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A bounded linear operator between normed spaces is approximable if it is an operator norm limit of bounded finite-rank operators. If the target is a Banach space, every approximable operator is a compact operator. For a given tolerance, approximate the operator in norm by one finite-rank operator, cover that operator's bounded unit-ball image by finitely many small balls in its finite-dimensional range, and transfer the cover across the operator-norm error. This makes the original image totally bounded; its closure is complete in the Banach target and therefore compact. The converse on arbitrary Banach spaces would require an additional approximation hypothesis and is not part of this assertion.

A bounded operator is Fredholm if and only if it has a bounded two-sided inverse modulo compact operators. For the forward implication invert its restriction from the orthogonal kernel complement to its closed range and extend by zero. Conversely a parametrix makes the identity compact on the kernel, forcing finite dimension. A failure of a lower bound on the kernel complement would yield unit vectors $x_n$ with $T{x}_n\to 0$; the compact error then forces a norm-convergent subsequence, whose nonzero limit lies both in the kernel and its complement. This proves closed range; applying the other parametrix identity to the adjoint gives finite cokernel.

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 ${\ell }^q$.

For an infinite-dimensional complex Hilbert space, the Calkin algebra is the unital quotient of the bounded operators by the closed two-sided ideal of compact operators. Its spectrum records invertibility up to compact errors. An operator is Fredholm exactly when its image in this quotient is invertible.

Bounded self-adjoint operators on separable infinite-dimensional complex Hilbert spaces are unitarily equivalent modulo compact operators exactly when they have the same essential spectrum in the Calkin algebra. The Weyl-von Neumann theorem diagonalizes each modulo a compact error. Their diagonal cluster sets are the essential spectra; cluster-set matching of bounded sequences pairs the two diagonal lists with differences tending to zero, giving a compact diagonal difference. The separability assumption is essential: projections with countable and uncountable range dimensions can have the same essential spectrum but cannot differ by a compact operator after unitary conjugation.

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.

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.

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 Fredholm operator $L$ on a Hilbert space, invert its restriction from the kernel complement to its closed range, extending by zero on the range complement. The resulting bounded $B$ has $BL=I-P$ and $LB=I-Q$ with finite-rank projections. Adding a compact operator $K$ gives two-sided inverses modulo compact errors. Such a parametrix forces finite kernel, a lower bound on the kernel complement, closed range and finite adjoint kernel; the latter is the cokernel. Repeating with $-K$ proves the equivalence without assuming self-adjointness.

A bounded linear operator between Banach spaces is completely continuous when it maps weakly convergent sequences to norm-convergent sequences. Equivalently, it maps weakly compact sets to norm-compact sets: the Eberlein-Šmulian theorem gives the forward direction, while a weakly convergent sequence together with its limit is weakly compact and gives the reverse direction. The inclusion $C(\mathbb{T})\to L^2(\mathbb{T})$ for finite Lebesgue measure is completely continuous by pointwise convergence, the Uniform boundedness principle and the dominated convergence theorem, but is not a compact operator because of the orthogonal Fourier modes.

For rigid walls $a<r<b$, a nonzero vertical wavenumber $k$ and radial Lagrangian fluid displacement $\xi$, set $y=r\xi$ and $\mathcal{D}={\kappa }^2-r(d/dr)(v_A^2/r^2)$. Axisymmetric modes satisfyThe operator on the left is a positive self-adjoint operator. Its Rayleigh quotient is ${\omega }^2=k^2\int \mathcal{D}|y{|}^2/rdr/\int (|y^{\prime }{|}^2+k^2|y{|}^2)/rdr$. All mode squares are real; a growing mode exists exactly when the continuous coefficient $\mathcal{D}$ is negative somewhere. Sufficiency follows by a test function supported in that negative interval and the compact operator and self-adjoint operator eigenproblem obtained by conjugating the multiplication operator with the inverse square root of the positive differential operator.

