# Functional Analysis/Special topics

This chapter collect some materials that didn't quite fit in the main development of the theory.

## Fredholm theory

We recall that the closed unit ball of a Banach space is compact if and only if the space is finite-dimensional. This is a special case of the next lemma:
7 Lemma Let $T:{\mathcal {X}}\to {\mathcal {Y}}$ be a closed densely defined operator. Then the following are equivalent.
* (i) $\dim \operatorname {ker} (T)<\infty$ and the range of T is closed.
* (ii) Every bounded sequence $f_{j}\in {\mathcal {X}}$ has a convergent subsequence when $Tf_{j}$ is convergent.
Proof: We may assume T has dense range. (i) $\Rightarrow$ (ii): Suppose $f_{j}$ is a bounded sequence such that $Tf_{j}$ is convergent. In view of the Hahn-Banach theorem, X is a direct sum of the kernel of $T$ and some other subspace, say, ${\mathcal {W}}$. Thus, we can write:
    $f_{j}=g_{j}+h_{j},(g_{j}\in \operatorname {ker} (T),h_{j}\in {\mathcal {W}})$
By the closed graph theorem, the inverse of $T:{\mathcal {W}}\to {\mathcal {Y}}$ is continuous. Since $Tf_{j}=Th_{j}$, the continuity implies that $h_{j}$ is convergent. Since $g_{j}$ contains a convergent subsequence by the paragraph preceding the theorem, $f_{j}$ has a convergent subsequence then. (ii) $\Rightarrow$ (i): (ii) implies the first condition of (i), again by the preceding paragraph. For the second, suppose $Tf_{j}$ is convergent. Then by (ii) $f_{j}$ has a subsequence $f_{j_{k}}$ converging to, say, $f$. Since the graph of T is closed, $Tf_{j_{k}}$ converges to $Tf$. $\square$
A bounded linear operator $T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}$ between Hilbert spaces is said to be Fredholm if T and $T^{*}$ both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of T and the quotient ${\mathfrak {H}}_{2}/T({\mathfrak {H}}_{1})$ are finite-dimensional. In fact, if ${\mathfrak {H}}_{2}/T({\mathfrak {H}}_{1})$ is finite-dimensional, then $T({\mathfrak {H}}_{1})$ is a complemented subspace; thus, closed. That $T$ has closed range implies that $T^{*}$ has closed range. For a Fredholm operator at least, it thus makes sense to define:
    $\operatorname {ind} (T)=\dim \operatorname {ker} (T)-\dim \operatorname {Coker} (T)$.
Because of the first isomorphism theorem, the index is actually independent of any operator T when T is a map between finite-dimensional spaces. This is no longer the case for operators acting on infinite-dimensional spaces.
7 Lemma Let $T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})$ and $S\in B({\mathfrak {H_{2}}},{\mathfrak {H_{3}}})$. If $T$ and $S$ are Fredholm operators, then $ST$ is a Fredholm operator with
    $\operatorname {ind} (ST)=\operatorname {ind} (T)+\operatorname {ind} (S)$.
Conversely, if ${\mathfrak {H_{3}}}={\mathfrak {H_{1}}}$, and both $TS$ and $ST$ are Fredholm operators, then $T$ is a Fredholm operator. Proof: Since
    $\dim \operatorname {ker} (ST)\leq \dim \operatorname {ker} (S)+\dim \operatorname {ker} (T)$, and $\dim \operatorname {Coker} (ST)\leq \dim \operatorname {Coker} (S)+\dim \operatorname {Coker} (T)$,
we see that $ST$ is Fredholm. Next, using the identity
    $\dim X+\dim X^{\bot }\cap Y=\dim Y+\dim Y^{\bot }\cap X$
we compute:
    $\dim \operatorname {ker} (ST)=\dim \operatorname {ker} (S)\cap \operatorname {ran} (T)+\dim \operatorname {ker} (T)=\dim \operatorname {ker} (S)\cap \operatorname {ker} (T^{*})^{\bot }+\dim \operatorname {ker} (T)$
    $=-\dim \operatorname {ker} (T^{*})+\dim \operatorname {ker} (S)+\dim \operatorname {ker} (S)^{\bot }\cap \operatorname {ker} (T^{*})+\dim \operatorname {ker} (T)$
    $=\operatorname {ind} (T)+\operatorname {ind} (S)+\dim \operatorname {ker} (T^{*}S^{*}).$
For the conversely, let $f_{j}$ be a bounded sequence such that $Tf_{j}$ is convergent. Then $STf_{j}$ is convergent and so $f_{j}$ has a convergent sequence when $ST$ is Fredholm. Thus, $\dim \operatorname {ker} T<\infty$ and $T$ has closed range. That $TS$ is a Fredholm operator shows that this is also true for $T^{*}$ and we conclude that $T$ is Fredholm. $\square$
