Compact operator

A bounded operator is compact when it maps the unit ball to a relatively compact set.

For a bounded operator $K$ between Hilbert spaces, compactness is equivalent to $f_n\rightharpoonup f$ implying $K{f}_n\to Kf$ in norm. The Uniform boundedness principle bounds a weakly convergent sequence. Relative compactness of its images and the unique possible weak limit then give norm convergence. Conversely, every bounded domain sequence has a weakly convergent subsequence, so the image of the unit ball is relatively compact.

An operator norm limit of compact operators between Banach spaces is compact. Approximate the image of the unit ball by the image under one compact operator and transfer a finite epsilon-net across the small operator-norm error.

A bounded operator $T:X\to Y$ between Banach spaces is compact if and only if its dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$ is compact.

If $T{x}_j={\lambda }_j{x}_j$ for an infinite orthonormal sequence $(x_j)$ and a compact operator $T$ on a Hilbert space, then ${\lambda }_j\to 0$. Otherwise a subsequence of the vectors $T{x}_j$ stays pairwise separated, contradicting compactness.

Every compact operator on a Hilbert space is an operator-norm limit of finite-rank operators. Cover the compact closure of the image of the unit ball by finitely many small balls, span their centres by a finite-dimensional space $E$, and approximate $T$ by $P_{E}T$.

If $T$ is compact and $\lambda \ne 0$, then $T-\lambda I$ is injective exactly when it is surjective. Equivalently, every nonzero point of the spectrum of $T$ is an eigenvalue. Compactness turns an approximate eigenvector sequence into a convergent eigenvector sequence.

A bounded linear operator on a Hilbert space is Hilbert-Schmidt when ${\sum }_n\Vert T{e}_n{\Vert }^2<\infty$ for one, equivalently every, orthonormal basis. Every Hilbert-Schmidt operator is a compact operator.

For $(Tg)(v)=\int k(v,w)g(w)dw$, the Cauchy-Schwarz inequality gives $\Vert Tg{\Vert }_2\le \Vert k{\Vert }_{L^2(v,w)}\Vert g{\Vert }_2$. The kernel norm is the Hilbert-Schmidt norm, which bounds the operator norm. Uniform kernel bounds apply also to spatially parametrized operators after integrating over position.

The Hilbert-Schmidt norm of a Hilbert-Schmidt operator is the square root of the displayed sum. The value is independent of the orthonormal basis, by the Parseval identity. For a finite matrix, it is the Frobenius norm.

The Frobenius norm is the finite-matrix Hilbert-Schmidt norm. It equals $\sqrt{\mathrm{tr}(A^{\ast }A)}$, is unchanged by unitary multiplication on either side, and is the square root of the sum of squared singular values. For unit vectors $u,v$, $\Vert u{u}^{\ast }-v{v}^{\ast }{\Vert }_F^2=2(1-|u^{\ast }v{|}^2)$, a useful sign-invariant measure of distance between rank-one orthogonal projection matrices.

For Hilbert-Schmidt operators $A$ and $B$, the Hilbert-Schmidt inner product is $⟨A,B{⟩}_{\mathrm{HS}}=\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm.

If $\rho$ is a density operator with decreasing eigenvalues ${\lambda }_j$ and $T$ has matrix rank at most $k$, let $\Pi$ project onto the image of $T$. The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product givesThe Hermitian effect variational principle bounds $\mathrm{Tr}(\rho \Pi )$ by ${\sum }_{j<k}{\lambda }_j$, proving the claimed estimate without assuming $\rho$ is invertible.

The Hilbert-Schmidt distance is $d_{\mathrm{HS}}(A,B)=\Vert A-B{\Vert }_{\mathrm{HS}}$.

A bounded operator $T$ on a separable Hilbert space is trace class when $|T{|}^{1/2}$ is Hilbert--Schmidt. Its trace norm is

An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\Vert T{\Vert }_1=\Vert A{\Vert }_{\mathrm{HS}}\Vert B{\Vert }_{\mathrm{HS}}$ by using the polar decomposition of a bounded operator.

The Schatten norm Hölder inequality gives $\Vert AB{\Vert }_r\le \Vert A{\Vert }_p\Vert B{\Vert }_q$ when $1/r=1/p+1/q$. In particular, the product of two Hilbert-Schmidt operators is trace class and

For a trace-class operator, the series $\mathrm{tr}T={\sum }_n⟨T{e}_n,e_n⟩$ converges absolutely and is independent of the orthonormal basis. It satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

Every bounded operator $S$ defines a functional $T\mapsto \mathrm{tr}(ST)$ on the trace-class operators, andFinite-rank density in ${\mathcal{S}}_1$ makes every continuous functional arise uniquely this way.

For vectors $x,y$ in a Hilbert space, the rank-one operator $(x\otimes y)z=⟨z,y⟩x$ has $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $\mathrm{tr}(x\otimes y)=⟨x,y⟩$.

A finite-rank operator has finite-dimensional image and is compact.

For a finite-rank operator $S$ and $\lambda \ne 0$, if $\lambda$ is not an eigenvalue then $S-\lambda I$ is surjective. Decompose the Hilbert space into $\mathrm{im}S$ and its orthogonal complement, then use injectivity and surjectivity equivalence on the finite-dimensional image.

A compact Hermitian operator has real nonzero eigenvalues of finite multiplicity, with zero as their only possible accumulation point, and an orthonormal eigenbasis after a basis of its kernel is included.

For a compact positive self-adjoint operator $T$ with decreasing eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$, the Courant–Fischer principle givesIn particular, $S\ge T$ in quadratic-form order implies ${\lambda }_k(S)\ge {\lambda }_k(T)$ for every $k$.

Ifthen truncating the sum after $N$ terms gives finite-rank Hermitian $T_N$ with

Let $P_N$ project ${\ell }^2$ onto its first $N$ coordinates. Although $P_N\to I$ only strongly on the whole unit ball, convergence is uniform on every compact subset. Hence compact $T$ satisfiesand each $P_{N}T$ has finite rank.

Every nonzero spectral point of a compact operator is an isolated eigenvalue of finite algebraic multiplicity, and zero is the only possible accumulation point of the spectrum.

If finite-rank orthogonal projections $P_n$ converge strongly to the identity and $A$ is compact, then $\Vert A-P_{n}A{P}_n\Vert \to 0$. The finite-section spectra converge to $\mathrm{Sp}(A)$ in Hausdorff distance, including for nonnormal $A$.

A closed densely defined operator has compact resolvent when $(A-zI{)}^{-1}$ is compact at one, equivalently every, resolvent point. Its spectrum consists only of isolated eigenvalues of finite multiplicity, with possible accumulation only at infinity.

On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

The imaginary Airy operator $-d^2/d{x}^2+ix$ on $L^2(\mathbb{R})$ has compact resolvent but empty spectrum. Real translations shift it by a purely imaginary scalar, forcing vertical translation invariance of its spectrum.