# 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 ${\displaystyle T:{\mathcal {X}}\to {\mathcal {Y}}}$ be a closed densely defined operator. Then the following are equivalent.

• (i) ${\displaystyle \dim \operatorname {ker} (T)<\infty }$ and the range of T is closed.
• (ii) Every bounded sequence ${\displaystyle f_{j}\in {\mathcal {X}}}$ has a convergent subsequence when ${\displaystyle Tf_{j}}$ is convergent.

Proof: We may assume T has dense range. (i) ${\displaystyle \Rightarrow }$ (ii): Suppose ${\displaystyle f_{j}}$ is a bounded sequence such that ${\displaystyle Tf_{j}}$ is convergent. In view of the Hahn-Banach theorem, X is a direct sum of the kernel of ${\displaystyle T}$ and some other subspace, say, ${\displaystyle {\mathcal {W}}}$. Thus, we can write:

${\displaystyle f_{j}=g_{j}+h_{j},(g_{j}\in \operatorname {ker} (T),h_{j}\in {\mathcal {W}})}$

By the closed graph theorem, the inverse of ${\displaystyle T:{\mathcal {W}}\to {\mathcal {Y}}}$ is continuous. Since ${\displaystyle Tf_{j}=Th_{j}}$, the continuity implies that ${\displaystyle h_{j}}$ is convergent. Since ${\displaystyle g_{j}}$ contains a convergent subsequence by the paragraph preceding the theorem, ${\displaystyle f_{j}}$ has a convergent subsequence then. (ii) ${\displaystyle \Rightarrow }$ (i): (ii) implies the first condition of (i), again by the preceding paragraph. For the second, suppose ${\displaystyle Tf_{j}}$ is convergent. Then by (ii) ${\displaystyle f_{j}}$ has a subsequence ${\displaystyle f_{j_{k}}}$ converging to, say, ${\displaystyle f}$. Since the graph of T is closed, ${\displaystyle Tf_{j_{k}}}$ converges to ${\displaystyle Tf}$. ${\displaystyle \square }$

A bounded linear operator ${\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}}$ between Hilbert spaces is said to be Fredholm if T and ${\displaystyle T^{*}}$ both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of T and the quotient ${\displaystyle {\mathfrak {H}}_{2}/T({\mathfrak {H}}_{1})}$ are finite-dimensional. In fact, if ${\displaystyle {\mathfrak {H}}_{2}/T({\mathfrak {H}}_{1})}$ is finite-dimensional, then ${\displaystyle T({\mathfrak {H}}_{1})}$ is a complemented subspace; thus, closed. That ${\displaystyle T}$ has closed range implies that ${\displaystyle T^{*}}$ has closed range. For a Fredholm operator at least, it thus makes sense to define:

${\displaystyle \operatorname {ind} (T)=\dim \operatorname {ker} (T)-\dim \operatorname {Coker} (T)}$.

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 ${\displaystyle T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})}$ and ${\displaystyle S\in B({\mathfrak {H_{2}}},{\mathfrak {H_{3}}})}$. If ${\displaystyle T}$ and ${\displaystyle S}$ are Fredholm operators, then ${\displaystyle ST}$ is a Fredholm operator with

${\displaystyle \operatorname {ind} (ST)=\operatorname {ind} (T)+\operatorname {ind} (S)}$.

Conversely, if ${\displaystyle {\mathfrak {H_{3}}}={\mathfrak {H_{1}}}}$, and both ${\displaystyle TS}$ and ${\displaystyle ST}$ are Fredholm operators, then ${\displaystyle T}$ is a Fredholm operator.
Proof: Since

${\displaystyle \dim \operatorname {ker} (ST)\leq \dim \operatorname {ker} (S)+\dim \operatorname {ker} (T)}$, and ${\displaystyle \dim \operatorname {Coker} (ST)\leq \dim \operatorname {Coker} (S)+\dim \operatorname {Coker} (T)}$,

we see that ${\displaystyle ST}$ is Fredholm. Next, using the identity

${\displaystyle \dim X+\dim X^{\bot }\cap Y=\dim Y+\dim Y^{\bot }\cap X}$

we compute:

${\displaystyle \dim \operatorname {ker} (ST)=\dim \operatorname {ker} (S)\cap \operatorname {ran} (T)+\dim \operatorname {ker} (T)=\dim \operatorname {ker} (S)\cap \operatorname {ker} (T^{*})^{\bot }+\dim \operatorname {ker} (T)}$
${\displaystyle =-\dim \operatorname {ker} (T^{*})+\dim \operatorname {ker} (S)+\dim \operatorname {ker} (S)^{\bot }\cap \operatorname {ker} (T^{*})+\dim \operatorname {ker} (T)}$
${\displaystyle =\operatorname {ind} (T)+\operatorname {ind} (S)+\dim \operatorname {ker} (T^{*}S^{*}).}$

For the conversely, let ${\displaystyle f_{j}}$ be a bounded sequence such that ${\displaystyle Tf_{j}}$ is convergent. Then ${\displaystyle STf_{j}}$ is convergent and so ${\displaystyle f_{j}}$ has a convergent sequence when ${\displaystyle ST}$ is Fredholm. Thus, ${\displaystyle \dim \operatorname {ker} T<\infty }$ and ${\displaystyle T}$ has closed range. That ${\displaystyle TS}$ is a Fredholm operator shows that this is also true for ${\displaystyle T^{*}}$ and we conclude that ${\displaystyle T}$ is Fredholm. ${\displaystyle \square }$

7 Theorem The mapping

${\displaystyle T\mapsto \operatorname {ind} (T)}$

is a locally constant function on the set of Fredhold operators ${\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}}$.
Proof: By the Hahn-Banach theorem, we have decompositions:

${\displaystyle {\mathfrak {H}}_{1}=C_{1}\oplus \operatorname {ker} (T),\quad {\mathfrak {H}}_{2}=\operatorname {ran} (T)\oplus C_{2}}$.

