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For a bounded operator on a complex Banach space, the resolvent set is $\rho (T)=\{\lambda :\lambda I-T\text{ is bijective with bounded inverse}\}$, and the resolvent operator is $R_T(\lambda )=(\lambda I-T{)}^{-1}$. The spectrum is its complement $\sigma (T)=\mathbb{C}\setminus \rho (T)$; the point spectrum consists of $\lambda$ with $\ker (\lambda I-T)\ne 0$. For a real Banach space these definitions use its complexification.

If $|\lambda |>\Vert T\Vert$, the norm-convergent Neumann series gives $R_T(\lambda )={\lambda }^{-1}{\sum }_{n\ge 0}(T/\lambda {)}^n$, so the spectrum is bounded. If ${\lambda }_0\in \rho (T)$, factorThe second factor is invertible by a Neumann series whenever $|\lambda -{\lambda }_0|\Vert R_T({\lambda }_0)\Vert <1$. Thus the resolvent set is open and the spectrum is closed and bounded.

The point spectrum need not be closed. On ${\ell }^2$, define $T{e}_n=e_n/n$. Every $1/n$ is an eigenvalue, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other eigenvalues occur, so ${\sigma }_p(T)=\{1/n:n\ge 1\}$ is not closed.

Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne 0$, then $⟨Tv,v⟩$ is real and equals $\lambda \Vert v{\Vert }^2$ under the convention linear in the first argument. Therefore every eigenvalue of a self-adjoint operator is real.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

Take a nonzero complex unital Banach algebra, with identity $1$. The spectrum of an element isThe standard Banach algebra spectrum theorem states that it is a nonempty compact subset of $\mathbb{C}$, contained in the disk $|\lambda |\le \Vert a\Vert$. Conversely, every nonempty compact set $K\subseteq \mathbb{C}$ occurs: in the Banach algebra $C(K)$, the coordinate function $a(z)=z$ has Banach algebra spectrum exactly $K$, since $\lambda -a$ has a continuous reciprocal precisely when $\lambda \notin K$. Thus the possible Banach algebra spectra are exactly the nonempty compact subsets of the complex plane.

For a nonconstant polynomial $p$, factor $p(z)-\lambda =c{\prod }_{j=1}^n(z-z_j)$ over $\mathbb{C}$. Substitution givesThese factors commute. A finite commuting product is invertible exactly when every factor is invertible: if $P={\prod }_j{x}_j$ is invertible, its inverse commutes with each $x_j$, and $x_j^{-1}=({\prod }_{k\ne j}{x}_k)P^{-1}$ is a two-sided inverse. Hence $\lambda$ belongs to the Banach algebra spectrum of $p(a)$ exactly when some root $z_j$ belongs to ${\sigma }_A(a)$, which is equivalent to $\lambda \in p({\sigma }_A(a))$. Constant polynomials give the same result directly, since ${\sigma }_A(c1)=\{c\}$. This proves the polynomial spectral mapping theorem:For a closed unital Banach subalgebra $B\subseteq A$, use the common identity $1_A\in B$. An inverse in $B$ is an inverse in $A$, so $\boxed{{\sigma }_A(b)\subseteq {\sigma }_B(b)}$. Let $\lambda \in \partial {\sigma }_B(b)$. Choose ${\lambda }_n\notin {\sigma }_B(b)$ tending to $\lambda$. Then ${\lambda }_n\notin {\sigma }_A(b)$, and $({\lambda }_{n}1-b{)}^{-1}\in B$. If $\lambda$ were outside ${\sigma }_A(b)$, continuity of inversion in $A$ would givewhere membership follows from closedness. This would make $\lambda \notin {\sigma }_B(b)$, contrary to the closed Banach algebra spectrum having its boundary inside it. Therefore $\lambda \in {\sigma }_A(b)$ and is approached by points outside that Banach algebra spectrum, provingContinuity of inversion here follows, for example, by writing a nearby element as an invertible element times $1+$ a norm-small perturbation and using the Neumann series. This is spectrum in a closed unital subalgebra.

To retain the original Banach algebra spectrum in a commutative subalgebra, take the resolvent-generated commutative algebraEvery generator commutes with $a$ and with every other generator: commuting invertible elements have commuting inverses, and their inverses commute with all elements commuting with them. commutativity persists on the algebra they generate and on its norm closure. For every $\lambda \notin {\sigma }_A(a)$, the required inverse is in $C$ by construction, so ${\sigma }_C(a)\subseteq {\sigma }_A(a)$. The inclusion for unital subalgebras gives the reverse containment. Hence $\boxed{{\sigma }_C(a)={\sigma }_A(a)}$.

