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 $T:𝒳\to 𝒴$ be a closed densely defined operator. Then the following are equivalent.

• (i) $\dim \ker (T)<\mathrm{\infty }$ and the range of T is closed.
• (ii) Every bounded sequence $f_j\in 𝒳$ has a convergent subsequence when $T{f}_j$ is convergent.

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

$f_j=g_j+h_j,(g_j\in \ker (T),h_j\in 𝒲)$

By the closed graph theorem, the inverse of $T:𝒲\to 𝒴$ is continuous. Since $T{f}_j=T{h}_j$, 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 $T{f}_j$ is convergent. Then by (ii) $f_j$ has a subsequence $f_{j_k}$ converging to, say, $f$. Since the graph of T is closed, $T{f}_{j_k}$ converges to $Tf$. $◻$

A bounded linear operator $T:{ℌ}_1\to {ℌ}_2$ between Hilbert spaces is said to be Fredholm if T and $T^{\ast }$ both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of T and the quotient ${ℌ}_2/T({ℌ}_1)$ are finite-dimensional. In fact, if ${ℌ}_2/T({ℌ}_1)$ is finite-dimensional, then $T({ℌ}_1)$ is a complemented subspace; thus, closed. That $T$ has closed range implies that $T^{\ast }$ has closed range. For a Fredholm operator at least, it thus makes sense to define:

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

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 $T\in B({ℌ}_1,{ℌ}_2)$ and $S\in B({ℌ}_2,{ℌ}_3)$. If $T$ and $S$ are Fredholm operators, then $ST$ is a Fredholm operator with

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

Conversely, if ${ℌ}_3={ℌ}_1$, and both $TS$ and $ST$ are Fredholm operators, then $T$ is a Fredholm operator.
Proof: Since

$\dim \ker (ST)\le \dim \ker (S)+\dim \ker (T)$, and $\dim \mathrm{Coker}(ST)\le \dim \mathrm{Coker}(S)+\dim \mathrm{Coker}(T)$,

we see that $ST$ is Fredholm. Next, using the identity

$\dim X+\dim X^{\mathrm{\perp }}\cap Y=\dim Y+\dim Y^{\mathrm{\perp }}\cap X$

we compute:

$\dim \ker (ST)=\dim \ker (S)\cap \mathrm{ran}(T)+\dim \ker (T)=\dim \ker (S)\cap \ker (T^{\ast }{)}^{\mathrm{\perp }}+\dim \ker (T)$

$=-\dim \ker (T^{\ast })+\dim \ker (S)+\dim \ker (S{)}^{\mathrm{\perp }}\cap \ker (T^{\ast })+\dim \ker (T)$

$=\mathrm{ind}(T)+\mathrm{ind}(S)+\dim \ker (T^{\ast }{S}^{\ast }).$

For the conversely, let $f_j$ be a bounded sequence such that $T{f}_j$ is convergent. Then $ST{f}_j$ is convergent and so $f_j$ has a convergent sequence when $ST$ is Fredholm. Thus, $\dim \ker T<\mathrm{\infty }$ and $T$ has closed range. That $TS$ is a Fredholm operator shows that this is also true for $T^{\ast }$ and we conclude that $T$ is Fredholm. $◻$

7 Theorem The mapping

$T\mapsto \mathrm{ind}(T)$

is a locally constant function on the set of Fredhold operators $T:{ℌ}_1\to {ℌ}_2$.
Proof: By the Hahn-Banach theorem, we have decompositions:

${ℌ}_1=C_1\oplus \ker (T),{ℌ}_2=\mathrm{ran}(T)\oplus C_2$.

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

$T=\left[\begin{array}{cc}{T}^{\prime }& 0\\ 0& 0\end{array}\right]$

where $T^{\prime }:C_1\to \mathrm{ran}T$. By the above lemma, $\mathrm{ind}$ is invariant under row and column operations. Thus, for any $S=\left[\begin{array}{cc}{S}_1& S_2\\ S_3& S_4\end{array}\right]$, we have:

$\mathrm{ind}(T+S)=\mathrm{ind}(\left[\begin{array}{cc}{T}^{\prime }+S_1& 0\\ 0& A\end{array}\right])=\mathrm{ind}(A)$,

since $T^{\prime }+S_1$ is invertible when $\Vert S\Vert$ 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. $◻$

7 Corollary If $T\in B({ℌ}_1,{ℌ}_2)$ is a Fredholm operator and K is a compact operator, then $T+K$ is a Fredholm operator with

$\mathrm{ind}(T+K)=\mathrm{ind}(T)$

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 $K{f}_{j_k}$ is convergent. $T{f}_{j_k}$ is then convergent and so $f_{j_k}$ contains a convergent subsequence. Since $K^{\ast }$ is compact, the same argument applies to $T^{\ast }+K^{\ast }$. 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. $◻$

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

7 Corollary If $K\in B({ℌ}_1,{ℌ}_2)$ is a compact, then

$\ker (K-\lambda I)$ and ${ℌ}_2/\mathrm{ran}(K-\lambda I)$

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

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

and the second is the immediate consequence. $◻$

7 Theorem Let $T\in B({ℌ}_1,{ℌ}_2)$. Then $T$ is a Fredholm operator if and only if $I-TS$ and $I-ST$ are finite-rank operators for some $S\in B({ℌ}_2,{ℌ}_1)$. Moreover, when $I-TS$ and $I-ST$ are of trace class (e.g., of finite-rank),

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

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:

$T=\left[\begin{array}{cc}{T}^{\prime }& 0\\ 0& 0\end{array}\right]$

where $T^{\prime }$ is invertible. If we set, for example, $S=\left[\begin{array}{cc}T{\text{'}}^{-1}& 0\\ 0& 0\end{array}\right]$, then $S$ has required properties. Next, suppose S is given arbitrary: $S=\left[\begin{array}{cc}{S}_1& S_2\\ S_3& S_4\end{array}\right]$. Then

$\mathrm{Tr}(I-ST)=\mathrm{Tr}(1-S_1{T}^{\prime })+\dim \ker T$.

Similarly, we compute:

$\mathrm{Tr}(I-TS)=\mathrm{Tr}(1-T^{\prime }{S}_1)+\dim \mathrm{Coker}T$.

Now, since $T^{\prime }(I-S_1{T}^{\prime })=(I-T^{\prime }{S}_1)T^{\prime }$, and $T^{\prime }$ is invertible, we have:

$\mathrm{Tr}(I-S_1{T}^{\prime })=\mathrm{Tr}(I-T^{\prime }{S}_1)$. $◻$

Representations of compact groups

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