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The Born series is the Neumann series expansion of the Lippmann-Schwinger equation. Its terms describe successive orders of wave scattering; convergence is ensured, for example, by a scattering operator norm smaller than one.

For a closed unital subalgebra $A\subseteq B$ with common identity, its group of invertible elements of a Banach algebra is relatively open in $A\cap G(B)$ by the Neumann series. It is relatively closed because inversion in $B$ is continuous and $A$ is closed. Consequently each connected component of $A\cap G(B)$ either consists entirely of elements invertible in $A$, or entirely of elements whose inverses do not belong to $A$.

Inversion is continuous on the group of invertible elements of a Banach algebra. The Neumann series gives a local bound on inverse norms, while $b^{-1}-a^{-1}=b^{-1}(a-b)a^{-1}$ makes the difference tend to zero as $b\to a$.

Conjugate the collocation matrix by the diagonal matrix with entries $(-1{)}^i$. The resulting negative-off-diagonal matrix has a nonnegative inverse, as a convergent Neumann series shows. The absolute inverse row sums are therefore the entries of its inverse applied to the all-ones vector. They solve $4u_i-u_{i-1}-u_{i+1}=6$ with $u_0=u_{n+1}=0$, whose solution is displayed. The maximum occurs at the central row or central two rows and equals the inverse operator norm on ${\ell }^{\infty }$. Consequently the associated spline interpolation operator has norm at most three independently of dimension.

If $\Vert \tau f(t)\Vert \le C{\int }_0^t\Vert f(s)\Vert ds$, repeated integration produces the time-ordered simplex of volume $t^n/n!$. Thus the Neumann series ${\sum }_{n\ge 0}{\tau }^n$ converges on every finite interval, without requiring the single-step operator norm to be less than one.

A uniform algebra is natural exactly when functions with no common zero generate the unit ideal. If a generated ideal is proper, it is contained in a maximal ideal. That maximal ideal is closed: the closure of a proper ideal stays proper, since an element within distance less than one of the identity is invertible by the Neumann series. The Gelfand-Mazur theorem identifies the maximal-ideal quotient with the complex numbers, yielding a character of an algebra. Naturality would make that character a common-zero evaluation, a contradiction. Conversely, the asserted unit-ideal property makes the zero sets of every finite collection in a character's kernel intersect. Compactness gives a point common to all these zero sets; applying this to $f-\phi (f)1$ proves $\phi (f)=f(x)$. For $C(K)$, one can directly take $g_j=\overline{f_j}/{\sum }_k|f_k{|}^2$.

The invertible elements of a unital Banach algebra form an open multiplicative group. For $a\in G(A)$, every $b$ satisfying $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ is invertible, because $b=a(1+a^{-1}(b-a))$ and the second factor has a Neumann series inverse.

For $E=L_{x,v}^1$, let $(\tau f)(t)={\int }_0^t{U}_{t-s}(P-I)f(s)ds$. The factorial bound for a Volterra iterate is $\Vert {\tau }^{n}f(t){\Vert }_1\le (2t{)}^n{\sup }_{s\le t}\Vert f(s){\Vert }_1/n!$. Thus the Neumann series converges in $C([0,T];E)$ for every finite $T$ and solves $f=U_{\cdot }{f}_0+\tau f$. The same estimate applied to a difference proves uniqueness among locally time-bounded integral solutions. The whole collision term remains inside this undamped integral, rather than absorbing loss into the free propagator.

On the fixed finite-dimensional space of polynomials of degree at most $d$, uniform convergence of $B_n$ on each monomial and equivalence of norms in finite dimensions imply ${\epsilon }_n=\Vert B_n-I{\Vert }_{\mathrm{op}}\to 0$. A Neumann series then gives $\Vert B_n^{-1}-I{\Vert }_{\mathrm{op}}\le {\epsilon }_n/(1-{\epsilon }_n)$ for sufficiently large $n$. Consequently the displayed convergence holds. A strictly positive polynomial has positive minimum on the compact interval, so its inverse Bernstein approximants are also strictly positive for all sufficiently large $n$.

In a unital Banach algebra, if a subsequence of inverse norms remained bounded, $a=a_n[1+a_n^{-1}(a-a_n)]$ would be invertible for large subsequence indices by the Neumann series. This contradicts noninvertibility of $a$. If $A$ is a closed unital subalgebra of a larger Banach algebra $B$ and the $a_n$ lie in $G(A)$, then such a limit cannot be invertible in $B$ either: continuity of inversion in a Banach algebra would make their inverses converge in $B$ to $a^{-1}$, which closedness puts in $A$.

In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the algebra character uniquely since the quotient is one-dimensional.