Finally, the Banach subalgebra generated by one element is $D=\overline{\{p(a):p\text{ a polynomial}\}}$, and it is commutative. Put $K={\sigma }_D(a)$ and suppose its complement has a bounded connected component $U$. The boundary of $U$ lies in $K$. For $\lambda \in U$, the maximum modulus principle on a bounded domain applied to any polynomial givesHere $r_D$ is the spectral radius, and the equality uses the already proved polynomial spectral mapping theorem. Define $\chi (p(a))=p(\lambda )$. The displayed bound makes this well defined, even when two polynomials represent the same algebra element, and bounded on the dense polynomial algebra. Its continuous extension to $D$ remains linear and multiplicative, with $\chi (1)=1$ and $\chi (a)=\lambda$, so it is a character of an algebra. If $\lambda 1-a$ were invertible in $D$, applying $\chi$ to its inverse equation would give $0=1$. Thus $\lambda \in K$, contradicting $\lambda \in U$.

There are therefore no bounded complementary components. A compact subset of the plane has exactly one unbounded complementary component: everything outside a sufficiently large disk is connected, and every unbounded component meets it. Consequently $\boxed{\mathbb{C}\setminus {\sigma }_D(a)\text{ is connected}.}$

Use additive notation and fix Haar measure $m$ on $G$. A continuous unitary character is a continuous homomorphism $\chi :G\to \mathbb{T}$. With the conventionthe Fubini theorem and the character identity imply $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Also $|\hat{f}(\chi )|\le \Vert f{\Vert }_1$, and choosing $f=\chi a$ for a nonzero nonnegative $a\in C_c(G)$ shows that evaluation at $\chi$ is a nonzero functional. Thus each character gives a character of an algebra on the L1 convolution algebra. Here a multiplicative linear functional means a nonzero multiplicative linear functional; the zero functional must be excluded for the claimed bijection.

Conversely, let $\phi$ be such a functional. It is automatically bounded with $|\phi (f)|\le \Vert f{\Vert }_1$: extend it to the unitization of an algebra by $\tilde{\phi }(f+z1)=\phi (f)+z$. If $|z|>\Vert f{\Vert }_1$, the Neumann series makes $z1-f$ invertible, so its image under the unital homomorphism cannot be zero. Hence $\phi (f)$ cannot lie outside the norm disc.

Choose $a\in L^1(G)$ with $\phi (a)\ne 0$, and let $T_{x}a(t)=a(t-x)$. The Abelian translation identities $(T_{x}a)\ast b=a\ast (T_{x}b)$ giveTherefore, defining $c(x)=\phi (T_{x}a)/\phi (a)$, one has $\phi (T_{x}b)=c(x)\phi (b)$ for every $b$. Applying this to two successive translations gives $c(x+y)=c(x)c(y)$ and $c(0)=1$. By translation continuity in Lp on a locally compact group, $c$ is continuous. It is uniformly bounded by $\Vert a{\Vert }_1/|\phi (a)|$; applying this bound to $c(nx)=c(x{)}^n$ for positive and negative integers forces $|c(x)|=1$. Thus $\chi (x)=\overline{c(x)}$ is a continuous unitary character.

The Bochner integral identity $a\ast b={\int }_{G}b(y)T_{y}adm(y)$ holds in $L^1$, since $\Vert T_{y}a{\Vert }_1=\Vert a{\Vert }_1$. Applying the bounded functional yieldsCancel $\phi (a)\ne 0$ to obtain the required correspondenceThe recovery formula $\overline{\chi (x)}=\phi (T_{x}a)/\phi (a)$ proves uniqueness. These functionals have norm exactly one, since $b=\chi a$ with $a\ge 0$ and $\int a=1$ has $\Vert b{\Vert }_1=\phi (b)=1$.

For the topology, identify the Pontryagin dual group with these functionals and give it the Gelfand topology, namely pointwise convergence of $\hat{f}(\chi )$ for every $f\in L^1$. If a net ${\chi }_{\lambda }$ converges to $\gamma$ uniformly on every compact subset of $G$, then for $f\in C_c(G)$,Density of $C_c$ in $L^1$ and the uniform functional norm bound one extend this convergence to every $f\in L^1$.

