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An algebra norm is a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra in which the original algebra embeds densely.

For a character of an algebra $\phi$ on a complex unital Banach algebra, $\phi (1)=1$ and $\phi (a)\in \sigma (a)$, so $|\phi (a)|\le \Vert a\Vert$ without any prior continuity assumption. Under the usual normalization $\Vert 1\Vert =1$, this gives $\Vert \phi \Vert =1$; for an unnormalized algebra norm, only the upper bound is automatic.

The exponential series converges absolutely in a unital Banach algebra. Multiplication of absolutely convergent series gives $e^x{e}^{-x}=1$, and gives $e^{x+y}=e^x{e}^y$ when $x,y$ commute. Termwise differentiation gives $e^{tx}=1+tx+O(t^2)$ near zero. In a C-star algebra, if $h=h^{\ast }$ and $t$ is real, taking adjoints termwise gives $(e^{ith}{)}^{\ast }=e^{-ith}$, so this exponential is a Unitary element of a C-star algebra and has norm one. These facts let a norm bound on a functional recover its real values on Hermitian elements.

A norm-closed subalgebra of a Banach algebra is itself a Banach algebra with the inherited norm: completeness follows from closedness, and the submultiplicative norm bound restricts to it. A unital inclusion uses the same identity as the ambient algebra. Inverse membership may still differ between the two algebras; spectrum in a closed unital subalgebra describes that difference.

The displayed formula is a submultiplicative norm on the algebra of entire functions: definiteness follows from the identity theorem. No submultiplicative norm on the whole algebra can be complete. The coordinate function $Z(z)=z$ has every complex number in its algebraic spectrum of an element, because $Z-\lambda$ vanishes somewhere and has no entire inverse. This contradicts compactness of spectra in a Banach algebra.

For a unital commutative complex Banach algebra $A$, the Gelfand transform sends $a\in A$ to the continuous function $\hat{a}(\phi )=\phi (a)$ on its character space of an algebra. It is a contractive unital algebra homomorphism into $C({\Phi }_A)$.

For a commutative complex unital Banach algebra, its Gelfand transform is a continuous unital algebra homomorphism into the continuous functions on its compact character space. The values of the transform of $a$ are exactly ${\sigma }_A(a)$, so $\Vert \hat{a}{\Vert }_{\infty }=r(a)$. Its kernel is the intersection of the kernels of the algebra characters. For a commutative C-star algebra, spectral radius norm equality for normal elements and the Stone-Weierstrass theorem upgrade this to an isometric C-star homomorphism onto $C(\Delta (A))$.

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.

A Hermitian Banach algebra is a Banach algebra with an algebra involution for which every Hermitian element of a star algebra has real spectrum of an element. A closed unital star-subalgebra has spectral permanence for Hermitian Banach algebras. The assumption is weaker than being a C-star algebra and does not require the C-star norm identity.

A Hermitian algebra element of a star-algebra is an element fixed by its algebra involution. The identity and zero are Hermitian algebra elements. For every $x$, both $x{x}^{\ast }$ and $x^{\ast }x$ are Hermitian algebra elements. A real spectrum for all such elements is an extra condition in a general Banach algebra, rather than a consequence of the involution identities alone.

Let $x_1,x_2$ belong to a commutative unital Banach algebra, have the same Gelfand transform, and have common spectrum $K$. If $f$ is holomorphic near $K$ and $f^{\prime }$ has no zero on $K$, then $f(x_1)=f(x_2)$ implies $x_1=x_2$. Apply the two-variable holomorphic functional calculus to the divided difference of $f$; its Gelfand transform never vanishes, so it is invertible.

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 $A$.

The complex dual numbers $\mathbb{C}[\epsilon ]/({\epsilon }^2)$, with norm $\Vert a+b\epsilon \Vert =|a|+|b|$ and involution $a+b\epsilon \mapsto \bar{a}+\bar{b}\epsilon$, is a commutative unital Banach algebra with an isometric involution. Its sole character of an algebra kills $\epsilon$, so the Gelfand transform is neither injective nor isometric. The C-star identity excludes this example and gives the Commutative Gelfand--Naimark theorem. Even preservation of the involution on characters can fail without that identity: on ${\mathbb{C}}^2$, the involution $(a,b{)}^{\ast }=(\bar{b},\bar{a})$ exchanges the two evaluation characters.

Every algebraically irreducible representation of a Banach algebra that is normed is continuous. The continuous orbit vectors form an invariant subspace. If all nonzero orbit maps were discontinuous, apply the countable annihilation sequence in an irreducible Banach module and the gliding-hump continuity principle to right multiplication on the algebra and evaluation of the operator-valued representation. Off-diagonal maps vanish, whereas diagonal maps are discontinuous, a contradiction. Once every orbit map is continuous, the Uniform boundedness principle on the Banach algebra gives the common operator-norm bound. Completing the original normed representation space supplies the Banach operator space needed in the argument.

With left Haar measure, $(f\ast g)(x)={\int }_{G}f(y)g(y^{-1}x)dm(y)$ makes $L^1(G)$ a Banach algebra, since $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$. For an Abelian group it is commutative, and the involution is $f^{\ast }(x)=\overline{f(x^{-1})}$. Its nonzero characters of an algebra are exactly integration against inverses of continuous unitary group characters.

// Target: geometry-and-topology.bigb

A left inverse of an element $a$ of a unital associative algebra is an element $b$ with $ba=1$. A right inverse instead satisfies $ab=1$. If both exist, they agree: $b=b(ac)=(ba)c=c$. A one-sided inverse alone need not imply invertibility in an infinite-dimensional Banach algebra.

The displayed series converges absolutely in a Banach algebra. It is the holomorphic functional calculus value of the analytic branch $z\mapsto \log (1-z)$ on the unit disk. The composition rule for holomorphic functional calculus and $\exp (\log (1-z))=1-z$ give $\exp (\log (1-x))=1-x$. Thus every sufficiently small norm perturbation of the identity is an exponential.

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.