Compact operator

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

The operator in the title is bounded and compact on the absolutely summable sequence space. The norm of a matrix operator on ${\ell }^1$ is the supremum of its absolute column sums: the triangle inequality gives one bound and coordinate unit vectors give the other. Here the column sum is $c_m={\sum }_{n=1}^{m-1}n!/m!\le 2/m$, using ${\sum }_{n=1}^{k}n!\le 2k!$. The latter inequality follows by induction. Retaining the first $N$ columns gives a finite-rank operator and a remainder of norm at most $2/(N+1)$. A norm limit of finite-rank operators is a compact operator, proving the assertion. This column-decay method can prove compactness even when each row has infinitely many nonzero entries.

For a bounded linear operator into a Banach space, compactness is equivalent to the displayed condition, where ${\pi }_Y$ is the quotient map by a finite-dimensional subspace of the target. A finite net for the image of the unit ball gives such a subspace by taking its span. Conversely, if the image lies within $\epsilon$ of a finite-dimensional subspace, choose approximants there; they remain bounded because the original image is bounded. A finite net for these approximants gives a slightly larger finite net for the image. This proves total boundedness and hence compactness. This condition does not require choosing the approximants by a linear map, and therefore does not assume an approximation property for the Banach space.

Suppose a linear map $S:c_0\to Y$ has compact image of the closed unit ball. It is bounded, and each $g\in Y^{\ast }$ attains its maximum on this compact image. Thus $S^{\ast }g$ attains its norm, so norm attainment on the vanishing sequence space implies $S^{\ast }(Y^{\ast })\subseteq c_{00}$. The Baire category theorem applied to the inverse images of the first $n$ coordinate spans makes this adjoint finite rank: these are closed linear subspaces covering the complete continuous dual space $Y^{\ast }$, so one has interior and is all of $Y^{\ast }$. Finally the Hahn-Banach theorem identifies $\ker S$ as the common kernel of $S^{\ast }(Y^{\ast })$, proving finite rank of $S$. Ordinary compact operators require only relative compactness of the ball image, so they need not have finite rank.

For $1<p<\infty$, let $q=p/(p-1)$ and partition the coordinates into consecutive finite blocks $B_k$ of lengths $m_k\to \infty$. The displayed map defines a norm-one operator on the l-p sequence space. By Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$, giving the upper norm bound. For the unit vector $e_j$ with $j\in B_k$, $T{e}_j=m_k^{-1/q}{e}_k$, so its norm tends to zero as $j\to \infty$.

Nevertheless, $x^{(k)}=m_k^{-1/p}{1}_{B_k}$ has norm one and maps to $e_k$. Those images are pairwise distance $2^{1/p}$ apart. Thus $T$ is not a compact operator, and has norm exactly one. Vanishing images of a chosen basis do not alone test compactness on ${\ell }^p$ for $p>1$. The map is the adjoint of the isometric block embedding $(Uy{)}_j=m_k^{-1/q}{y}_k$, $j\in B_k$, on ${\ell }^q$.

A nonlinear compact operator between Banach spaces is a continuous map that sends every bounded set to a relatively compact subset. Thus every bounded sequence of inputs has a subsequence whose images converge in the target norm. No linearity is assumed. On a bounded domain of definition it suffices that the entire image be relatively compact. For linear operators, this recovers the usual compact operator notion.

On a bounded open set, let $Tv\in W_0^{1,2}$ be the weak solution of $\Delta Tv=v$. The energy estimate bounds $T:L^2\to W_0^{1,2}$. For $p>n$, put $q=2p/(p-2)<2n/(n-2)$. The zero-boundary Rellich-Kondrachov compactness theorem gives compact embedding into $L^q$, and Hölder's inequality gives $\Vert wz{\Vert }_2\le \Vert w{\Vert }_p\Vert z{\Vert }_q$. Hence $Kv=wTv$ is compact on $L^2$, without a regularity assumption on the boundary. Writing $v=\Delta u$ makes $\Delta u-wu=f$ equivalent to $(I-K)v=f$. The Fredholm alternative for a compact operator then identifies unique solvability with triviality of the homogeneous kernel.

A nondegenerate star-algebra of compact operators decomposes its Hilbert space into irreducible closed reducing subspaces, each equivalence class occurring finitely many times. Pass to its norm closure. A nonzero compact positive element supplies a finite-rank spectral projection; a minimal projection in its finite-dimensional corner has scalar corner algebra. A cyclic vector in its range generates an irreducible reducing subspace. A maximal orthogonal family exhausts the space by nondegeneracy. Infinitely many equivalent copies would give orthogonal unit vectors whose images under some compact algebra element have equal nonzero norms, contradicting compactness.