Conversely, suppose convergence in the Gelfand topology and choose $a$ with $\hat{a}(\gamma )\ne 0$. For compact $K\subseteq G$, the set $\{T_{x}a:x\in K\}$ is compact in $L^1$ by translation continuity in Lp on a locally compact group. Evaluation convergence is uniform on this compact set: a finite $\delta$-net reduces its supremum to finitely many convergent evaluations plus $2\delta$, using the common functional norm one. ThusNow $\widehat{T_{x}a}(\chi )=\overline{\chi (x)}\hat{a}(\chi )$, and $\hat{a}({\chi }_{\lambda })\to \hat{a}(\gamma )\ne 0$. Dividing this identity proves ${\sup }_{x\in K}|{\chi }_{\lambda }(x)-\gamma (x)|\to 0$. Hence the Gelfand topology is exactly the compact-open topology, with the requested neighbourhood basisFor compact $K$, the continuous difference attains its maximum, so this strict pointwise inequality on $K$ is equivalent to a strict uniform bound. Finite intersections contain another such set by taking the union of the compact sets and the minimum of the tolerances.

Because $A$ is a closed subalgebra of a Banach algebra, it is itself complete. If $\Vert a-1\Vert <1$, the Neumann series converges in $A$:Multiplying the partial sums by $a$ on either side gives $1-(1-a{)}^{N+1}$, which tends to $1$. Thus every such $a$ is invertible in $A$, and $1$ is an interior point of $G(A)$.

For $a\in G(A)$, any $b\in A$ with $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ has the factorizationThe second factor has a Neumann series inverse in $A$, so $b\in G(A)$. Hence the group of invertible elements of a Banach algebra $G(A)$ is open in $A$.

For $a\in G_B(A)$ use its inverse in $B$ instead. The same small-perturbation argument proves invertibility of $b$ in $B$ whenever $b\in A$ and $\Vert a^{-1}{\Vert }_B\Vert b-a{\Vert }_B<1$. Therefore both $G(A)$ and $G_B(A)$ are open subsets of $A$; in the second argument the inverse need not lie in $A$.

Suppose the inverse norms did not tend to infinity. Some subsequence would satisfy $\Vert a_n^{-1}\Vert \le C$. Since $a_n\to a$, for its large indices $\Vert a_n^{-1}(a-a_n)\Vert <1$. Thenis invertible in $A$ by the Neumann series, contradicting the hypothesis. Hence $\boxed{\Vert a_n^{-1}\Vert \to \infty }$, the inverse norm divergence at noninvertible boundary points property.

If $a$ were invertible in $B$, then continuity of inversion in a Banach algebra would give $a_n^{-1}\to a^{-1}$ in $B$. Explicitly, the same factorization around $a$ gives a Neumann series and the bound $\Vert a_n^{-1}-a^{-1}\Vert \le \Vert a_n^{-1}\Vert \Vert a-a_n\Vert \Vert a^{-1}\Vert$, with the inverse norms locally bounded in $B$. Since every $a_n^{-1}$ belongs to the closed subalgebra $A$, so would $a^{-1}$. That would make $a\in G(A)$, again a contradiction. Therefore $a\notin G_B(A)$.

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.

The midpoint quadratic spline collocation matrix isIts strict diagonal dominance margin is at least $3/4-2/8=1/2$. The inverse infinity-norm bound from diagonal dominance thus already proves $\Vert A_x^{-1}{\Vert }_{\infty }\le 2$: choosing an index where a vector $c$ attains its maximum absolute entry gives $\Vert A_{x}c{\Vert }_{\infty }\ge \frac{1}{2}\Vert c{\Vert }_{\infty }$, which also proves invertibility.

To evaluate the finite-dimensional norm exactly, set $D_{ii}=(-1{)}^i$ and $B=D{A}_{x}D$. If $T$ is the matrix with ones on its two adjacent diagonals and zeros elsewhere, then $B=(3/4)(I-T/6)$. Since $\Vert T/6{\Vert }_{\infty }\le 1/3$, the Neumann serieshas nonnegative entries. Thus $A_x^{-1}=D{B}^{-1}D$ has absolute entry values equal to those of $B^{-1}$. Its absolute row-sum vector is $v=B^{-1}1$, characterized byThe homogeneous roots are $3\pm 2\sqrt{2}$, and the constant particular solution is $v_i=2$. Writing $q=3-2\sqrt{2}\in (0,1)$ and imposing the two boundary values givesThis is largest at the middle row or rows. Therefore, with $i_{\ast }=\lfloor (n+1)/2\rfloor$, the exact norm isFor example, it is $4/3$ for $n=1$ and $8/5$ for $n=2$, and it increases towards two as the system grows. Thus two is a uniform bound, rather than the exact finite-$n$ inverse norm.

