# 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.