Compact Operators | Mathematics

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 $T$).

• Bounded Linear Operators: Provides the framework of bounded linear maps between Banach spaces within which compact operators are defined.

• The Fundamental Theorems: The bounded inverse theorem and closed graph theorem are used in the Fredholm alternative for compact operators.

• Weak and Weak* Convergence: Compact operators map weakly convergent sequences to strongly convergent sequences, linking these two modes of convergence.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Statistical Learning

• Statistical Mechanics

flowchart TD
    A[5_Compact Operators] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Compact operators are the infinite-dimensional analogue of finite matrices. In finite dimensions, the unit ball is compact and every bounded operator maps it to a relatively compact set. In infinite dimensions this fails, and compact operators are precisely those that restore this finite-dimensional behaviour: they squeeze the unit ball into a nearly finite-dimensional image. The spectral theorem for compact self-adjoint operators decomposes them as weighted sums of rank-one projections, with eigenvalues decaying to zero. The Fredholm alternative then says that for compact operators, solving a linear equation reduces to a finite-dimensional problem: either the homogeneous equation has only the trivial solution, or it has a finite-dimensional space of solutions.

• 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 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).

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

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