Fix a Fredholm operator $T$. Split its domain as $X\oplus N=(\ker T{)}^{\perp }\oplus \ker T$ and its target as $Y\oplus M=\mathrm{ran}T\oplus \ker T^{\ast }$. For a sufficiently small norm perturbation $R$, writewhere $A:X\to Y$ remains invertible by the Neumann series. Invertible triangular row and column operations reduce this matrix toThe second block maps the finite-dimensional space $N$ into $M$. Hence the perturbed operator has closed range, finite kernel and cokernel, andsince the rank of that finite block cancels from the difference. Thus the Fredholm set is open and its index is locally constant, hence norm continuous.

If $S,T$ are Fredholm, their product is Fredholm by multiplying their invertible images in the Calkin algebra. For index additivity use the following finite-dimensional exact sequence:Here $q$ is reduction modulo $\mathrm{ran}T$, the induced $S$-map sends $[x]$ to $[Sx]$, and the last map is the quotient by the larger range $\mathrm{ran}S$. These maps are well-defined. For example, $\ker q=\ker S\cap \mathrm{ran}T$ is the image of the preceding map. If $Sx\in \mathrm{ran}ST$, subtracting a suitable $Ty$ makes $x-Ty\in \ker S$, proving exactness at $\mathrm{coker}T$. The other positions follow directly from the definitions. Alternating dimensions in this exact sequence gives

Let $X$ be the character space of an algebra, consisting of nonzero multiplicative complex linear functionals on $A$. We construct the required Gelfand transform and prove its properties.

Characters exist. A maximal proper ideal exists by the axiom of choice and is closed: if it were dense it would contain an element within distance one of the identity, hence an invertible element, contradicting properness. The quotient is a complex unital Banach division algebra, and the Gelfand-Mazur theorem identifies it with $\mathbb{C}$, producing a character. The same argument applied to the proper ideal generated by $a-\lambda$ provesEvery character has norm one, since its value belongs to the spectrum and is bounded by $\Vert a\Vert$, while $\chi (1)=1$. The character equations are weak-star closed in the dual unit ball. Thus Banach-Alaoglu theorem makes $X$ compact Hausdorff; separability of $A$ makes it metrizable.

The algebra is inverse-closed in $B(H)$. If $a\in A$ is invertible there, $a^{\ast }a\ge cI$ for some $c>0$. For $M=\Vert a^{\ast }a\Vert$, the Neumann series for $(a^{\ast }a{)}^{-1}=M^{-1}[I-(I-a^{\ast }a/M){]}^{-1}$ converges in $A$, since the positive contraction in brackets has norm at most $1-c/M<1$. Therefore $a^{-1}=(a^{\ast }a{)}^{-1}{a}^{\ast }\in A$. This proves spectral permanence for C-star algebras in the present concrete setting. In particular a self-adjoint element has real spectrum in $A$, so characters take real values on it. Decomposing $a$ into self-adjoint real and imaginary parts gives $\chi (a^{\ast })=\overline{\chi (a)}$.

Define $\hat{a}(\chi )=\chi (a)$. This is a unital star-homomorphism to $C(X)$. Every element of the abelian star-algebra is normal, and the C-star identity gives $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral radius formula therefore gives $r_A(a)=\Vert a\Vert$. The displayed spectrum identity now yields $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$, proving isometry and injectivity. Its image is closed because $A$ is complete, contains the constants, is closed under conjugation, and separates distinct characters. The Stone-Weierstrass theorem makes that image dense in $C(X)$, hence equal to it. $\boxed{A\cong C(X)}$ isometrically as a star-algebra.

For the Fredholm convention, defineThe equality follows from the Atkinson theorem. The quotient is a unital complex Banach algebra when $H$ is infinite-dimensional. Its invertible set is open by the Neumann series, so the spectrum is closed, and it lies in the disk $|\lambda |\le \Vert [T]\Vert$.

