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.

Fredholm theory[edit]

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 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 \\

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 \\

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[edit]

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