Compact Operators

5.1 Definition

A linear operator $T : X \to Y$ is compact if the image of the closed unit ball, $T(B_X)$, is relatively compact (its closure is compact) in $Y$.

Proposition 5.1. Every compact operator is bounded. Every finite-rank operator is compact.

Proposition 5.2. If $T$ is compact and $S$ is bounded, then $TS$ and $ST$ are compact.

Proposition 5.3. If $T_n$ are compact and $T_n \to T$ in operator norm, then $T$ is compact.

5.2 Spectral Theory for Compact Operators

Theorem 5.4 (Spectral Theorem for Compact Self-Adjoint Operators). Let $T$ be a compact self-adjoint operator on a Hilbert space $H$. Then:

1. All eigenvalues of $T$ are real.
2. Eigenvectors corresponding to distinct eigenvalues are orthogonal.
3. There exists an orthonormal basis of $H$ consisting of eigenvectors of $T$.
4. If $\{\lambda_n\}$ are the nonzero eigenvalues with orthonormal eigenvectors $\{e_n\}$, then $Tx = \sum_n \lambda_n \langle x, e_n\rangle e_n$.

Corollary 5.5. A compact self-adjoint operator on an infinite-dimensional Hilbert space has at most countably many nonzero eigenvalues, and $0$ is the only possible accumulation point.

5.3 Spectral Theorem for Normal Operators

A bounded operator $T$ on a Hilbert space $H$ is normal if $T^*T = TT^*$, and unitary if $T^*T = TT^* = I$.

Proposition 5.6. If $T$ is normal, then $\|Tx\| = \|T^*x\|$ for all $x \in H$.

Proposition 5.7. If $T$ is normal, then $\ker T = \ker T^*$ and eigenvectors corresponding to distinct eigenvalues are orthogonal.

Theorem 5.8 (Spectral Theorem for Normal Compact Operators). Let $T$ be a compact normal operator on a Hilbert space $H$. Then there exists an orthonormal basis of $H$ consisting of eigenvectors of $T$, and the eigenvalues satisfy $|\lambda_n| \to 0$.

This generalises Theorem 5.4: self-adjoint operators are normal, and unitary operators are normal (with eigenvalues on the unit circle in $\mathbb{C}$).

Theorem 5.9 (Spectral Theorem for Bounded Normal Operators). Let $T$ be a bounded normal operator on $H$. There exists a unique projection-valued measure $E$ on the Borel subsets of $\sigma(T) \subseteq \mathbb{C}$ such that

$T = \int_{\sigma(T)} \lambda\, dE(\lambda)$

This integral representation implies: if $f$ is a bounded Borel function on $\sigma(T)$, then $f(T) = \int f(\lambda)\, dE(\lambda)$ defines a bounded normal operator satisfying the functional calculus relations.

5.4 Fredholm Alternative

Theorem 5.6 (Fredholm Alternative). Let $T$ be a compact operator on a Banach space $X$ and $\lambda \neq 0$. Then exactly one of the following holds:

1. $(\lambda I - T)$ is bijective (hence invertible by the bounded inverse theorem).
2. $(\lambda I - T)x = 0$ has a nontrivial solution (i.e., $\lambda$ is an eigenvalue of $T$).