For nonemptiness, suppose the quotient resolvent $R(\lambda )=(\lambda 1-[T]{)}^{-1}$ existed for every complex $\lambda$. It is analytic everywhere, and outside a sufficiently large disk its Neumann expansion gives $\Vert R(\lambda )\Vert \le 1/(|\lambda |-\Vert [T]\Vert )$. Continuity bounds it on the remaining compact disk. Every continuous linear functional applied to $R$ is therefore a bounded entire scalar function, constant by Liouville's theorem and zero because it tends to zero at infinity. The Hahn-Banach theorem, or separation by the dual, forces $R(\lambda )=0$, contradicting its inverse identity. The essential spectrum is nonempty and closed.

Infinite dimension is necessary: every operator on a finite-dimensional Hilbert space is Fredholm, so its essential spectrum in this convention is empty. The proof works on nonseparable infinite-dimensional spaces as well; it needs neither self-adjointness nor a sequence basis.

Assume the finite-generator identity holds whenever there is no common zero. Let $\phi \in {\Phi }_A$. For every finite family $f_1,\dots ,f_n\in \ker \phi$, their zero sets must have nonempty intersection: otherwise an identity $\sum f_j{g}_j=1$ would give $0=\phi (1)=1$. These zero sets are closed in compact $K$. The finite intersection property therefore supplies $x\in K$ at which every member of $\ker \phi$ vanishes. For any $f\in A$, the element $f-\phi (f)1$ is in that kernel, so $f(x)=\phi (f)$. Thus $\phi ={\delta }_x$, proving

For $C(K)$, the identity is explicit. If the $f_j$ have no common zero, the continuous function $q={\sum }_j|f_j{|}^2$ is strictly positive on compact $K$. SetThen $\sum f_j{g}_j=1$. The equivalence just proved shows that $C(K)$ is natural.

Next consider the disc algebra $A(\overline{\mathbb{D}})$, the continuous functions on the closed unit disk that are holomorphic in its interior. We show directly that each algebra character is evaluation. Polynomials are uniformly dense in this algebra: for $0<r<1$, $f_r(z)=f(rz)$ tends uniformly to $f$ as $r\uparrow 1$, by uniform continuity. The function $f_r$ is holomorphic on $|z|<1/r$, so its Taylor polynomials converge uniformly on $|z|\le 1$.

Every algebra character of a unital Banach algebra is bounded, with $|\phi (f)|\le \Vert f\Vert$: if $f-\phi (f)1$ were invertible, applying $\phi$ would contradict invertibility, so $\phi (f)$ belongs to the spectrum of an element, which is contained in the disk of radius $\Vert f\Vert$ by the Neumann series. For the coordinate function $z$, put $\lambda =\phi (z)$; then $|\lambda |\le 1$. Multiplicativity gives $\phi (p)=p(\lambda )$ for every polynomial. By polynomial density and continuity, $\phi (f)=f(\lambda )$ for every member of the disc algebra. Hence

Finally consider the algebra $C$ in the last request. For each $f\in C$, its holomorphic extension $g$ from the boundary is unique by the maximum modulus principle. Define $T(f)=g\in A(\overline{\mathbb{D}})$. This is a surjective unital algebra homomorphism, since it restricts to the identity on the disc algebra. AlsoThe algebra $C$ is closed: if $f_n\to f$ uniformly, the associated $T(f_n)$ are uniformly Cauchy by this bound, so their limit belongs to the closed disc algebra and agrees with $f$ on the boundary. Constants and the coordinate function belong to $C$, so it is itself a uniform algebra on the disk. Its idealis contained in $C$, and $C/I$ is the disc algebra.

If a algebra character $\chi$ on $C$ vanishes on $I$, it factors through $T$ and the preceding result givesIf $\chi$ does not vanish on $I$, choose $h\in I$ with $\chi (h)\ne 0$. For $F\in C(\overline{\mathbb{D}})$, defineThe product $hF$ is in $I$, so this is well defined. It is unital and linear, andshows multiplicativity. Since $C(\overline{\mathbb{D}})$ is natural, $\tilde{\chi }$ is evaluation at a point $x$ of the closed disk. The equality $h(x)=\tilde{\chi }(h)=\chi (h)\ne 0$ places $x$ in the open disk. On $C$, $\tilde{\chi }(f)=\chi (f)$, so $\chi$ itself is evaluation at that interior point.

