Functional Analysis/Special topics

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

[edit] 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 $T f j$ 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, $\mathcal{W}$ . 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 $T f j = T h j$ , 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 $T f$ . $\square$

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 $S T$ is a Fredholm operator with

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

Conversely, if $\mathfrak{H_3} = \mathfrak{H_1}$ , and both $T S$ and $S T$ are Fredholm operators, then $T$ is a Fredholm operator.
Proof: Since

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

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

$\dim X + \dim X^\bot \cap Y = \dim Y + \dim Y^\bot \cap X$

we compute:

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

For the conversely, let $f j$ be a bounded sequence such that $T f j$ is convergent. Then $S T f j$ is convergent and so $f j$ has a convergent sequence when $S T$ is Fredholm. Thus, $\dim\operatorname{ker}T < \infty$ and $T$ has closed range. That $T S$ 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 $T + λ K$ is Fredholm for any complex number $λ$ , and the index of $T + λ K$ 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 $λ$ , 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 $I - T S$ and $I - S T$ are finite-rank operators for some $S \in B(\mathfrak{H_2}, \mathfrak{H_1})$ . Moreover, when $I - T S$ and $I - S T$ 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

S T = I 1 + (I 1 - S T)

T S = I 2 + (I 2 - T S)

$S T$ and $T S$ 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:

$\operatorname{Tr}(I - S_1 T') = \operatorname{Tr}(I - T' S_1)$ . $\square$

[edit] Representations of compact groups

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

Functional Analysis/Special topics

From Wikibooks, the open-content textbooks collection

[edit] Fredholm theory

[edit] Representations of compact groups

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