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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.

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 between Hilbert spaces, the index is the difference of the finite dimensions of its kernel and cokernel. Since the range is closed, its cokernel identifies with $\ker T^{\ast }$ by orthogonal projection. An invertible operator has index zero. Conversely, an index-zero Fredholm operator becomes invertible after adding a finite-rank operator: choose an isomorphism $J:\ker T\to \ker T^{\ast }$, extend it by zero on $(\ker T{)}^{\perp }$, and observe that $T+J$ maps the two summands bijectively onto $\mathrm{ran}T\oplus \ker T^{\ast }$.

Write the integral part as $Ky=12b⟨a,y⟩-30a⟨b,y⟩$, where $a(x)=e^{-x}$ and $b(x)=e^{-2x}$. The kernel is real, so its adjoint transposes the two arguments. HenceThis follows by exchanging the two integrations in the $L^2(0,\infty )$ inner product; the exponential factors make the finite-rank operators bounded.

Every image of $L$ is a linear combination $a\cos 2x+b\sin 2x$. The function $\cos x$ is not such a combination. Indeed these functions are continuous, so equality almost everywhere would imply equality everywhere; evaluation at the endpoints would require simultaneously $a=1$ and $a=-1$. Therefore the forcing is outside the image and there is no solution. Orthogonality to the kernel is the equivalent image test for this self-adjoint finite-rank operator.

A nonlinear compact operator is a continuous map that sends bounded sets of its domain to relatively compact subsets of the target space; linearity is not part of this general definition.

For the specified diagonal operator on sequence space on l2 sequence space,so it is well-defined, linear and bounded. Let $T_N$ retain only its first $N$ output coordinates. This is a finite-rank operator, andThe upper bound follows by estimating each tail multiplier by $1/(N+1)$; equality follows on the unit coordinate vector $e_{N+1}$. Finite-rank operators between Banach spaces are compact, and a limit of compact linear operators in operator norm is compact, the norm-closedness of compact operators. Therefore $\boxed{T\text{ is compact}}$.

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

Define a diagonal operator on sequence space byIt is bounded because $\Vert Sx{\Vert }_2^2\le \Vert x{\Vert }_{\infty }^2{\sum }_{n\ge 1}{4}^{-n}=\Vert x{\Vert }_{\infty }^2/3$. Its truncation $S_{m}x=(2^{-1}{x}_1,\dots ,2^{-m}{x}_m,0,\dots )$ is a finite-rank operator, andEquality follows by using finitely supported sequences equal to $1$ on longer and longer finite tails. An operator norm limit of compact operators into the Banach space ${\ell }_2$ is compact; therefore $S$ is a compact operator. The vectors $S{e}_n=2^{-n}{e}_n$ are linearly independent, so its rank is infinite.

There is no conflict with the preceding result about a compact closed-ball image. Indeed, take $x^{(m)}=(1,\dots ,1,0,\dots )$ with $m$ initial ones. Then $S{x}^{(m)}\to (2^{-n}{)}_{n\ge 1}$ in ${\ell }_2$, but that limit has no preimage in $c_0$: its only possible preimage is the constant sequence $1$. Thus $S(B_{c_0})$ is relatively compact but is not closed and hence is not compact.

Define the finite-rank operatorIt is linear and bounded, with $\Vert P_n\Vert \le {\sum }_{i=1}^n\Vert x_i^{\ast }\Vert \Vert x_i\Vert$. Biorthogonality gives exactly the two required values on $x_j$. It also givesThus $P_n$ is a bounded projection onto $\mathrm{span}\{x_1,\dots ,x_n\}$, with kernel ${\bigcap }_{i=1}^n\ker x_i^{\ast }$. Its action on the dense span determines it uniquely among bounded linear operators. These are the partial-sum projections; no uniform bound in $n$ has yet been assumed.

A bounded operator on a complex Hilbert space is self-adjoint when $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$. The spectral theorem for compact self-adjoint operators says its nonzero spectrum consists of real eigenvalues of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their eigenspaces are mutually orthogonal; their closed span is the orthogonal complement of the kernel. On that span the operator has the norm-convergent spectral action $Tx={\sum }_j{\lambda }_j⟨x,e_j⟩e_j$.