We therefore obtain two families: actual evaluations $f\mapsto f(x)$ on one closed disk, and analytic-extension evaluations $f\mapsto T(f)(\lambda )$ on a second closed disk. They agree on the boundary. They are distinct at interior points: functions in $I$ distinguish actual interior evaluations from every analytic-extension evaluation, and the coordinate function distinguishes points within each family. ThusThis is also a topological identification. Each disk family is continuous in the Gelfand topology, and they agree along their boundary, giving a continuous bijection from the glued disks to ${\Phi }_C$. The glued disks form a compact space, explicitly homeomorphic to the sphere by sending $(x,y)$ in the two copies to $(x,y,\pm \sqrt{1-x^2-y^2})$. Since the character space is Hausdorff, the continuous bijection is a homeomorphism. This boundary-analytic disk algebra with doubled character space has more algebra characters than evaluations on its original disk.

We prove the forward direction of the finite-generator criterion for a natural uniform algebra. Suppose that $A$ is a natural uniform algebra, and let $f_1,\dots ,f_n$ have no common zero on $K$. The idealis all of $A$. Otherwise it is contained in a maximal ideal $M$, whose associated character of an algebra vanishes on all the $f_j$. Since every algebra character is an evaluation character under naturality, this would give a common zero, a contradiction. Therefore $1\in J$, meaning that

Here are the Banach-algebra details behind the algebra character associated with $M$. A proper ideal cannot have dense closure: if it contained an element within distance less than one of $1$, the Neumann series would make that element invertible, forcing the ideal to contain $1$. Thus the closure of a proper ideal is proper. A maximal ideal, obtained from Zorn's lemma, is therefore closed. Its quotient is a complex unital Banach algebra in which every nonzero element is invertible. The Gelfand-Mazur theorem identifies this quotient with $\mathbb{C}$, so the quotient map is the required algebra character. No closedness of the originally generated ideal $J$ is needed.

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:

A bounded linear operator is compact if it maps the unit ball to a set with compact closure; equivalently, every bounded sequence has a subsequence whose images converge.

If $S,T$ are compact operators, select a subsequence on which $S{x}_n$ converges and then a further subsequence on which $T{x}_n$ converges. This proves compactness of any linear combination. For norm closure, suppose compact $T_j$ converge to $T$ in operator norm. Given $\epsilon >0$, choose $j$ with $\Vert T-T_j\Vert <\epsilon /3$. A finite $\epsilon /3$-net for $T_j$ of the unit ball is then a finite $2\epsilon /3$-net for its image under $T$. This image is totally bounded, and its closure is compact because $X$ is complete. Thus $B_0(X)$ is a closed linear subspace.

For bounded $S$ and compact $T$, the image of a bounded sequence under $S$ is bounded, so $TS$ is compact; and a convergent subsequence of $T{x}_n$ remains convergent after applying the continuous map $S$, so $ST$ is compact. Therefore the compact operators form a closed two-sided ideal in $B(X)$.

For the weighted backward shift, truncate its output after coordinate $N$, obtaining a finite-rank operator $T_N$. The exact norm estimate isHence $T$ is compact. Its iterates areThus for every nonzero complex $\lambda$ the Neumann seriesconverges absolutely in operator norm. Zero is an eigenvalue, since $T{e}_1=0$, whereas every nonzero value is in the resolvent set. Consequently

For an explicit Runge-Kutta method, write its strictly lower-triangular stage matrix as $A$, weight vector as $b$, and $1=(1,\dots ,1{)}^T$. Applied to the scalar test equation, its stage vector satisfiesSince $A^s=0$, inversion is a finite Neumann series. Putting $z=\lambda h$ givesThis proves degree at most $s$. The printed statement of degree exactly $s$ needs a nondegeneracy assumption: a two-stage explicit scheme with $b=(1,0)$ is simply Euler's method with an unused second stage, and has polynomial $1+z$ of degree one.

If $|z|>\Vert L\Vert$, the Neumann series givesThe series converges in operator norm because $\Vert L/z\Vert <1$. HenceThe same argument applied to small perturbations of an invertible operator shows that the resolvent set is open and the spectrum is closed.

Suppose $L=L^{\ast }$ and $\mathrm{Im}z\ne 0$. The quadratic form $⟨Lh,h⟩$ is real, so the Cauchy-Schwarz inequality impliesThus $L-zI$ is bounded below by $|\mathrm{Im}z|$: it is injective and has closed range. Its adjoint operator $L-\bar{z}I$ has the same lower bound and trivial kernel. Sincethe range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\mathrm{Im}z{|}^{-1}$. For a bounded self-adjoint operator, $\Sigma (L)\subset \mathbb{R}$. The closed-range and adjoint steps are essential: merely excluding nonreal eigenvalues would not exclude all nonreal spectral points.