View source for 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 <math>T: \mathcal{X} \to \mathcal{Y}</math> be a closed densely defined operator. Then the following are equivalent.
*''(i) <math>\dim \operatorname{ker}(T) < \infty</math> and the range of ''T'' is closed.''
*''(ii) Every bounded sequence <math>f_j \in \mathcal{X}</math> has a convergent subsequence when ''<math>T f_j</math> is convergent.
Proof: We may assume ''T'' has dense range. (i) <math>\Rightarrow</math> (ii): Suppose <math>f_j</math> is a bounded sequence such that <math>T f_j</math> is convergent. In view of the Hahn-Banach theorem, ''X'' is a direct sum of the kernel of <math>T</math> and some other subspace, say, <math>\mathcal{W}</math>. Thus, we can write:
:<math>f_j = g_j + h_j, (g_j \in \operatorname{ker}(T), h_j \in \mathcal{W})</math>
By the closed graph theorem, the inverse of <math>T: \mathcal{W} \to \mathcal{Y}</math> is continuous. Since <math>T f_j = T h_j</math>, the continuity implies that <math>h_j</math> is convergent. Since <math>g_j</math> contains a convergent subsequence by the paragraph preceding the theorem, <math>f_j</math> has a convergent subsequence then. (ii) <math>\Rightarrow</math> (i): (ii) implies the first condition of (i), again by the preceding paragraph. For the second, suppose <math>T f_j</math> is convergent. Then by (ii) <math>f_j</math> has a subsequence <math>f_{j_k}</math> converging to, say, <math>f</math>. Since the graph of ''T'' is closed, <math>T f_{j_k}</math> converges to <math>T f</math>. <math>\square</math>

A bounded linear operator <math>T: \mathfrak{H}_1 \to \mathfrak{H}_2</math> between Hilbert spaces is said to be ''Fredholm'' if ''T'' and <math>T^*</math> both satisfy the condition of (i) in the lemma. The definition is equivalent to requiring that the kernel of ''T'' and the quotient <math>\mathfrak{H}_2 / T(\mathfrak{H}_1)</math> are finite-dimensional. In fact, if <math>\mathfrak{H}_2 / T(\mathfrak{H}_1)</math> is finite-dimensional, then <math> T(\mathfrak{H}_1) </math> is a complemented subspace; thus, closed. That <math>T</math> has closed range implies that <math>T^*</math> has closed range. For a Fredholm operator at least, it thus makes sense to define:
:<math>\operatorname{ind}(T) = \dim \operatorname{ker}(T) - \dim \operatorname{Coker}(T)</math>.
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 <math>T \in B(\mathfrak{H_1}, \mathfrak{H_2})</math> and <math>S \in B(\mathfrak{H_2}, \mathfrak{H_3})</math>. If <math>T</math> and <math>S</math> are Fredholm operators, then <math>ST</math> is a Fredholm operator with
:<math>\operatorname{ind}(ST) = \operatorname{ind}(T) + \operatorname{ind}(S)</math>.
Conversely, if <math>\mathfrak{H_3} = \mathfrak{H_1}</math>, and both <math>TS</math> and <math>ST</math> are Fredholm operators, then <math>T</math> is a Fredholm operator.''<br />
Proof:
Since
:<math>\dim\operatorname{ker}(ST) \le \dim\operatorname{ker}(S) + \dim\operatorname{ker}(T)</math>, and <math>\dim\operatorname{Coker}(ST) \le \dim\operatorname{Coker}(S) + \dim\operatorname{Coker}(T)</math>,
we see that <math>ST</math> is Fredholm. Next, using the identity
:<math>\dim X + \dim X^\bot \cap Y = \dim Y + \dim Y^\bot \cap X</math>
we compute:
:<math>
\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) </math>
:<math>= - \dim \operatorname{ker}(T^*) + \dim \operatorname{ker}(S) + \dim \operatorname{ker}(S)^\bot \cap \operatorname{ker}(T^*) + \dim \operatorname{ker}(T) </math>
:<math>= \operatorname{ind}(T) + \operatorname{ind}(S) + \dim \operatorname{ker}(T^* S^*).
</math>
For the conversely, let <math>f_j</math> be a bounded sequence such that <math>T f_j</math> is convergent. Then <math>ST f_j</math> is convergent and so <math>f_j</math> has a convergent sequence when <math>ST</math> is Fredholm. Thus, <math>\dim\operatorname{ker}T < \infty</math> and <math>T</math> has closed range. That <math>TS</math> is a Fredholm operator shows that this is also true for <math>T^*</math> and we conclude that <math>T</math> is Fredholm.
<math>\square</math>

