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Compact operator

Type of continuous linear operator From Wikipedia, the free encyclopedia

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

Any bounded operator 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 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 Alexander Grothendieck and Stefan 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.

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Definitions

TVS case

Let be topological vector spaces and a linear operator.

The following statements are equivalent, and different authors may pick any one of these as the principal definition for " is a compact operator":[3]

  • there exists a neighborhood of the origin in and is a relatively compact subset of ;
  • there exists a neighborhood of the origin in and a compact subset such that ;
  • there exists a nonempty open set in and is a relatively compact subset of .

Normed case

If in addition are normed spaces, these statements are also equivalent to:[4]

  • the image of the unit ball of under is relatively compact in ;
  • the image of any bounded subset of under is relatively compact in ;
  • for any bounded sequence in , the sequence contains a converging subsequence.

Banach case

If in addition is Banach, these statements are also equivalent to:

  • the image of any bounded subset of under is totally bounded in .
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Properties

In the following, are Banach spaces, is the space of bounded operators under the operator norm, and denotes the space of compact operators . denotes the identity operator on , , and .

  • If a linear operator is compact, then it is continuous.
  • is a closed subspace of (in the norm topology). Equivalently,[5]
    • given a sequence of compact operators mapping (where are Banach) and given that converges to with respect to the operator norm, is then compact.
  • In particular, the limit of a sequence of finite rank operators is a compact operator.
  • Conversely, if are Hilbert spaces, then every compact operator from is the limit of finite rank operators. Notably, this "approximation property" is false for general Banach spaces and .[2][4]
  • where the composition of sets is taken element-wise. In particular, forms a two-sided ideal in .
  • 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).[7]
    • If is bounded and compact, then:[5][7]
      • the closure of the range of is separable.
      • if the range of is closed in , then the range of is finite-dimensional.
  • If is a Banach space and there exists an invertible bounded compact operator then is necessarily finite-dimensional.[7]

Now suppose that is a Banach space and is a compact linear operator, and is the adjoint or transpose of T.

  • For any , is a Fredholm operator of index 0. In particular, is closed. This is essential in developing the spectral properties of compact operators. One can notice the similarity between this property and the fact that, if and are subspaces of where is closed and is finite-dimensional, then is also closed.
  • If is any bounded linear operator then both and are compact operators.[5]
  • If then the range of is closed and the kernel of is finite-dimensional.[5]
  • If then the following are finite and equal: [5]
  • The spectrum of is compact, countable, and has at most one limit point, which would necessarily be the origin.[5]
  • If is infinite-dimensional then .[5]
  • If and then is an eigenvalue of both and .[5]
  • For every the set is finite, and for every non-zero the range of is a proper subset of .[5]
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Origins in integral equation theory

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Perspective

A crucial property of compact operators is the Fredholm alternative in the solution of linear equations. Let be a compact operator, a given function, and the unknown function to be solved for. Then the Fredholm alternative states that the equationbehaves much like as in finite dimensions.

The spectral theory of compact operators then follows, and it is due to Frigyes Riesz (1918). It shows that a compact operator on an infinite-dimensional Banach space has spectrum that is either a finite subset of which includes 0, or the spectrum is a countably infinite subset of which has as its only limit point. Moreover, in either case the non-zero elements of the spectrum are eigenvalues of with finite multiplicities (so that has a finite-dimensional kernel for all complex ).

An important example of a compact operator is compact embedding of Sobolev spaces, which, along with the Gårding inequality and the Lax–Milgram theorem, can be used to convert an elliptic boundary value problem into a Fredholm integral equation.[8] Existence of the solution and spectral properties then follow from the theory of compact operators; in particular, an elliptic boundary value problem on a bounded domain has infinitely many isolated eigenvalues. One consequence is that a solid body can vibrate only at isolated frequencies, given by the eigenvalues, and arbitrarily high vibration frequencies always exist.

The compact operators from a Banach space to itself form a two-sided ideal in the algebra of all bounded operators on the space. Indeed, the compact operators on an infinite-dimensional separable Hilbert space form a maximal ideal, so the quotient algebra, known as the Calkin algebra, is simple. More generally, the compact operators form an operator ideal.

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Compact operator on Hilbert spaces

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For Hilbert spaces, another equivalent definition of compact operators is given as follows.

An operator on an infinite-dimensional Hilbert space ,

,

is said to be compact if it can be written in the form

,

where and are orthonormal sets (not necessarily complete), and is a sequence of positive numbers with limit zero, called the singular values of the operator, and the series on the right hand side converges in the operator norm. The singular values can accumulate only at zero. If the sequence becomes stationary at zero, that is for some and every , then the operator has finite rank, i.e., a finite-dimensional range, and can be written as

.

An important subclass of compact operators is the trace-class or nuclear operators, i.e., such that . While all trace-class operators are compact operators, the converse is not necessarily true. For example tends to zero for while .

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Completely continuous operators

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Perspective

Let be Banach spaces. A bounded linear operator is called completely continuous if, for every weakly convergent sequence from , the sequence is norm-convergent in (Conway 1985, §VI.3). Compact operators on a Banach space are always completely continuous. If is a reflexive Banach space, then every completely continuous operator is compact.

Somewhat confusingly, compact operators are sometimes referred to as "completely continuous" in older literature, even though they are not necessarily completely continuous by the definition of that phrase in modern terminology.

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Examples

  • Every finite rank operator is compact.
  • The scaling operator for any nonzero is compact if and only if the space is finite-dimensional. This can be proven directly, or as a corollary of Riesz's lemma.[9]
  • The multiplication operator on sequence space with fixed , defined as and sequence converging to zero, is compact.
  • Every Hilbert–Schmidt operator is compact.
    • In particular, every Hilbert–Schmidt integral operator is compact. That is, if is any domain in and the integral kernel satisfies , then the integral operator on defined by is a compact operator.
  • The integral transform on (i.e. the continuous function space on a closed bounded real interval), defined by for any fixed , is a compact operator by the Arzelà–Ascoli theorem.
  • The inclusion map compactly embedding the Sobolev space in the Lebesgue space for every and , is a compact operator by the Rellich–Kondrachov theorem.
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See also

Notes

References

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