// Target: astrophysical-fluid-dynamics.bigb

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.

A left singular vector $u_j$ is a unit vector in the output space satisfying $A{A}^{\ast }{u}_j={\sigma }_j^2{u}_j$. For positive singular values, the pairing also satisfies $A^{\ast }{u}_j={\sigma }_j{v}_j$. Thus $u_1=(1,0{)}^T$ and $u_2=(0,1{)}^T$ describe the output axes of the preceding ellipse, while the right singular vectors describe its input directions. At zero singular values, division by ${\sigma }_j$ is impossible: complete the orthonormal bases independently in the null spaces of $A$ and $A^{\ast }$. Repeated singular values allow different matched orthonormal bases within the corresponding eigenspaces. The same positive pairing extends to a compact operator through its singular value system.

Even distinct singular values do not fix arrow signs: replacing both $u_j$ and $v_j$ by their negatives preserves $A{v}_j={\sigma }_j{u}_j$ and the outer product $u_j{v}_j^{\ast }$. Reversing only one arrow would change that term's sign. The paired directions, rather than independently chosen arrow orientations, encode the matrix.

For the Volterra integration operator $Af(x)={\int }_0^{x}f(t)dt$, its adjoint is $A^{\ast }g(x)={\int }_x^{1}g(t)dt$. With $a_n=(n-\frac{1}{2})\pi$, the input singular vectors are $\sqrt{2}\cos (a_{n}x)$ and the output singular vectors $\sqrt{2}\sin (a_{n}x)$, with singular values $1/a_n$. The associated Sturm-Liouville problem has Neumann condition at zero and Dirichlet condition at one for the cosine basis; the sine basis has the reversed conditions. These complete orthonormal bases give a singular value system for an injective compact operator with dense range.

For a bounded operator $A:X\to Y$ between Hilbert spaces, the Moore–Penrose inverse maps admissible data to the unique minimum-norm least-squares solution. If $({\sigma }_j,u_j,v_j)$ is a singular system of a compact operator, thenon the data for which this series converges.

More precisely, its domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$. It vanishes on $(\mathrm{ran}A{)}^{\perp }$ and takes values in $(\ker A{)}^{\perp }$. An infinite-rank compact operator on infinite-dimensional spaces has singular values approaching zero, so its inverse on its range is generally unbounded; Picard criterion specifies admissible data.

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.

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.

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.

For compact metric $X$, odd topological K-theory is the homotopy set of norm-continuous maps into unitaries equal to the identity modulo compact operators on a fixed separable infinite-dimensional Hilbert space. Uniform finite-rank approximation over $X$ and over homotopies gives the equivalent stable model ${\mathrm{colim}}_N[X,G{L}_N(\mathbb{C})]$. Block sum gives an abelian group: a finite-dimensional unitary path swaps blocks, and a two-block rotation contracts $\mathrm{diag}(u,u^{-1})$ to the identity. On the circle the determinant winding identifies $K^1(S^1)$ with $\mathbb{Z}$; Bott periodicity gives $K^1(S^n)=\mathbb{Z}$ in odd dimensions and zero in even dimensions.

For Banach spaces $X,Y$, a nonlinear compact operator $T:X\to Y$ is a continuous map that sends every bounded set to a relatively compact subset of $Y$: the closure of its image is a compact set. For a continuous map, the image condition is equivalent to requiring that every bounded sequence $(x_j)$ has a subsequence along which $T(x_j)$ converges in the norm topology of $Y$. For a linear operator, this is the usual compact operator definition; the continuity requirement is included explicitly here because the subsequent applications are nonlinear.

The compact-domain version of the Schauder fixed-point theorem saysA useful equivalent version replaces $K$ by a nonempty closed bounded convex set $C$ and assumes $T:C\to C$ is a continuous map with relatively compact image. This latter form will be used for the partial differential equation. Both forms assert existence of a fixed point, without asserting uniqueness.