the second countability axiom is not satisfied. \endproof \theorem {\bf \red All Hilbert spaces are isometric}. \proof Each Hilbert space has a countable orthonormal basis. \endproof \remark The concept of ``Hilbert space'' and this theorem is due to John von Neumann (1927); he's also responsible for the spectral theorem for compact operators. \newpage {\bf \blue John von Neumann, 1903-1957}\\ \centerline{\epsfig{file=john-von-neumann-p-cropped.jpg,width=0.70\linewidth}} {\small Dr. John Von Neumann (right) stands with Dr. J. Robert Oppenheimer in front of an early computer. {\it \green ...Although Max insisted von Neumann attend school at the grade level appropriate to his age, he agreed to hire private tutors to give him advanced instruction in those areas in which he had displayed an aptitude. At the age of 15, he began to study advanced calculus under the renowned analyst G\'abor Szeg\H o. On their first meeting, Szeg\H o was so astounded with the boy's mathematical talent that he was brought to tears... }} \newpage {\bf \blue Real Hilbert spaces} \definition A Euclidean space is a vector space over $\R$ equipped with a positive definite scalar product $g$. \definition {\bf \blue Real Hilbert space} is a complete, infinite-dimensional Euclidean space which is second countable (that is, has a countable dense set). \definition {\bf \blue Orthonormal basis} in a Hilbert space $H$ is a set of pairwise orthogonal vectors $\{x_\alpha\}$ which satisfy $|x_\alpha|=1$, and such that $H$ is the closure of the subspace generated by the set $\{x_\alpha\}$. \theorem {\bf \red Any real Hilbert space has a basis, and all such bases are countable.} \proof A basis is found using Zorn lemma. If it's not countable, open balls with centers in $x_\alpha$ and radius $\epsilon < 2^{-1/2}$ don't intersect, which means that the second countability axiom is not satisfied. \endproof \theorem {\bf \red All real Hilbert spaces are isometric}. \proof Each Hilbert space has a countable orthonormal basis. \endproof \newpage {\bf\blue Adjoint operators (reminder)} \claim Let $V$ be a Hilbert space (real or complex), $g$ a scalar product on $V$, and $A \in \End(V)$. {\bf \red Then there exists a unique operator $A^*\in \End(V)$ such that $g(A (x), y)= g(x, A^* (y))$ for all $x, y\in V$. } \proof Let $x_1, ..., x_n, ...$ be an orthonormal basis in $V$, $A=(a_{ij})$ the matrix of $A$. Then $g(A(x_i), x_j)=a_{ij}$ and $g(x_i, A^*(x_j))= \bar a_{ij}$. This gives the existence. Uniqueness is clear, because if $g(x, (A^*_1-A^*_2) (y))=0$ for all $x, y$, we have $A^*_1-A^*_2=0$ {\bf \purple (prove it).} \endproof \definition In this situation, the operator $A^*$ is called {\bf\blue orthogonal adjoint} or {\bf \blue Hermitian adjoint} to $A$. In orthonormal basis, {\bf \purple this operator is represented by the transposed matrix $A^t$} for real Hilbert spaces and by $\bar A^t$ for complex Hilbert spaces. \newpage {\bf\blue Self-adjoint operators} \definition Let $V$ be a vector space and $g\in \Sym^2 V$ a scalar product. An operator $A:\; V \arrow V$ is called {\bf \blue self-adjoint} if $A=A^*$. \remark If we work in a real vector space orthonormal basis, a self-adjoint operator is given by a matrix that satisfies $A=A^t$, that is, {\bf \purple symmetric}. The self-adjoint operators are often called {\bf \blue symmetric operators}. For complex spaces, $A$ is Hermitian if and only if $A^t=\bar A$; such operators are called {\bf \blue Hermitian}. Assume that $V$ is finitely-dimensional. \claim Let $A$ be a (real) self-adjoint operator on $(V,g)$, and $g_A(x, y):= g(A(x), y)$. {\bf \red Then $g_A$ is a bilinear symmetric form on $V$.} Moreover, {\bf \red the map $A \mapsto g_A$ gives a bijective correspondence between self-adjoint operators and bilinear symmetric forms on $V$.} \proof Using $g$ to identify $V$ and $V^*$, we obtain that the spaces $V^* \otimes V^*$ of bilinear symmetric forms and $\End(V)=V\otimes V^*$ are also identified. This identification is given by a map $A\mapsto g(A(\cdot), \cdot)$. By definition, the form $g_A(\cdot, \cdot):=g(A(\cdot), \cdot)$. is symmetric if and only if $A$ is self-adjoint. \endproof \remark This is just another way to construct the well-known {\bf \purple bijective correspondence between symmetric matrices and bilinear symmetric forms.