Take $H={\ell }^2(\mathbb{N})$ and $T{e}_n=e_n/n$. It is self-adjoint and has infinite-dimensional range. Truncation to the first $N$ coordinates is a finite-rank operator and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.

The resolvent set comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the spectrum, and the point spectrum consists of eigenvalues. In this exampleFor $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda {)}^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has eigenvector $e_n$. Zero is not an eigenvalue, but $\Vert T{e}_n\Vert \to 0$ prevents a bounded inverse; also $(1/n{)}_n\in {\ell }^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.

The Fredholm index measures the finite-dimensional obstruction to inverting an operator. For a bounded operator $T:H_1\to H_2$ between Hilbert spaces, the Fredholm conditions are that its range of a bounded linear operator is closed and its kernel and cokernel are finite-dimensional. The orthogonal complement of the range is $\ker T^{\ast }$, soAn invertible linear map has Fredholm index zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large kernel and cokernel.

First construct an inverse up to finite-rank operators. On $(\ker T{)}^{\perp }$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the bounded inverse theorem. Extend that inverse by zero on $(\mathrm{ran}T{)}^{\perp }$, obtaining $S:H_2\to H_1$. ThenConversely, suppose $ST-I$ and $TS-I$ are compact operators. On $\ker T$, the first identity expresses the identity as a compact operator, so this kernel is finite-dimensional. If $T$ were not bounded below on $(\ker T{)}^{\perp }$, there would be unit vectors $x_j$ in that complement with $T{x}_j\to 0$. Write $ST=I+K$. Compactness of $K$ gives a subsequence for which $K{x}_j$ converges, hence $x_j=ST{x}_j-K{x}_j$ converges to a unit vector $x$ in $(\ker T{)}^{\perp }$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^{\ast }$, using $S^{\ast }{T}^{\ast }=I+K^{\prime \ast }$, proves that its kernel is finite-dimensional. This proves the Atkinson theorem: a bounded operator is Fredholm exactly when it has an inverse modulo compact operators. For operators on one Hilbert space this says that its image in the Calkin algebra is invertible.

The Fredholm index is stable. Decompose the domain as $N\oplus N^{\perp }$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^{\ast }$ and $R=\mathrm{ran}T$. A sufficiently small norm perturbation has block matrixMultiplying by invertible triangular block matrices reduces this matrix to the direct sum of $D$ and the Schur complement $a-b{D}^{-1}c:N\to C$. The latter's kernel dimension minus cokernel dimension is $\dim N-\dim C$, whatever its rank. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo compact operators for every $T+tK$, $0\le t\le 1$. Local constancy along this path gives

For Fredholm operators $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an exact sequence of finite-dimensional spacesThe map from $\ker B$ takes a vector to its class modulo $\mathrm{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\mathrm{ran}A$ lies in the kernel of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in \ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the dimensions givesAlso $\mathrm{ind}{T}^{\ast }=-\mathrm{ind}T$, and indices add under direct sums. For $I+K$ with $K$ compact, the index is zero, so injectivity is equivalent to surjectivity. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^{\ast }$: these are the finitely many compatibility conditions in the Fredholm alternative.

Toeplitz operators exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the Hardy space of the circle, spanned by $1,z,z^2,\dots$, and let $P$ be its orthogonal projection. For a continuous symbol $g$, define $T_g=P{M}_g{|}_{H^2}$, where $M_g$ is multiplication by $g$. Then $\Vert T_g\Vert \le \Vert g{\Vert }_{\infty }$. The product defect isFor trigonometric polynomials the crossing of negative and nonnegative Fourier modes has finite rank. Uniform approximation of continuous symbols by trigonometric polynomials, together with the displayed norm estimate, therefore makes this defect a compact operator for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo compact operators. Thus $T_g$ is Fredholm.

