Hilbert–Schmidt integral operator

In mathematics, a Hilbert–Schmidt integral operator is a type of integral transform. Specifically, given a domain $&Omega;$ in $n$-dimensional Euclidean space $R^{n}$, then the square-integrable function $k : &Omega; &times; &Omega; &rarr; C$ belonging to $L^{2}(&Omega;×&Omega;)$ such that
 * $$\int_{\Omega} \int_{\Omega} | k(x, y) |^{2} \,dx \, dy < \infty ,$$

is called a Hilbert–Schmidt kernel and the associated integral operator $T : L^{2}(&Omega;) &rarr; L^{2}(&Omega;)$ given by
 * $$(Tf) (x) = \int_{\Omega} k(x, y) f(y) \, dy, \quad f \in L^2(\Omega),$$

is called a Hilbert–Schmidt integral operator. Then $T$ is a Hilbert–Schmidt operator with Hilbert–Schmidt norm


 * $$\Vert T \Vert_\mathrm{HS} = \Vert k \Vert_{L^2}.$$

Hilbert–Schmidt integral operators are both continuous and compact.

The concept of a Hilbert–Schmidt operator may be extended to any locally compact Hausdorff spaces. Specifically, let $L^{2}(X)$ be a separable Hilbert space and $X$ a locally compact Hausdorff space equipped with a positive Borel measure. The initial condition on the kernel $k$ on $&Omega; ⊆ R^{n}$ can be reinterpreted as demanding $k$ belong to $L^{2}(X × X)$. Then the operator
 * $$(Tf)(x) = \int_{X} k(x,y)f(y)\,dy,$$

is compact. If
 * $$k(x,y) = \overline{k(y,x)},$$

then $T$ is also self-adjoint and so the spectral theorem applies. This is one of the fundamental constructions of such operators, which often reduces problems about infinite-dimensional vector spaces to questions about well-understood finite-dimensional eigenspaces.