With respect to these, we represent T by a block matrix:

${\displaystyle T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}}$

where ${\displaystyle T':C_{1}\to \operatorname {ran} T}$. By the above lemma, ${\displaystyle \operatorname {ind} }$ is invariant under row and column operations. Thus, for any ${\displaystyle S={\begin{bmatrix}S_{1}&S_{2}\\S_{3}&S_{4}\\\end{bmatrix}}}$, we have:

${\displaystyle \operatorname {ind} (T+S)=\operatorname {ind} ({\begin{bmatrix}T'+S_{1}&0\\0&A\\\end{bmatrix}})=\operatorname {ind} (A)}$,

since ${\displaystyle T'+S_{1}}$ is invertible when ${\displaystyle \|S\|}$ is small. A depends on S but the point is that ${\displaystyle A}$ is a linear operator between finite-dimensional spaces. Hence, the index of A is independent of A; thus, of S. ${\displaystyle \square }$

7 Corollary If ${\displaystyle T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})}$ is a Fredholm operator and K is a compact operator, then ${\displaystyle T+K}$ is a Fredholm operator with

${\displaystyle \operatorname {ind} (T+K)=\operatorname {ind} (T)}$

Proof: Let ${\displaystyle f_{j}}$ be a bounded sequence such that ${\displaystyle (T+K)f_{j}}$ is convergent. By compactness, ${\displaystyle f_{j}}$ has a convergent subsequence ${\displaystyle f_{j_{k}}}$ such that ${\displaystyle Kf_{j_{k}}}$ is convergent. ${\displaystyle Tf_{j_{k}}}$ is then convergent and so ${\displaystyle f_{j_{k}}}$ contains a convergent subsequence. Since ${\displaystyle K^{*}}$ is compact, the same argument applies to ${\displaystyle T^{*}+K^{*}}$. The invariance of the index follows from the preceding theorem since ${\displaystyle T+\lambda K}$ is Fredholm for any complex number ${\displaystyle \lambda }$, and the index of ${\displaystyle T+\lambda K}$ is constant. ${\displaystyle \square }$

The next result, known as Fredholm alternative, is now easy but is very important in application.

7 Corollary If ${\displaystyle K\in B({\mathfrak {H}}_{1},{\mathfrak {H}}_{2})}$ is a compact, then

${\displaystyle \operatorname {ker} (K-\lambda I)}$ and ${\displaystyle {\mathfrak {H}}_{2}/\operatorname {ran} (K-\lambda I)}$

have the same (finite) dimension for any nonzero complex number ${\displaystyle \lambda }$, and ${\displaystyle \sigma (K)\backslash \{0\}}$ consists of eigenvalues of K.
Proof: The first assertion follows from:

${\displaystyle \operatorname {ind} (K-\lambda I)=\operatorname {ind} (I-\lambda ^{-1}K)=0}$,

and the second is the immediate consequence. ${\displaystyle \square }$

7 Theorem Let ${\displaystyle T\in B({\mathfrak {H_{1}}},{\mathfrak {H_{2}}})}$. Then ${\displaystyle T}$ is a Fredholm operator if and only if ${\displaystyle I-TS}$ and ${\displaystyle I-ST}$ are finite-rank operators for some ${\displaystyle S\in B({\mathfrak {H_{2}}},{\mathfrak {H_{1}}})}$. Moreover, when ${\displaystyle I-TS}$ and ${\displaystyle I-ST}$ are of trace class (e.g., of finite-rank),

${\displaystyle \operatorname {ind} (T)=\operatorname {Tr} (I-ST)-\operatorname {Tr} (I-TS)}$

Proof: Since the identities are Fredholm operators (in fact, any invertible operator) and since

${\displaystyle ST=I_{1}+(I_{1}-ST)}$
${\displaystyle TS=I_{2}+(I_{2}-TS)}$

${\displaystyle ST}$ and ${\displaystyle TS}$ are Fredholm operators, which implies T is a Fredholm operator. Conversely, suppose T is a Fredholm operator. Then, as before, we can write:

${\displaystyle T={\begin{bmatrix}T'&0\\0&0\\\end{bmatrix}}}$

where ${\displaystyle T'}$ is invertible. If we set, for example, ${\displaystyle S={\begin{bmatrix}T'^{-1}&0\\0&0\\\end{bmatrix}}}$, then ${\displaystyle S}$ has required properties. Next, suppose S is given arbitrary: ${\displaystyle S={\begin{bmatrix}S_{1}&S_{2}\\S_{3}&S_{4}\\\end{bmatrix}}}$. Then

${\displaystyle \operatorname {Tr} (I-ST)=\operatorname {Tr} (1-S_{1}T')+\operatorname {dim} \operatorname {ker} T}$.

Similarly, we compute:

${\displaystyle \operatorname {Tr} (I-TS)=\operatorname {Tr} (1-T'S_{1})+\operatorname {dim} \operatorname {Coker} T}$.

Now, since ${\displaystyle T'(I-S_{1}T')=(I-T'S_{1})T'}$, and ${\displaystyle T'}$ is invertible, we have:

${\displaystyle \operatorname {Tr} (I-S_{1}T')=\operatorname {Tr} (I-T'S_{1})}$. ${\displaystyle \square }$

## Representations of compact groups

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