The converse is instructive. If $g(\zeta )=0$ at $\zeta \in S^1$, consider the normalized Hardy kernelsTheir norms are one, their individual Fourier coefficients tend to zero as $r\uparrow 1$, and hence they converge weakly to zero. Their squared moduli are the Poisson kernel, so continuity of $g$ givesA Fredholm operator is bounded below off its finite-dimensional kernel. The projection of these weakly vanishing vectors onto that kernel tends to zero, contradicting this lower bound. Thus a continuous-symbol Toeplitz operator is Fredholm exactly when its symbol never vanishes. Applying the same argument to $g-\lambda$ gives its essential spectrum as $g(S^1)$, with essential spectrum defined by failure of the Fredholm property.

For nonvanishing $g$, let $m=\mathrm{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\le t\le 2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a continuous function $h:S^1\to \mathbb{C}$ with $g(z)=z^m{e}^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m{e}^{th}$ reduces the index to that of $T_{z^m}$. For $m\ge 0$ this is the $m$th power of the unilateral shift operator, whose kernel is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose kernel has dimension $-m$ and whose range is all of $H^2$. ConsequentlyThe minus sign comes from counting missing nonnegative Fourier modes. This is a concrete bridge between the Fredholm index, stability under compact perturbations, and the winding number of a loop.

For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

A bounded operator $T:H\to H$ is a Fredholm operator when its range is closed and both $\ker T$ and $\mathrm{ran}(T{)}^{\perp }=\ker T^{\ast }$ have finite dimension. Its Fredholm index isFor index zero, choose an isomorphism $J:\ker T\to \ker T^{\ast }$ and extend it by zero on $(\ker T{)}^{\perp }$. The restricted map $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ is a bounded isomorphism, by the bounded inverse theorem. The finite-rank operator $J$ fills exactly the missing kernel-to-cokernel block. Thus $T+J$ is invertible and $T=(T+J)-J$ is an invertible operator plus a compact operator.

Conversely suppose $T=A+K$ with $A$ invertible and $K$ compact. The Atkinson theorem proved in part (b) makes every $A+tK$, $0\le t\le 1$, Fredholm: its quotient image is always $[A]$. The block argument in part (c) proves local constancy of the Fredholm index, so its index is constant along this interval and equals $\mathrm{ind}A=0$. This invokes the actual parametrix and block proofs below, not an unproved perturbation assertion. The claimed equivalence holds.

Let $u:X\to U_{\mathcal{K}}$ be continuous. The family $\{u(x)-I:x\in X\}$ is norm compact in the compact-operator ideal. For any $\epsilon >0$ there is one finite-rank orthogonal projection $P$ withTo prove it, take a finite $\epsilon$-net of compact operators, approximate each by a finite-rank operator, and include both their ranges and adjoint ranges in $P\mathcal{H}$. Contractivity of compression transfers the estimate to the whole family.

Take $\epsilon <1$ and set $g(x)=I+P(u(x)-I)P$. It is invertible because it is within distance one of the unitary $u(x)$. It has the form $g_V(x)\oplus I$ on $V\oplus V^{\perp }$, where $V=P\mathcal{H}$ is finite-dimensional and $g_V:X\to GL(V)$ is continuous. The straight segment from $u$ to $g$ stays invertible; taking its unitary polar factors supplies a homotopy in $U_{\mathcal{K}}$, since inverse square root depends continuously on the positive invertible factors and preserves identity modulo compacts.

Conversely a map $g:X\to GL(V)$ defines such a class by the continuous deformation $g(g^{\ast }g{)}^{-t/2}$ to its unitary polar factor, then extension by the identity on $V^{\perp }$. Apply the same uniform finite-rank approximation to a homotopy on the compact space $X\times [0,1]$. Choosing $P$ to include the original endpoint supports keeps those endpoints unchanged under compression. This proves that two finite-dimensional maps give the same class exactly when, after adding identity blocks, they are homotopic through a common finite-dimensional general linear group. Hencewith stabilization $g\mapsto \mathrm{diag}(g,1)$.

