Compact Operators

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.

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.

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.

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 $T$).

• Assuming all bounded operators are compact. The identity operator on an infinite-dimensional Hilbert space is bounded but not compact: $B_H$ is closed but not compact (since the unit ball is not compact in infinite dimensions).
• Thinking finite-rank operators are the only compact operators. The limit of finite-rank operators is compact, and on Hilbert spaces every compact operator is the norm limit of finite-rank operators. But there exist compact operators that are not themselves finite-rank.
• Forgetting that $0$ is always in the spectrum of a compact operator on an infinite-dimensional space. Even if $0$ is not an eigenvalue, it belongs to the spectrum (as an accumulation point of eigenvalues or as essential spectrum).
• Confusing the Fredholm alternative with the Fredholm index. The alternative deals with solvability of $(\lambda I - T)x = y$ for compact $T$. The Fredholm index $\text{ind}(T) = \dim\ker T - \text{codim}\ \text{ran}\ T$ applies to Fredholm operators more generally.

Problem 1. Show that the Volterra operator $(Vf)(x) = \int_0^x f(t)\,dt$ on $C([0,1])$ is compact.

Solution. $V$ is bounded: $|(Vf)(x)| \leq \int_0^1 |f(t)|\,dt \leq \|f\|_\infty$. The image $V(B_{C([0,1])})$ is equicontinuous (by the fundamental theorem of calculus, derivatives are bounded by $\|f\|_\infty$), and uniformly bounded. By the Arzela-Ascoli theorem, it is relatively compact in $C([0,1])$. Therefore $V$ is compact. $\blacksquare$

Problem 2. Let $T$ be a compact self-adjoint operator on $H$ with eigenvalues $\lambda_n \to 0$. Show that $T$ has a maximum eigenvalue (in absolute value).

Solution. Since $\lambda_n \to 0$, the set $\{|\lambda_n|\}$ has a maximum (attained at some finite index) if there are finitely many nonzero eigenvalues; otherwise, the maximum of $|\lambda_n|$ is attained at the first eigenvalue (since they converge to $0$). In either case, $\max_n |\lambda_n| = \|T\|$ by the spectral radius formula for self-adjoint operators. $\blacksquare$

• Integral equations: Fredholm integral equations of the second kind $f - Kf = g$ are solved using the Fredholm alternative. Compact integral operators arise naturally in potential theory and scattering.
• Quantum mechanics: Position and momentum operators are unbounded, but their resolvents $(H - z)^{-1}$ are often compact. The spectral theorem for compact operators underlies the solution of the Schrodinger equation for bound states.
• Differential equations: The inverse of a differential operator with compact resolvent (e.g., $-\Delta + V$ on a bounded domain) is compact, ensuring discrete spectrum — the basis for Sturm-Liouville theory.
• Signal processing: The Karhunen-Loeve transform uses the spectral decomposition of compact covariance operators to find optimal bases for signal representation and compression (PCA).