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.

For a symmetric uniformly elliptic divergence-form operator with zero conormal flux and zero reaction, Lax-Milgram theorem constructs the shifted inverse $K:L^2\to H^1$. The compact Rellich-Kondrachov theorem embedding makes it compact on $L^2$, and symmetry makes it self-adjoint. If the homogeneous kernel consists of constants, then $\ker (I-K)$ is the constant line and $K1=1$. The Fredholm alternative makes solvability equivalent to $Kf$ being orthogonal to constants, or $⟨Kf,1⟩=⟨f,K1⟩=\int f=0$. The weak solution is unique up to an additive constant. For a disconnected domain the identical argument gives one compatibility condition per connected component.

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.

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.

An everywhere-defined linear operator $T$ on an infinite-dimensional separable Hilbert space need not be bounded merely because ${\sum }_k\Vert T{e}_k{\Vert }^2<\infty$ for a Hilbertian basis $(e_k)$. Let $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$, extend $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis, and let the linear functional $\varphi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\varphi (x)e_1$ has $T{e}_k=0$ for every $k$ but $Tv=e_1$. For $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k$, $v-v_N\to 0$ while $T(v-v_N)=e_1$, proving discontinuity. Thus the definition of a Hilbert-Schmidt operator must require boundedness, or first define the operator on finite linear combinations and take its unique continuous extension.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

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 $B$, define the Hilbert-Schmidt inner product by $⟨A,B{⟩}_{\mathrm{HS}}={\sum }_k⟨A{e}_k,B{e}_k⟩$ for a Hilbertian basis $(e_k)$. With an inner product linear in its first argument this equals $\mathrm{tr}(B^{\ast }A)$; with an inner product conjugate-linear in its first argument it equals $\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm, independently of that convention.

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

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

For bounded operators with trace-class commutator, the multiplicative commutator differs from identity by a trace-class operator. Along $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$, the trace of its logarithmic derivative is the constant $\mathrm{Tr}[A,B]$. This follows by writing $e^{B}A{e}^{-B}-A$ as the integral of conjugates of $[B,A]$ and using trace invariance under bounded similarities. Integrating the Fredholm determinant derivative from $F(0)=I$ proves the formula. Individual traces of $A$ and $B$ need not exist.

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 $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

The trace-class operator condition makes this operator trace well-defined; the class of eligible operators and the trace functional are distinct concepts.

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 $\mathrm{tr}(x\otimes y)=⟨x,y⟩$.

For the rank-one operator $Tf=⟨f,v⟩u$ on a Hilbert space with a countable Hilbertian basis $(e_k)$, the operator norm and Hilbert-Schmidt norm both equal $\Vert u\Vert \Vert v\Vert$. The Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, with equality at $f=v/\Vert v\Vert$ when $v\ne 0$. The Parseval identity for a Hilbertian basis gives ${\sum }_k\Vert T{e}_k{\Vert }^2=\Vert u{\Vert }^2{\sum }_k|⟨e_k,v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$. The zero-vector cases are immediate.

Let $Kx=a\phi (x)$ on a vector space, where $\phi$ is a linear functional. Its image of a linear map lies in the span of $a$. Applying $K$ twice gives $K^{2}x=a\phi (a)\phi (x)=\phi (a)Kx$. Mathematical induction then proves the displayed power formula. If $\phi (a)=0$, $K^2=0$, even when $K\ne 0$; this is a nilpotent operator. If $\phi (a)\ne 0$, then $K/\phi (a)$ is an idempotent linear map. In a real inner product space, the special case $\phi (x)=b\cdot x$ gives $\phi (a)=a\cdot b$.

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.

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.

This is the discrete-eigenbasis case of the spectral theorem for a compact operator that is self-adjoint.

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.

For a nonnegative self-adjoint operator with compact resolvent, this function counts eigenvalues up to a given level with their multiplicities. For the positive Laplace-Beltrami operator on a closed $n$-dimensional Riemannian manifold, $N(\Lambda )\le C(1+\Lambda {)}^{n/2}$. To prove this bound, multiply each vector in the low-eigenvalue subspace by a finite set of chart cutoffs and record its local Fourier coefficients of frequency at most $R$. The total number of retained coefficients is $O(R^n)$, while the sum of discarded squared norms is at most $C{R}^{-2}(1+\Lambda )\Vert u{\Vert }_2^2$ by Parseval identity and the energy bound. For $R$ a sufficiently large multiple of $\sqrt{1+\Lambda }$, vanishing of all recorded coefficients forces $u=0$. The recording map is injective and proves the dimension bound. Inverting this estimate gives ${\lambda }_j\ge c(j+1{)}^{2/n}-1$.

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.