Suppose first that the bounded Fredholm operator $T:H\to H$ has Fredholm index zero. Put $N=\ker T$ and $C=(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$. These finite-dimensional linear subspaces have the same dimension. The restricted map $T:N^{\perp }\to \mathrm{ran}T$ is a bounded bijection, with bounded inverse by the bounded inverse theorem. Choose an isomorphism $J:N\to C$ and extend it by zero on $N^{\perp }$, writing the resulting finite-rank operator as $K$.

Relative to the orthogonal decompositions $H=N\oplus N^{\perp }$ and $H=C\oplus \mathrm{ran}T$, the operator $T+K$ is the direct sum of $J$ and the invertible restricted map. Thus $T+K$ is invertible, and $T=(T+K)-K$ is the sum of an invertible operator and a compact operator.

Conversely write $T=U+K$ with $U$ invertible and $K$ compact. It suffices to treat $I+C$ with $C=U^{-1}K$. A compact operator on a Hilbert space admits arbitrary operator norm approximation by finite-rank operators: take a finite epsilon-net of its unit-ball image and project onto the span of the net. Choose $F$ of finite rank with $\Vert C-F\Vert <1$, put $R=C-F$, and factorThe first factor is invertible by the Neumann series. Set $G=(I+R{)}^{-1}F$ and $E=\mathrm{ran}G$, a finite-dimensional linear subspace. On $E\oplus E^{\perp }$, the second factor has block formLeft multiplication by the invertible block matrix $\left(\begin{array}{cc}I& -B\\ 0& I\end{array}\right)$ reduces it to $A\oplus I$. Its range is closed, and its kernel and cokernel have equal dimension by finite-dimensional rank-nullity theorem for $A:E\to E$. Invertible factors preserve all these properties. Hence $T$ is a Fredholm operator of Fredholm index zero, proving both directions:

The real Stone-Weierstrass theorem says that a subalgebra of $C(X,\mathbb{R})$ on a compact metric space is uniformly dense if it contains the constants and separates points. The finite sums ${\sum }_i{f}_i(x)g_i(y)$ form such an algebra on $[0,1{]}^2$: multiplication preserves this form, the constant one belongs to it, and the coordinate functions distinguish every pair of different points. Thus they approximate the continuous kernel uniformly.

A finite-rank operator has finite-dimensional range. A compact operator maps the unit ball to a set with compact closure, equivalently sends every bounded sequence to a sequence with a convergent subsequence. The identity on $C[0,1]$ is bounded but not compact: the functions $x^n$ have supremum norm one, and every subsequence has the discontinuous pointwise limit which is zero on $[0,1)$ and one at one. It therefore has no uniformly convergent subsequence.

A bounded finite-rank operator is compact because bounded sets in its finite-dimensional range have compact closure. Suppose $T_n\to T$ in operator norm. For any $\epsilon >0$, choose $n$ with $\Vert T-T_n\Vert <\epsilon /3$ and a finite $\epsilon /3$-net for $T_n$ of the unit ball. The same centres give an $\epsilon$-net for $T$ of that ball. Its closure is totally bounded and complete in the Banach space $Y$, hence compact. Thus an operator-norm limit of compact operators is compact.

For $K_n(x,y)={\sum }_i{f}_i(x)g_i(y)$ uniformly approximating $K$, the integral operator $T_n$ has range in $\mathrm{span}\{f_i\}$ and is bounded. MoreoverConsequently $T_n\to T$ in operator norm, proving that the stated integral operator is compact.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

Compactness gives a stronger statement than the one requested. Since $T(B_H)$ has compact closure, finitely many open balls of radius $\delta$ cover it. Choose their centres from $T(B_H)$ and let $E$ be their finite-dimensional linear span. ThenFor the orthogonal projection $P_E$, best approximation in a Hilbert space givesIn particular this holds for every vector $e_n$ in the given orthonormal basis.

The operator $P_{E}T$ has image in $E$ and hence has finite rank, whileThus every compact operator is an operator norm limit of finite-rank operators. This is the finite-rank approximation theorem for compact operators on a Hilbert space.