Compact operator

TVS case Let X,Y be topological vector spaces and T: X \to Y a linear operator. The following statements are equivalent, and different authors may pick any one of these as the principal definition for "T is a compact operator": • there exists a neighborhood U of the origin in X and T(U) is a relatively compact subset of Y; • there exists a neighborhood U of the origin in X and a compact subset V\subseteq Y such that T(U)\subseteq V; • there exists a nonempty open set U in X and T(U) is a relatively compact subset of Y. Normed case If in addition X,Y are normed spaces, these statements are also equivalent to: • the image of the unit ball of X under T is relatively compact in Y; • the image of any bounded subset of X under T is relatively compact in Y; • for any bounded sequence (x_n)_{n\in \N} in X, the sequence (Tx_n)_{n\in\N} contains a converging subsequence. Banach case If in addition Y is Banach, these statements are also equivalent to: • the image of any bounded subset of X under T is totally bounded in Y. == Properties ==

In the following, X, Y, Z, W are Banach spaces, B(X,Y) is the space of bounded operators X \to Y under the operator norm, and K(X,Y) denotes the space of compact operators X \to Y. \operatorname{Id}_X denotes the identity operator on X, B(X) = B(X,X), and K(X) = K(X,X). • If a linear operator is compact, then it is continuous. • K(X,Y) is a closed subspace of B(X,Y) (in the norm topology). Equivalently, • given a sequence of compact operators (T_n)_{n \in \mathbf{N}} mapping X \to Y (where X,Yare Banach) and given that (T_n)_{n \in \mathbf{N}} converges to T with respect to the operator norm, T is then compact. • In particular, the limit of a sequence of finite rank operators is a compact operator. • Conversely, if X,Y are Hilbert spaces, then every compact operator from X \to Y is the limit of finite rank operators. Notably, this "approximation property" is false for general Banach spaces X and Y. • A bounded linear operator between Banach spaces is compact if and only if its adjoint is compact (''Schauder's theorem''). • If T: X \to Y is bounded and compact, then: • the closure of the range of T is separable. • if the range of T is closed in Y, then the range of T is finite-dimensional. • If X is a Banach space and there exists an invertible bounded compact operator T: X \to X then X is necessarily finite-dimensional. Now suppose that X is a Banach space and T\colon X \to X is a compact linear operator, and T^* \colon X^* \to X^* is the adjoint or transpose of T. • For any T\in K(X), {\operatorname{Id}_X} - T is a Fredholm operator of index 0. In particular, \operatorname{Im}({\operatorname{Id}_X} - T) 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 M and N are subspaces of X where M is closed and N is finite-dimensional, then M+N is also closed. • If S\colon X \to X is any bounded linear operator then both S \circ T and T \circ S are compact operators. • If \lambda \neq 0 then the range of T - \lambda \operatorname{Id}_X is closed and the kernel of T - \lambda \operatorname{Id}_X is finite-dimensional. • If \lambda \neq 0 then the following are finite and equal: \dim \ker \left( T - \lambda \operatorname{Id}_X \right) = \dim\big(X / \operatorname{Im}\left( T - \lambda \operatorname{Id}_X \right) \big) = \dim \ker \left( T^* - \lambda \operatorname{Id}_{X^*} \right) = \dim\big(X^* / \operatorname{Im}\left( T^* - \lambda \operatorname{Id}_{X^*} \right) \big) • The spectrum \sigma(T) of T is compact, countable, and has at most one limit point, which would necessarily be the origin. • If X is infinite-dimensional then 0 \in \sigma(T). • If \lambda \neq 0 and \lambda \in \sigma(T) then \lambda is an eigenvalue of both T and T^{*}. • For every r > 0 the set E_r = \left\{ \lambda \in \sigma(T) : | \lambda | > r \right\} is finite, and for every non-zero \lambda \in \sigma(T) the range of T - \lambda \operatorname{Id}_X is a proper subset of X. ==Origins in integral equation theory==