In a commutative complex unital Banach algebra, a maximal ideal is closed: its closure remains proper, since an element within distance less than one of the identity is invertible by the Neumann series. The quotient is a complex division Banach algebra, hence is the complex field by the Gelfand-Mazur theorem. Its quotient map is a character of an algebra. Conversely a nonzero multiplicative complex linear functional has value one at the identity, is onto the complex field, and has maximal kernel. It is automatically continuous because its value at each element belongs to that element's spectrum. The kernel determines the character uniquely since the quotient is one-dimensional.

A maximal left ideal is a proper left ideal contained in no larger proper left ideal. In a unital Banach algebra it is norm closed: its closure is a left ideal, and if that closure were the entire algebra it would contain elements of $L$ within distance less than one of $1$, hence invertible by the Neumann series. A proper left ideal contains no invertible element. The quotient $A/L$ is consequently a quotient Banach space and a simple module.

Sampling order-three Cardinal B-splines $N_j$ at $x_i=i+3/2$ gives $N_i(x_i)=3/4$, $N_{i-1}(x_i)=N_{i+1}(x_i)=1/8$, and all other entries zero. The B-spline collocation matrix therefore has row strict diagonal dominance margin at least $1/2$, giving inverse operator norm at most two. More precisely, with $q=3-2\sqrt{2}$, its absolute inverse row sums are $v_i=2[1-(q^i+q^{n+1-i})/(1+q^{n+1})]$. To see this, change signs by $D_{ii}=(-1{)}^i$: $DAD$ has negative off-diagonal entries and a nonnegative inverse given by a convergent Neumann series. Its inverse row sums solve $6v_i-v_{i-1}-v_{i+1}=8$, $v_0=v_{n+1}=0$. Thus $\Vert A^{-1}{\Vert }_{\infty }=v_{\lfloor (n+1)/2\rfloor }<2$, tending to two as the dimension grows.

Suppose $a_n\to a$ in a unital Banach algebra, with each $a_n$ invertible. If $ba=1$, then $b{a}_n\to 1$, so $b{a}_n$ is invertible eventually by the Neumann series. Then $b=(b{a}_n)a_n^{-1}$ is invertible, and $a=b^{-1}$. The analogous argument with $a_{n}b\to 1$ treats a right inverse of an algebra element. Thus a noninvertible such limit admits neither type of one-sided inverse.

The complement of a compact planar set has a unique unbounded component: outside a large disc lies a connected exterior, and every unbounded component meets that exterior. The Neumann series for $|\lambda |>\Vert x\Vert$ converges in both algebras, so this exterior belongs to ${\rho }_B(x)$. Part i retains the whole component of ${\rho }_A(x)$ containing it. That retained component is unbounded and therefore is also the unique unbounded component of ${\rho }_B(x)$. The two unbounded components coincide.

Use the common nonzero identity $1$. If $\Vert h\Vert <1$, completeness and submultiplicativity give the Neumann seriesFor $a\in G(A)$ and $\Vert a^{-1}\Vert \Vert b-a\Vert <1$, write $b=a(1+a^{-1}(b-a))$ and apply that series. This proves that the group of invertible elements of a Banach algebra is open. Locally the same series bounds $\Vert b^{-1}\Vert$, andproves continuity of inversion. Inversion is its own inverse, hence it is a homeomorphism of $G(A)$ onto itself.

The spectrum of an element is ${\sigma }_A(x)=\{\lambda :\lambda 1-x\notin G(A)\}$, and its complement is the algebra resolvent set ${\rho }_A(x)$. Openness of $G(A)$ makes the algebra spectrum closed. For $|\lambda |>\Vert x\Vert$,so it is bounded and thus compact. For nonemptiness, suppose $R$ existed everywhere. The local Neumann expansion around each point makes $R$ a Banach-space-valued entire function. For every bounded linear functional $\ell$ on $A$, $\ell (R(\lambda ))$ is entire, bounded on compact discs, and tends to zero at infinity by the displayed series. The Liouville theorem makes it identically zero. The Hahn-Banach theorem separates points, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the algebra spectrum is nonempty and compact.

If $B$ is a closed unital subalgebra with the same identity, then ${\rho }_B(x)\subseteq {\rho }_A(x)$. It is relatively open by the Neumann series in $B$. If ${\lambda }_j\in {\rho }_B(x)$ tends to $\lambda \in {\rho }_A(x)$, continuity of inversion in $A$ gives $({\lambda }_{j}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Closedness puts this last inverse in $B$, so $\lambda \in {\rho }_B(x)$. Therefore ${\rho }_B(x)$ is relatively open and closed in ${\rho }_A(x)$. This proves the required resolvent components in a closed unital subalgebra directly, with no spectral-boundary assumption.

For a self-adjoint C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $x$ is a normal C-star element, thenAll powers are normal C-star elements, so induction gives $\Vert x^{2^j}\Vert =\Vert x{\Vert }^{2^j}$. The general spectral radius formula now gives $r(x)=\Vert x\Vert$.

