Compact operators behave like infinite-dimensional shadows of matrices. They send bounded sets into sets with convergent subsequences, recovering part of finite-dimensional spectral behavior.
Compact linear maps
Compactness is not boundedness; it is a stronger regularizing property. A compact operator turns bounded input sets into sets with convergent subsequences.
Definition. A bounded linear map \(T:X\to Y\) is compact if \(T(B_X)\) has compact closure in \(Y\), equivalently if every bounded sequence \((x_n)\) in \(X\) has a subsequence \((Tx_{n_k})\) converging in norm in \(Y\).
Proposition. Every compact linear map is bounded.
Proof. If \(T\) were unbounded, choose \(x_n\) with \(\|x_n\|\le1\) and \(\|Tx_n\|\to\infty\). Then \((Tx_n)\) cannot have a convergent subsequence, contradicting compactness.
Finite-rank operators
Finite-rank maps are compact because bounded sets in finite-dimensional spaces have compact closure. They are the matrix-like examples inside infinite-dimensional analysis.
Proposition. Every finite-rank bounded operator is compact.
Proof. If \(\operatorname{Ran}T\) is finite-dimensional, then \(T(B_X)\) is bounded in a finite-dimensional normed space. Closed bounded sets in finite dimension are compact, so \(\overline{T(B_X)}\) is compact.
Norm limits
Compact operators form a closed class under operator-norm limits. This makes approximation by finite-rank operators a viable strategy whenever it is available.
Theorem. The norm limit of compact operators is compact.
Proof. Let \(T_n\to T\) in operator norm and let \((x_k)\) be bounded. Choose a subsequence on which \(T_1x_k\) converges. From that subsequence choose a further subsequence on which \(T_2x_k\) converges, and continue. The diagonal subsequence makes \(T_jx_{k_m}\) converge for each fixed \(j\).
Let \(\|x_k\|\le M\). Given \(\varepsilon>0\), choose \(j\) with \(\|T-T_j\|M<\varepsilon/3\). Since \(T_jx_{k_m}\) is Cauchy, \(Tx_{k_m}\) is Cauchy up to the two approximation errors. Completeness of \(Y\) gives convergence.
Composition ideal
Compactness is stable under composition with bounded maps. This ideal property explains why compact operators behave like a distinguished small class of operators.
Proposition. If \(T:X\to Y\) is compact and \(A:W\to X\), \(B:Y\to Z\) are bounded, then \(TA\) and \(BT\) are compact.
Proof. The image of a bounded set under \(A\) is bounded, so compactness of \(T\) makes \(TA\) compact. If \(K=\overline{T(B_X)}\) is compact in \(Y\), then \(B(K)\) is compact in \(Z\) by continuity, and \(BT(B_X)\subset B(K)\).
Compactness and weak convergence
Compact operators convert weak convergence into norm convergence under suitable hypotheses. This is one reason compactness is so useful in PDE and variational arguments.
Proposition. If \(X\) is reflexive, \(T:X\to Y\) is compact, and \(x_n\rightharpoonup x\) weakly in \(X\), then \(Tx_n\to Tx\) in norm provided every subsequence has the same possible norm limit. In particular compact operators turn weakly convergent sequences into norm-convergent sequences.
Proof. The sequence \((x_n)\) is bounded. Compactness gives every subsequence a further subsequence such that \(Tx_{n_k}\to y\) in norm. Since bounded linear maps are weak-to-weak continuous, \(Tx_{n_k}\rightharpoonup Tx\) weakly. Norm convergence implies weak convergence to the same limit, so \(y=Tx\). If the full sequence did not converge in norm to \(Tx\), some subsequence would stay \(\varepsilon\)-away, contradicting the existence of a further norm-convergent subsequence to \(Tx\).
Integral operators
Integral operators show how compactness can arise from smoothing or equicontinuity rather than finite rank. They are the analytic prototype behind many compact embeddings.
Example. If \(K\in C([0,1]^2)\), the operator \(Tf(s)=\int_0^1K(s,t)f(t)\,\mathrm{d}t\) maps \(C[0,1]\) compactly into itself. Bounded sets are sent to uniformly bounded and equicontinuous families; Arzel?-Ascoli gives relative compactness.