Compact operators are crucial in functional analysis, bridging finite and infinite dimensions. They map bounded sets to relatively compact ones, ensuring convergent subsequences. This property makes them invaluable for solving integral equations and studying spectral theory.

Examples include integral operators and finite-rank operators. Compact operators form a closed subspace of bounded linear operators. They're closely related to completely continuous operators, coinciding in reflexive Banach spaces.

Compact Operators

Definition of compact operators

• A bounded linear operator $T: X \to Y$ between Banach spaces ($X$ and $Y$) is compact if it maps bounded sequences $(x_n)$ in $X$ to sequences $(Tx_n)$ in $Y$ that have convergent subsequences
• Compact operators transform bounded sets into relatively compact sets, meaning the closure of the image of the unit ball $B_X$ under $T$ is compact in $Y$
• For Hilbert spaces ($H$), a compact operator $T: H \to H$ is characterized by the property that every bounded sequence $(x_n)$ in $H$ has a subsequence $(Tx_n)$ that converges in $H$

Examples of compact operators

• Integral operators on function spaces ($L^2([a,b])$) are often compact, such as the operator $(Tf)(x) = \int_a^b K(x,y)f(y)dy$ with a continuous kernel function $K$ on $[a,b] \times [a,b]$
  • The Arzelà-Ascoli theorem can be used to prove the compactness of such integral operators
• Finite-rank operators, which have a finite-dimensional range, are always compact
  • These operators can be expressed as $Tx = \sum_{i=1}^n f_i(x)y_i$, where $f_i \in X^$ and $y_i \in Y$ for $i=1,\ldots,n$
  • The compactness of finite-rank operators follows from the compactness of the unit ball in finite-dimensional spaces
• The composition of a bounded linear operator with a compact operator results in a compact operator
  • If $T: X \to Y$ is compact and $S: Y \to Z$ is bounded, then their composition $ST: X \to Z$ is compact

Compact operators as closed subspace

• The set of compact operators $\mathcal{K}(X,Y)$ forms a closed subspace of the bounded linear operators $\mathcal{B}(X,Y)$
• To prove this, we need to show that:
  1. The sum $T+S$ and scalar multiple $\alpha T$ of compact operators $T,S \in \mathcal{K}(X,Y)$ are also compact for any scalar $\alpha \in \mathbb{C}$
  2. If a sequence of compact operators $(T_n)$ converges to a bounded linear operator $T \in \mathcal{B}(X,Y)$, then $T$ is also compact
• The first property follows from the linearity of limits and the definition of compact operators
• The second property can be proven using the characterization of compact operators in terms of relatively compact sets and the fact that the limit of a sequence of compact sets is compact

Compact vs completely continuous operators

• A bounded linear operator $T: X \to Y$ between Banach spaces is completely continuous if it maps weakly convergent sequences $(x_n)$ to strongly convergent sequences $(Tx_n)$
• Every compact operator is completely continuous
  • If $T$ is compact and $(x_n)$ converges weakly to $x$, then $(Tx_n)$ has a convergent subsequence, and the limit of this subsequence must be $Tx$ by the uniqueness of weak limits
• In reflexive Banach spaces, the concepts of compact operators and completely continuous operators coincide
  • A bounded linear operator $T: X \to Y$ is compact if and only if it is completely continuous when $X$ is reflexive