7 Theorem The mapping
    $T\mapsto \operatorname {ind} (T)$
is a locally constant function on the set of Fredhold operators $T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}$. Proof: By the Hahn-Banach theorem, we have decompositions:
    ${\mathfrak {H}}_{1}=C_{1}\oplus \operatorname {ker} (T),\quad {\mathfrak {H}}_{2}=\operatorname {ran} (T)\oplus C_{2}$.
With respect to these, we represent T by a block matrix:
    $T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}$
where $T':C_{1}\to \operatorname {ran} T$. By the above lemma, $\operatorname {ind}$ is invariant under row and column operations. Thus, for any $S={\begin{bmatrix}S_{1}&S_{2}\\S_{3}&S_{4}\\\end{bmatrix}}$, we have:
    $\operatorname {ind} (T+S)=\operatorname {ind} ({\begin{bmatrix}T'+S_{1}&0\\0&A\\\end{bmatrix}})=\operatorname {ind} (A)$,
since $T'+S_{1}$ is invertible when $\|S\|$ is small. A depends on S but the point is that $A$ is a linear operator between finite-dimensional spaces. Hence, the index of A is independent of A; thus, of S. $\square$
7 Corollary If $T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})$ is a Fredholm operator and K is a compact operator, then $T+K$ is a Fredholm operator with
    $\operatorname {ind} (T+K)=\operatorname {ind} (T)$
Proof: Let $f_{j}$ be a bounded sequence such that $(T+K)f_{j}$ is convergent. By compactness, $f_{j}$ has a convergent subsequence $f_{j_{k}}$ such that $Kf_{j_{k}}$ is convergent. $Tf_{j_{k}}$ is then convergent and so $f_{j_{k}}$ contains a convergent subsequence. Since $K^{*}$ is compact, the same argument applies to $T^{*}+K^{*}$. The invariance of the index follows from the preceding theorem since $T+\lambda K$ is Fredholm for any complex number $\lambda$, and the index of $T+\lambda K$ is constant. $\square$
The next result, known as Fredholm alternative, is now easy but is very important in application.
7 Corollary If $K\in B({\mathfrak {H}}_{1},{\mathfrak {H}}_{2})$ is a compact, then
    $\operatorname {ker} (K-\lambda I)$ and ${\mathfrak {H}}_{2}/\operatorname {ran} (K-\lambda I)$
have the same (finite) dimension for any nonzero complex number $\lambda$, and $\sigma (K)\backslash \{0\}$ consists of eigenvalues of K. Proof: The first assertion follows from:
    $\operatorname {ind} (K-\lambda I)=\operatorname {ind} (I-\lambda ^{-1}K)=0$,
and the second is the immediate consequence. $\square$
7 Theorem Let $T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})$. Then $T$ is a Fredholm operator if and only if $I-TS$ and $I-ST$ are finite-rank operators for some $S\in B({\mathfrak {H_{2}}},{\mathfrak {H_{1}}})$. Moreover, when $I-TS$ and $I-ST$ are of trace class (e.g., of finite-rank),
    $\operatorname {ind} (T)=\operatorname {Tr} (I-ST)-\operatorname {Tr} (I-TS)$
Proof: Since the identities are Fredholm operators (in fact, any invertible operator) and since
    $ST=I_{1}+(I_{1}-ST)$
    $TS=I_{2}+(I_{2}-TS)$
$ST$ and $TS$ are Fredholm operators, which implies T is a Fredholm operator. Conversely, suppose T is a Fredholm operator. Then, as before, we can write:
    $T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}$
where $T'$ is invertible. If we set, for example, $S={\begin{bmatrix}T'^{-1}&0\\0&0\\\end{bmatrix}}$, then $S$ has required properties. Next, suppose S is given arbitrary: $S={\begin{bmatrix}S_{1}&S_{2}\\S_{3}&S_{4}\\\end{bmatrix}}$. Then
    $\operatorname {Tr} (I-ST)=\operatorname {Tr} (1-S_{1}T')+\operatorname {dim} \operatorname {ker} T$.
Similarly, we compute:
    $\operatorname {Tr} (I-TS)=\operatorname {Tr} (1-T'S_{1})+\operatorname {dim} \operatorname {Coker} T$.
Now, since $T'(I-S_{1}T')=(I-T'S_{1})T'$, and $T'$ is invertible, we have:
    $\operatorname {Tr} (I-S_{1}T')=\operatorname {Tr} (I-T'S_{1})$. $\square$

## Representations of compact groups

Theorem Every irreducible unitary representation of a compact group is finite-dimensional.