A crucial property of compact operators is the Fredholm alternative in the solution of linear equations. Let K be a compact operator, f a given function, and u the unknown function to be solved for. Then the Fredholm alternative states that the equation(\lambda K + I)u = f behaves 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 K on an infinite-dimensional Banach space has spectrum that is either a finite subset of \mathbb{C} which includes 0, or the spectrum is a countably infinite subset of \mathbb{C} which has 0 as its only limit point. Moreover, in either case the non-zero elements of the spectrum are eigenvalues of K with finite multiplicities (so that K - \lambda I has a finite-dimensional kernel for all complex \lambda \neq 0). 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. 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. ==Compact operator on Hilbert spaces==

For Hilbert spaces, another equivalent definition of compact operators is given as follows. An operator T on an infinite-dimensional Hilbert space (\mathcal{H}, \langle \cdot, \cdot \rangle), :T\colon\mathcal{H} \to \mathcal{H}, is said to be compact if it can be written in the form :T = \sum_{n=1}^\infty \lambda_n \langle f_n, \cdot \rangle g_n, where \{f_1,f_2,\ldots\} and \{g_1,g_2,\ldots\} are orthonormal sets (not necessarily complete), and \lambda_1,\lambda_2,\ldots 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 \lambda_{N+k}=0 for some N \in \N and every k = 1,2,\dots, then the operator has finite rank, i.e., a finite-dimensional range, and can be written as :T = \sum_{n=1}^N \lambda_n \langle f_n, \cdot \rangle g_n. An important subclass of compact operators is the trace-class or nuclear operators, i.e., such that \operatorname{Tr}(|T|). While all trace-class operators are compact operators, the converse is not necessarily true. For example \lambda_n = \frac{1}{n} tends to zero for n \to \infty while \sum_{n=1}^{\infty} |\lambda_n| = \infty. == Completely continuous operators ==

Let X, Y be Banach spaces. A bounded linear operator T: X \to Y is called completely continuous if, for every weakly convergent sequence (x_n) from X, the sequence (Tx_n) is norm-convergent in Y. Compact operators between Banach spaces are always completely continuous, but the converse is false, because there exists a completely continuous operator that is not compact. However, the converse is true if X is a reflexive Banach space: then every completely continuous operator T: X \to Y is compact. Somewhat confusingly, compact operators are sometimes referred to as "completely continuous" in older literature, even though the latter is a weaker condition in modern terminology. == Examples ==

• Every finite rank operator is compact. • The scaling operator x \mapsto kx for any nonzero k is compact if and only if the space is finite-dimensional. This can be proven directly, or as a corollary of Riesz's lemma. • The multiplication operator on sequence space \ell^p with fixed p \in [1, \infty], defined as (Tx)_n = t_n x_n and sequence (t_n) converging to zero, is compact. • Every Hilbert–Schmidt operator is compact. • In particular, every Hilbert–Schmidt integral operator is compact. That is, if \Omega is any domain in \mathbf{R}^n and the integral kernel k: \Omega \times \Omega \to \mathbf{R} satisfies \iint |k|^2 , then the integral operator T on L^2(\Omega; \mathbf{R}) defined by (T f)(x) = \int_{\Omega} k(x, y) f(y) \, \mathrm{d} y is a compact operator. • The integral transform on C([0, 1]; \mathbf{R}) (i.e. the continuous function space on a closed bounded real interval), defined by(Tf)(x) = \int_0^x f(t)g(t) \, \mathrm{d} t, for any fixed g \in C([0, 1]; \mathbf{R}), is a compact operator by the Arzelà–Ascoli theorem. • The inclusion map I:W^{1, p} (\Omega) \hookrightarrow L^{q} (\Omega), \quad I(x)=x, compactly embedding the Sobolev space W^{1, p} (\Omega) in the Lebesgue space L^{q} (\Omega) for every 1 \leq q and 1 \leq p , is a compact operator by the Rellich–Kondrachov theorem. • The forward and backward unilateral shift operators are not compact. ==See also==