} \remark The same argument produces an {\bf \purple equivalence between Hermitian matrices $A^t=\bar A$ and Hermitian forms} on a complex vector space. \newpage {\bf\blue Normal form of a pair of bilinear symmetric forms} {\bf \green Theorem 1:} {\bf \blue (spectral theorem for self-adjoint operators)}\\ Let $A$ be a self-adjoint operator on a finite-dimensional space $(V, g)$. {\bf \red Then $A$ can be diagonalized in an orthonormal basis.} {\bf \green Theorem $1'$:} {\bf \blue (``principal axis theorem'')} Let $V=\R^n$, and $h, h'\in \Sym^2 V^*$ be two bilinear symmetric forms, with $h$ positive definite. {\bf \red Then there exists a basis $x_1, ..., x_n$ which is orthonormal with respect to $h$, and orthogonal with respect to $h'$. } {\bf \purple These theorems are clearly equivalent;} I will give a proof later. \newpage {\bf\blue ``Principal axis theorem''} \remark Theorem $1'$ implies the following statement about ellipsoids: for any positive definite quadratic form $q$ in $\R^n$, consider the ellipsoid \[ S= \{v\in V\ \ |\ \ q(v)=1\}. \] The group $SO(n)$ acts on $\R^n$ preserving the standard scalar product. {\bf \red Then for some $g\in SO(n)$, $g(S)$ is given by equation $\sum a_i x_i^2=1$, where $a_i >0$.} This is called {\bf \blue finding principal axes of an ellipsoid}. \centerline{\epsfig{file=Ellipsoid.png,width=0.40\linewidth}} \newpage {\bf\blue Maximum of a quadratic form on a sphere} \lemma Let $V=\R^n$, and $h, h'\in \Sym^2 V^*$ be two bilinear symmetric forms, $h$ positive definite, and $q(v)=h(v,v), q'(v)=h'(v,v)$ the corresponding quadratic forms. Consider $q'$ as a function on a sphere $S= \{v\in V\ \ |\ \ q(v)=1\}$, and let $x\in S$ be the point where $q'$ attains maximum. Denote by $x^{\bot_h}$ and $x^{\bot_{h'}}$ the orthogonal complements with respect to $h, h'$. {\bf \red Then $x^{\bot_h}=x^{\bot_{h'}}$.} \proof The tangent space $T_xS$ to a sphere $S$ is $x^{\bot_h}$. Since $q'(x)$ reaches maximum on a sphere, one has $\frac d {d\epsilon} q'(x+\epsilon v)= 2 h'(x, v)= 0$ for any $v\in T_x S=x^{\bot_h}$. This gives $h'(x, x^{\bot_h})=0$. \endproof {\bf \green Theorem $1'$:} {\bf \blue (``principal axis theorem'')} Let $V=\R^n$, and $h, h'\in \Sym^2 V^*$ be two bilinear symmetric forms, with $h$ positive definite. {\bf \red Then there exists a basis $x_1, ..., x_n$ which is orthonormal with respect to $h$, and orthogonal with respect to $h'$. } \proof Let $q(v)=h(v,v), q'(v)=h'(v,v)$ the corresponding quadratic forms. Consider $q'$ as a function on a sphere $S= \{v\in V\ \ |\ \ q(v)=1\}$, and let $x_1\in S$ be the point where $q'$ attains maximum. Then $x_1^{\bot_h}=x_1^{\bot_{h'}}$. Using induction, we may assume that on $x_1^{\bot_h}$, Theorem 1 is already proven, and there exists a basis $x_2, ..., x_n$ orthonormal for $h$ and orthogonal for $h'$. {\bf \purple Then $x_1, x_2, ..., x_n$ is orthonormal for $h$ and orthogonal for $h'$.} \endproof \newpage {\bf \blue Weak convergence} \definition Let $x_i \in H$ be a sequence of points in a Hilbert space $H$. We say that $x_i$ {\bf \blue weakly converges} to $x\in H$ if for any $z\in H$ one has $\lim_i g(x_i, z) = g(x, z)$. \remark Let $y(i)= \alpha_j(i) e_j$ be a sequence of points in a a Hilbert space with orthonormal basis $e_i$. {\bf \purple Then $y(i)$ converges to $y= \sum_j \alpha_j e_j $ if and only if $\lim_i \alpha_j(i)=\alpha_i$.} \claim For any sequence $\{y(i)=\sum_j \alpha_j(i) e_j\}$ of points in a unit ball, {\bf \red there exists a subsequence $\{\tilde y(i)=\tilde \alpha_j(i) e_i\}$ weakly converging to $y\in H$.} \proof Indeed, $|\alpha_j(i)|\leq 1$, hence there exist a subsequence $\tilde y(i)= \tilde \alpha_j(i) x_j$ with $\tilde \alpha_j(i)$ converging for each $j$. The limit belongs to the unit ball because otherwise $\left|\sum_{j=1}^n \tilde \alpha_j(i) e_j\right|>1$, which is impossible. \endproof \remark Note that {\bf \red the function $x\arrow |x |$ is not continuous in weak topology.