In functional analysis, a branch of mathematics, a compact operator is a linear operator ${\displaystyle T:X\to Y}$, where ${\displaystyle X,Y}$ are normed vector spaces, with the property that ${\displaystyle T}$ maps bounded subsets of ${\displaystyle X}$ to relatively compact subsets of ${\displaystyle Y}$ (subsets with compact closure in ${\displaystyle Y}$). Such an operator is necessarily a bounded operator, and so continuous.^[1] Some authors require that ${\displaystyle X,Y}$ are Banach, but the definition can be extended to more general spaces.

Any bounded operator ${\displaystyle T}$ that has finite rank is a compact operator; indeed, the class of compact operators is a natural generalization of the class of finite-rank operators in an infinite-dimensional setting. When ${\displaystyle Y}$ is a Hilbert space, it is true that any compact operator is a limit of finite-rank operators,^[1] so that the class of compact operators can be defined alternatively as the closure of the set of finite-rank operators in the norm topology. Whether this was true in general for Banach spaces (the approximation property) was an unsolved question for many years; in 1973 Per Enflo gave a counter-example, building on work by Grothendieck and Banach.^[2]

The origin of the theory of compact operators is in the theory of integral equations, where integral operators supply concrete examples of such operators. A typical Fredholm integral equation gives rise to a compact operator K on function spaces; the compactness property is shown by equicontinuity. The method of approximation by finite-rank operators is basic in the numerical solution of such equations. The abstract idea of Fredholm operator is derived from this connection.

Equivalent formulations

A linear map ${\displaystyle T:X\to Y}$ between two topological vector spaces is said to be compact if there exists a neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle T(U)}$ is a relatively compact subset of ${\displaystyle Y}$.^[3]

Let ${\displaystyle X,Y}$ be normed spaces and ${\displaystyle T:X\to Y}$ a linear operator. Then the following statements are equivalent, and some of them are used as the principal definition by different authors^[4]

• ${\displaystyle T}$ is a compact operator;
• the image of the unit ball of ${\displaystyle X}$ under ${\displaystyle T}$ is relatively compact in ${\displaystyle Y}$;
• the image of any bounded subset of ${\displaystyle X}$ under ${\displaystyle T}$ is relatively compact in ${\displaystyle Y}$;
• there exists a neighbourhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ and a compact subset ${\displaystyle V\subseteq Y}$ such that ${\displaystyle T(U)\subseteq V}$;
• for any bounded sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle X}$, the sequence ${\displaystyle (Tx_{n})_{n\in \mathbb {N} }}$ contains a converging subsequence.

If in addition ${\displaystyle Y}$ is Banach, these statements are also equivalent to:

• the image of any bounded subset of ${\displaystyle X}$ under ${\displaystyle T}$ is totally bounded in ${\displaystyle Y}$.

If a linear operator is compact, then it is continuous.

Important properties

In the following, ${\displaystyle X,Y,Z,W}$ are Banach spaces, ${\displaystyle B(X,Y)}$ is the space of bounded operators ${\displaystyle X\to Y}$ under the operator norm, and ${\displaystyle K(X,Y)}$ denotes the space of compact operators ${\displaystyle X\to Y}$. ${\displaystyle \operatorname {Id} _{X}}$ denotes the identity operator on ${\displaystyle X}$, ${\displaystyle B(X)=B(X,X)}$, and ${\displaystyle K(X)=K(X,X)}$.

• ${\displaystyle K(X,Y)}$ is a closed subspace of ${\displaystyle B(X,Y)}$ (in the norm topology). Equivalently,^[5]
  • given a sequence of compact operators ${\displaystyle (T_{n})_{n\in \mathbf {N} }}$ mapping ${\displaystyle X\to Y}$ (where ${\displaystyle X,Y}$are Banach) and given that ${\displaystyle (T_{n})_{n\in \mathbf {N} }}$ converges to ${\displaystyle T}$ with respect to the operator norm, ${\displaystyle T}$ is then compact.
• Conversely, if ${\displaystyle X,Y}$ are Hilbert spaces, then every compact operator from ${\displaystyle X\to Y}$ is the limit of finite rank operators. Notably, this "approximation property" is false for general Banach spaces X and Y.^[4]
• ${\displaystyle B(Y,Z)\circ K(X,Y)\circ B(W,X)\subseteq K(W,Z).}$  In particular, ${\displaystyle K(X)}$ forms a two-sided ideal in ${\displaystyle B(X)}$.
• Any compact operator is strictly singular, but not vice versa.^[6]
• A bounded linear operator between Banach spaces is compact if and only if its adjoint is compact (Schauder's theorem).
  • If ${\displaystyle T:X\to Y}$ is bounded and compact, then:
    • the closure of the range of ${\displaystyle T}$ is separable.^[5][7]
    • if the range of ${\displaystyle T}$ is closed in Y, then the range of ${\displaystyle T}$ is finite-dimensional.^[5][7]
• If $$Zdroj:https://en.wikipedia.org?pojem=Compact_operator
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