'''7 Theorem''' ''The mapping''
:<math>T \mapsto \operatorname{ind}(T)</math>
''is a locally constant function on the set of Fredhold operators <math>T: \mathfrak{H}_1 \to \mathfrak{H}_2</math>.''<br />
Proof: By the Hahn-Banach theorem, we have decompositions:
:<math>\mathfrak{H}_1 = C_1 \oplus \operatorname{ker}(T), \quad \mathfrak{H}_2 = \operatorname{ran}(T) \oplus C_2</math>.
With respect to these, we represent ''T'' by a block matrix:
:<math>T = \begin{bmatrix}
T' & 0 \\
0 & 0 \\
\end{bmatrix}</math>
where <math>T': C_1 \to \operatorname{ran}T</math>. By the above lemma, <math>\operatorname{ind}</math> is invariant under row and column operations. Thus, for any <math>S = \begin{bmatrix}
S_1 & S_2 \\
S_3 & S_4 \\
\end{bmatrix}</math>, we have:
:<math>\operatorname{ind}(T + S) = \operatorname{ind}(\begin{bmatrix}
T' + S_1 & 0 \\
0 & A \\
\end{bmatrix}) = \operatorname{ind}(A)</math>,
since <math> T' + S_1</math> is invertible when <math>\|S\|</math> is small. ''A'' depends on ''S'' but the point is that <math>A</math> is a linear operator between finite-dimensional spaces. Hence, the index of ''A'' is independent of ''A''; thus, of ''S''. <math>\square</math>

'''7 Corollary''' ''If <math>T \in B(\mathfrak{H_1}, \mathfrak{H_2})</math> is a Fredholm operator and ''K'' is a compact operator, then <math>T + K</math> is a Fredholm operator with
:<math>\operatorname{ind}(T + K) = \operatorname{ind}(T)</math>
Proof:
Let <math>f_j</math> be a bounded sequence such that <math>(T+K)f_j</math> is convergent. By compactness, <math>f_j</math> has a convergent subsequence <math>f_{j_k}</math> such that <math>Kf_{j_k}</math> is convergent. <math>Tf_{j_k}</math> is then convergent and so <math>f_{j_k}</math> contains a convergent subsequence. Since <math>K^*</math> is compact, the same argument applies to <math>T^*+K^*</math>. The invariance of the index follows from the preceding theorem since <math>T+\lambda K</math> is Fredholm for any complex number <math>\lambda</math>, and the index of <math>T+\lambda K</math> is constant. <math>\square</math>

The next result, known as [[w:Fredholm alternative|Fredholm alternative]], is now easy but is very important in application.

'''7 Corollary''' ''If <math>K \in B(\mathfrak{H}_1, \mathfrak{H}_2)</math> is a compact, then
:<math>\operatorname{ker}(K - \lambda I)</math> and <math>\mathfrak{H}_2 / \operatorname{ran}(K - \lambda I)</math>
have the same (finite) dimension for any nonzero complex number <math>\lambda</math>, and <math>\sigma(K) \backslash \{0\}</math> consists of eigenvalues of ''K''.<br />
Proof: The first assertion follows from:
:<math>\operatorname{ind} (K - \lambda I) = \operatorname{ind} (I - \lambda^{-1} K) = 0</math>,
and the second is the immediate consequence. <math>\square</math>

'''7 Theorem''' ''Let <math>T \in B(\mathfrak{H_1}, \mathfrak{H_2})</math>. Then <math>T</math> is a Fredholm operator if and only if <math>I - TS</math> and <math>I - ST</math> are finite-rank operators for some <math>S \in B(\mathfrak{H_2}, \mathfrak{H_1})</math>. Moreover, when <math>I - TS</math> and <math>I - ST</math> are of trace class (e.g., of finite-rank),
:<math>\operatorname{ind}(T) = \operatorname{Tr}(I - ST) - \operatorname{Tr}(I - TS)</math>
Proof: Since the identities are Fredholm operators (in fact, any invertible operator) and since
:<math>ST = I_1 + (I_1 - ST)</math>
:<math>TS = I_2 + (I_2 - TS)</math>
<math>ST</math> and <math>TS</math> are Fredholm operators, which implies ''T'' is a Fredholm operator. Conversely, suppose ''T'' is a Fredholm operator. Then, as before, we can write:
:<math>T = \begin{bmatrix}
T' & 0 \\
0 & 0 \\
\end{bmatrix}</math>
where <math>T'</math> is invertible. If we set, for example, <math>S = \begin{bmatrix}
T'^{-1} & 0 \\
0 & 0 \\
\end{bmatrix}</math>, then <math>S</math> has required properties. Next, suppose ''S'' is given arbitrary: <math>S = \begin{bmatrix}
S_1 & S_2 \\
S_3 & S_4 \\
\end{bmatrix}</math>. Then
:<math>\operatorname{Tr}(I - ST) = \operatorname{Tr}(1 - S_1 T') + \operatorname{dim}\operatorname{ker}T</math>.
Similarly, we compute:
:<math>\operatorname{Tr}(I - TS) = \operatorname{Tr}(1 - T' S_1) + \operatorname{dim}\operatorname{Coker}T</math>.
Now, since <math>T'(I - S_1 T') = (I - T' S_1) T'</math>, and <math>T'</math> is invertible, we have:
:<math>\operatorname{Tr}(I - S_1 T') = \operatorname{Tr}(I - T' S_1)</math>. <math>\square</math>

== Representations of compact groups ==

'''Theorem''' ''Every irreducible unitary representation of a compact group is finite-dimensional.''<br />

{{BookCat}}