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. 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:
 * (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.
 * $$f_j = g_j + h_j, (g_j \in \operatorname{ker}(T), h_j \in \mathcal{W})$$

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

we see that $$ST$$ 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$$ be a bounded sequence such that $$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\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.