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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}