One can also see the needed growth estimate directly. For $R>r(x)$ the resolvent Cauchy coefficient formula givesThe formula follows by the Neumann series on a larger circle and contour deformation through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to \infty$ to get $\Vert x\Vert \le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the norm bound on the algebra spectrum. ThusThis proves spectral radius norm equality for normal elements from the norm identity.

Work in a nonzero complex unital Banach algebra $B$, with identity $1$. Write $R(x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is invertible}\}$ for the resolvent set, $\sigma (x)=\mathbb{C}\setminus R(x)$ for the spectrum of an element, and $Q(\lambda )=(\lambda 1-x{)}^{-1}$ for its Banach algebra resolvent. Distinguishing the set from the inverse-valued function avoids a notational ambiguity.

The printed formula omits an essential modulus. Complex spectral values cannot be ordered to take the indicated supremum. Even for $x=-1$ in $B=\mathbb{C}$, the spectrum is $\{-1\}$ while the spectral radius is one. The correctly interpreted assertion, proved below, isAll required analytic facts about the Banach algebra resolvent will be derived from norm-convergent series and scalar complex analysis.

For $\Vert a\Vert <1$, the Neumann series ${\sum }_{n\ge 0}{a}^n$ converges in $B$ and is a two-sided inverse of $1-a$: multiplying its finite sums gives $1-a^{N+1}$ and passing to the limit gives $1$. Therefore for $|\lambda |>\Vert x\Vert$,The series converges locally uniformly on that exterior region and tends to zero in norm as $|\lambda |\to \infty$. This statement does not require the optional normalization $\Vert 1\Vert =1$; the zeroth term is $1/\lambda$, and the other terms have the usual geometric bound.

At any ${\lambda }_0\in R(x)$, factor ${\lambda }_{0}1-x+h1=({\lambda }_{0}1-x)(1+hQ({\lambda }_0))$. If $|h|\Vert Q({\lambda }_0)\Vert <1$, the Neumann series givesThus the resolvent set is open, $Q$ is norm-continuous there, and the series proves norm differentiability with $Q^{\prime }(\lambda )=-Q(\lambda {)}^2$. Repeated differentiation gives $Q^{(j)}(\lambda )=(-1{)}^{j}j!Q(\lambda {)}^{j+1}$. Direct subtraction of the inverse equations also yields the resolvent identityIn particular, every continuous linear functional $\varphi$ makes $\varphi (Q(\lambda ))$ a scalar holomorphic function; this has been proved, not assumed as an algebra-valued analytic theorem.

The spectrum of an element is closed, and is contained in $\{\lambda :|\lambda |\le \Vert x\Vert \}$ by the exterior series, so it is compact. It is nonempty. Otherwise $Q$ would be defined and norm-continuous on the entire plane, bounded on every fixed closed disk and tending to zero at infinity. Each $\varphi (Q)$ would be a bounded entire scalar function, hence identically zero by Liouville's theorem. Continuous complex linear functionals separate points of $B$, so $Q(\lambda )=0$, impossible for an inverse. This separation follows from the real Hahn-Banach theorem: a continuous real functional $h$ not vanishing on a given vector gives the complex linear functional $\varphi (v)=h(v)-ih(iv)$, which also does not vanish there. We have therefore proved thatis well-defined and finite.

Next establish the existence of the power-norm limit. If $x^m=0$ for some $m$, the limit is zero. Otherwise fix $m$ and write $n=qm+r$, $0\le r<m$. Submultiplicativity givesThe finitely many remainder factors have bounded size, and $q/n\to 1/m$. Hence ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$ for every $m$. The opposite lower bound by the infimum is automatic, soFor $|\lambda |>r_0$, the root test proves convergence of $\sum x^n/{\lambda }^{n+1}$ even if $|\lambda |\le \Vert x\Vert$. Its finite sums telescope against $\lambda 1-x$, and their remainder tends to zero, so this sum is an inverse. Thus $s\le r_0$.

For the reverse inequality, fix $R>s$ and first choose $T>\max (R,\Vert x\Vert )$. The exterior Neumann series converges uniformly on $|\lambda |=T$, so termwise vector integration givesOnly its $x^n{\lambda }^{-n-1}$ term contributes. The annulus between radii $R$ and $T$ is in the resolvent set. For every continuous complex linear $\varphi$, the scalar Cauchy integral theorem applied to ${\lambda }^n\varphi (Q(\lambda ))$ makes the two contour integrals equal. Continuous linear functionals commute with vector integration and separate points, so the vector integrals themselves agree. This proves the resolvent Cauchy coefficient formula without presupposing any Banach-algebra-valued analytic-function theorem:Let $M_R={\max }_{|\lambda |=R}\Vert Q(\lambda )\Vert$, finite by continuity. The contour-length estimate gives $\Vert x^n\Vert \le M_R{R}^{n+1}$. Taking $n$th roots gives $r_0\le R$, and letting $R\downarrow s$ gives $r_0\le s$, including $s=0$. Combined with the earlier inequality, this proves the spectral radius formula and the corrected printed identity.