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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 $b\to a$.

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.

In a unital Banach algebra, the Neumann series gives $(1-u{)}^{-1}={\sum }_{k=0}^{\infty }{u}^k$ whenever $\Vert u\Vert <1$: the series converges in norm by submultiplicativity and completeness, and multiplication of its partial sums on either side gives $1-u^{N+1}\to 1$. Consequently, if $g$ is invertible and $\Vert g^{-1}\Vert \Vert h-g\Vert <1$, writeThe second factor has a Neumann series inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the group of invertible elements of a Banach algebra, proving that group is an open set.

The same series proves continuity of inversion in a Banach algebra, with the useful quantitative estimateThe right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a bijection whose inverse is the same continuous map. Thus inversion is a homeomorphism of the open group of invertibles onto itself.

For the final unheaded request, defineIt is open by the preceding argument and contains $0$. If ${\lambda }_0$ were a boundary point, a sequence ${\lambda }_n\in E$ would converge to ${\lambda }_0$, while openness gives ${\lambda }_0\notin E$. Then $1-{\lambda }_{n}a\to 1-{\lambda }_{0}a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a left inverse of an algebra element nor a right inverse of an algebra element, contradicting the hypothesis. Therefore $E$ has empty boundary; it is both open and closed in the connected complex plane, and $E=\mathbb{C}$.

For every nonzero $\mu$, put $\lambda ={\mu }^{-1}$. The identity $\mu 1-a=\mu (1-\lambda a)$ proves invertibility, so the spectrum of an element $a$ is contained in $\{0\}$. It is nonempty: if the resolvent of an element existed for all $\mu \in \mathbb{C}$, it would be an entire algebra-valued function tending to zero at infinity by the Neumann series. Applying any bounded linear functional and then the Liouville theorem would make it zero; the Hahn-Banach theorem separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence

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 $A$.

Invertibility is stable under sufficiently small perturbations, and inversion is continuous. Let $1$ be the identity of the complex unital Banach algebra $A$, with its submultiplicative algebra norm. For $\Vert h\Vert <1$, completeness makes the Neumann seriesconverge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.

Fix $a\in G(A)$, the group of invertible elements of a Banach algebra. If $\Vert b-a\Vert <\Vert a^{-1}{\Vert }^{-1}$, thenso the Neumann series makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local boundWe retain $\Vert 1\Vert$ here so the estimate does not silently assume a normalized identity. The inverse identitythen impliesThis proves continuity of inversion in a Banach algebra.

For a unital Banach algebra, the spectrum of an element isFor a nonunital Banach algebra, use its unitization of an algebra $A^{+}=A\oplus \mathbb{C}$ with multiplication and normIt is a unital Banach algebra, with identity $(0,1)$, and define ${\sigma }_A(x)={\sigma }_{A^{+}}((x,0))$. In this nonunital convention, $0$ belongs to the spectrum: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^{+}$.

The spectrum is nonempty and compact. It suffices to work in a nonzero unital Banach algebra, since unitization reduces the other case to this one. The group of invertible elements of a Banach algebra is open, so the complement of the spectrum of an element is open. For $|\lambda |>\Vert x\Vert$, the Neumann series givesThus ${\sigma }_A(x)$ is closed and lies in $\{\lambda :|\lambda |\le \Vert x\Vert \}$, hence is compact.

For nonemptiness of the Banach-algebra spectrum, suppose that $R$ were defined on all of $\mathbb{C}$. At any ${\lambda }_0$, factoring $\lambda 1-x=({\lambda }_{0}1-x)(1+(\lambda -{\lambda }_0)R({\lambda }_0))$ gives a locally convergent power seriesFor every bounded linear functional $\ell \in A^{\ast }$, the scalar function $\ell (R(\lambda ))$ is therefore entire. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the Liouville theorem, it is constant; since it tends to zero at infinity, it is identically zero. The version of the Hahn-Banach theorem used here says that bounded linear functionals separate points of a normed vector space: if $z\ne 0$, there is $\ell$ with $\ell (z)\ne 0$. Consequently $R(\lambda )=0$ for every $\lambda$, contradicting $(\lambda 1-x)R(\lambda )=1$. This proves the assertion.

Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb{C}$. This is the Gelfand-Mazur theorem; completeness is not needed in its statement. Let $D$ be a normed division algebra, and take its completion of a normed space using the algebra norm to obtain a unital Banach algebra $\hat{D}$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.

For $x\in D$, the spectrum of an element of $x$ in $\hat{D}$ contains some $\lambda$. If $x-\lambda 1\ne 0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\hat{D}$. This contradicts $\lambda \in {\sigma }_{\hat{D}}(x)$. Therefore $x=\lambda 1$. The map $\lambda \mapsto \lambda 1$ is a bijective complex algebra homomorphism, andIt and its inverse are continuous; when $\Vert 1\Vert =1$, it is an isometry.

A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known. Let $A$ be the given algebra of functions. Form the same artificial unitization $A^{+}$ even if $A$ already has an identity. For each $t\in K$, the algebraic mapis a unital multiplicative complex linear functional. No continuity has been assumed. Since ${\chi }_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying ${\chi }_t$ to an inverse equation would give $0=1$. Hence $f(t)\in {\sigma }_{A^{+}}(f)$, and the spectrum bound already proved yieldsTaking the supremum proves the supremum bound for a complete function-algebra norm:In particular every $f$ is bounded, and every point evaluation is a bounded linear functional of norm at most $1$. This is an instance of automatic continuity of characters: a character of an algebra on a Banach algebra is bounded because its value at any element belongs to that element's spectrum.