} Indeed, weak limit of $\{e_i\}$ is 0. The proof above shows that $|\cdot|$ is semicontinuous. \newpage {\bf \blue Compact operators} \definition {\bf \blue Precompact set} is a set which has compact closure. {\bf \blue A compact operator} is an operator which maps bounded sets to precompact. \example Let $A\in \Hom(H,H)$ be an operator on Hilbert spaces and $\{e_i\}$ an orthonormal basis in $H$. Let $A(e_i)=z_i$; assume that $\sum |z_i|^2 <\infty$. {\bf \red Then $A$ is compact.} \pstep Let $y(i)= \alpha_j(i) e_j$ be a sequence of points in a unit ball. Replacing $y(i)$ by a subsequence, we may assume that $y(i)$ weakly converges to $y$. {\bf \green Step 2:} Then \[ \lim_i A(y(i))=\lim_i\lim_n A\left(\sum_{j=1}^n \alpha_j(i)\right)= \lim_n\sum_{j=1}^n \alpha_j A(e_i) \] and this sequence converges in the usual topology on $H$, because $\alpha_j$ are bounded and $\sum_i |A(e_i)|^2$ is bounded. \endproof \newpage {\bf \blue Compact operators and weak convergence} \theorem Let $A:\; H \arrow H$ be a compact operator. {\bf \red Then $A$ maps any weakly convergent sequence to a convergent one.} \proof Let $\{y_i\}$ be a sequence which weakly converges to $y$. Replacing $\{y_i\}$ by a subsequence, we may assume that $A(y_i)$ converges to $z$. Then $\lim_i g(A(y_i), v)= g(z, v)$ for any $v\in H$. However, \[ \lim_i g(A(y_i), v)= \lim_i g(y_i, A^*(v))= g(y, A^*(v))= g(A(y), v).\] Then $g(z, v)= g(A(y), v)$ for all $v\in H$, giving $z= A(y)$. \endproof \remark Converse is also true: {\bf \purple you can characterize a compact operator as one which maps weakly convergent sequences to convergent.} Indeed, unit ball is weakly compact, as we have shown, hence its image is precompact for any map which takes the weakly convergent sequences to convergent. \newpage {\bf \blue Spectral theorem} \theorem {\bf \blue (Spectral theorem for self-adjoint operators)} \\ Let $A:\; H \arrow H$ be a compact self-adjoint operator on a Hilbert space. {\bf \red Then $A$ can be diagonalized} in a certain orthonormal basis $e_1, e_2, ...$, with $\lim_i\alpha_i =0$. \pstep The eigenvalues converge to 0 because $A$ is compact. Let $B\subset H$ be the unit ball, and $X$ the closure of $A(B)$. Denote by $x\in X$ the vector where $|x|$ is maximal. We shall prove that $x= A(z)$. To finish the proof of Spectral Theorem {\bf \purple it would suffice to show that $z$ is an eigenvector and $A(z^\bot)\subset z^\bot$.} {\bf \green Step 2:} Let $z_i\in B$ be a sequence such that $\lim_i A(z_i)=x$. Replacing $z_i$ by a subsequence, we may assume that $z_i$ weakly converges to $z\in B$. Then $A(z)=x$, because $A$ maps weakly convergent sequences to convergent. {\bf \purple This implies that $x\in \im A$.} {\bf \green Step 3:} Let $z\in H$ be an element of the unit sphere such that $A(z)=x$. Then $|A(z)|= \| A\|$. Since $A$ is self-adjoint, $g(A(z), A(z))= g(A^2(z), z) = \| A\|^2$. This gives $\|A^2\|= \|A\|^2$, hence $|A^2(z)|=g(A^2(z), z)|$. Since $g(A^2(z), z)=|z||A^2(z)| \cos \phi$, where $\phi$ is an angle between $z$ and $A^2(z)$, the equality $g(A^2(z), z)= |z||A^2(z)|$ implies that $z$ and $A^2(z)$ are proportional, hence $x$ is an eigenvector for $A^2$. \newpage {\bf \blue Spectral theorem (2)} \theorem {\bf \blue (Spectral theorem for self-adjoint operators)} \\ Let $A:\; H \arrow H$ be a compact self-adjoint operator on a Hilbert space. {\bf \red Then $A$ can be diagonalized} in a certain orthonormal basis $e_1, e_2, ...$, with $\lim_i\alpha_i =0$. {\bf \green Steps 2-3:} We have shown that there exists a vector $z\in H$ in a unit sphere such that $|A(z)| = \|A\|$. Moreover, $z$ is an eigenvector of $A^2$. {\bf \green Step 4:} Since $A^2$ is self-adjoint, for any space $W\subset H$ satisfying $A^2(W)\subset W$, one has $A^2(W^\bot)\subset W^\bot$. {\bf \purple We proved that $A^2$ is diagonal in an orthonormal basis.} {\bf \green Step 5:} This implies that $H$ is an orthogonal direct sum of eigenspaces for $A^2$, which are finite-dimensional for non-zero eigenvalues, because $A^2$ is compact. Since $A$ and $A^2$ commute, on each of these eigenspaces $A$ acts as an adjoint operator, and we can apply the finite-dimensional spectral theorem. \endproof \remark {\bf \red The same statement is true for Hermitian self-adjoint operators;} the proof is the same. \end{document}