Banach algebra

A Banach algebra is an associative algebra with a complete submultiplicative norm, $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$.

A Banach algebra is semisimple when its Jacobson radical is zero. This does not assert that every quasinilpotent element is zero. For example, the algebra of bounded operators on a Banach space of dimension greater than one is semisimple, but contains nonzero square-zero rank-one operators. This notion also does not require the finite-dimensional direct-sum decomposition associated with a semisimple algebra.

An algebra homomorphism onto a semisimple Banach algebra is continuous. To see the spectral mechanism, suppose $a_n\to 0$ and $T(a_n)\to b$. For any $c\in B$, choose $u,v\in A$ with $T(u)=b$ and $T(v)=c$, and set $d=cb$, $d_n=T(v{a}_n)$. Apply the spectral-radius three-circle inequality to $T((1-z)v{a}_n+zvu)$. On the outer circle its norm tends uniformly to $\Vert d\Vert$; on the circle of radius $1/R$ its spectral radius is bounded by $(1+1/R)\Vert v{a}_n\Vert +\Vert vu\Vert /R$. Hence $r(d{)}^2\le \Vert d\Vert \Vert vu\Vert /R$ for every $R>1$, so $r(cb)=0$. The unit criterion for the Jacobson radical gives $b\in J(B)=0$, and the closed graph theorem finishes the proof.

Truncated convolution makes $L^1(0,1)$ a commutative nonunital Banach algebra. The Tonelli theorem gives $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$, and the Fubini theorem gives associativity over the integration simplex. Its unitization is $\mathbb{C}\oplus L^1(0,1)$ with norm $|\alpha |+\Vert f{\Vert }_1$ and product $(\alpha ,f)(\beta ,g)=(\alpha \beta ,\alpha g+\beta f+f\ast g)$. The constant function one has convolution powers $t^{n-1}/(n-1)!$, of norm $1/n!$. Thus it is a nonnilpotent quasinilpotent element in this unitization.

If an associative unital algebra has a complete norm and separately continuous multiplication, the Uniform boundedness principle applied to $L_x(y)=xy$, for $\Vert x\Vert \le 1$, yields $\Vert xy\Vert \le C\Vert x\Vert \Vert y\Vert$. The operator norm of $L_x$ then defines an equivalent algebra norm: $\Vert x\Vert /\Vert e\Vert \le \Vert L_x\Vert \le C\Vert x\Vert$, and $L_{xy}=L_x{L}_y$ proves submultiplicativity. It normalizes the identity to norm one.

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.

Let $B$ be a closed star-subalgebra of a unital Hermitian Banach algebra $A$, containing the same identity. For Hermitian $h\in B$, the complement of ${\sigma }_A(h)\subseteq \mathbb{R}$ is connected. The resolvent belongs to $B$ at large modulus by the Neumann series; its membership in $B$ is both open and closed in that complement, so ${\sigma }_B(h)={\sigma }_A(h)$. If $b$ is invertible in $A$, both $b{b}^{\ast }$ and $b^{\ast }b$ are invertible there and Hermitian algebra elements, hence invertible in $B$. Their inverses supply respectively a right and a left inverse for $b$ in $B$, proving invertibility there. Applying this to $b-\lambda 1$ proves the spectrum equality.

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

For a locally compact Hausdorff Abelian group, each continuous unitary character defines an algebra character by $f\mapsto \int f(x)\chi (x)dx$. Conversely, if $\Phi$ is a nonzero algebra character and $\Phi (g)\ne 0$, the ratio $\Phi (T_{x}g)/\Phi (g)$ is independent of $g$, continuous and multiplicative. Translation isometries bound all its integer powers, so its modulus is one. The convolution formula $f\ast g=\int f(x)T_{x}gdx$ recovers $\Phi (f)=\int f\chi$. The Gelfand topology corresponds to the compact-open topology on the Pontryagin dual group; conjugating the characters gives the alternative Fourier sign convention.

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.

With complex scalars and pointwise multiplication, the product rule gives a submultiplicative norm. For a Cauchy sequence, uniform limits $f$ of the functions and $g$ of their derivatives satisfy $f(t)-f(0)={\int }_0^{t}g(s)ds$, proving $f^{\prime }=g$ and completeness. Polynomials are dense in this norm: approximate $f^{\prime }$ uniformly by polynomials and integrate, fixing the initial value. This also identifies the character space of C1 on a compact interval.

The coordinate function has spectrum $[0,1]$: outside the interval its difference from a scalar has a continuously differentiable reciprocal, and on the interval it has a zero. Every character of an algebra therefore takes the coordinate to some $t\in [0,1]$. Polynomial density in the norm containing both function and derivative, together with automatic continuity of characters, forces evaluation at $t$ on every function. The maximal ideals are exactly $\{f:f(t)=0\}$.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra of $C(K)$, with the supremum norm, containing the constants and separating points. Evaluation embeds $K$ continuously and injectively in the character space of an algebra. The embedding is a homeomorphism onto a closed subset because its domain is compact and its target is Hausdorff. Extra characters need not be point evaluations on the specified $K$.

Let $C$ consist of continuous functions on the closed unit disk whose boundary values extend to a member of the disk algebra. The extension is unique by the maximum modulus principle; the extension map $T:C\to A(\overline{\mathbb{D}})$ is a surjective unital algebra homomorphism of norm at most one. Its kernel $I$ is the ideal of functions vanishing on the boundary. A character vanishing on $I$ factors through $T$, so it is $f\mapsto T(f)(\lambda )$ for some closed-disk point $\lambda$. For a character $\chi$ not vanishing on $I$, choose $h\in I$ with $\chi (h)\ne 0$. The rule $F\mapsto \chi (hF)/\chi (h)$ extends $\chi$ to a character of all continuous functions on the disk: multiplicativity follows from $\chi (hF)\chi (hG)=\chi (h)\chi (hFG)$. It is therefore evaluation at an interior point, since its value at $h$ is nonzero. These two closed-disk families agree exactly on their boundaries. Their glued compact space maps continuously and bijectively to the Hausdorff character space of an algebra, giving the stated homeomorphism. The algebra is closed because uniform convergence of boundary data gives uniform convergence of their analytic extensions.

The disk algebra consists of functions continuous on the closed unit disk and holomorphic in its interior, with the supremum norm. Polynomials are dense: dilations $f_r(z)=f(rz)$ converge uniformly to $f$ as $r\uparrow 1$, and each $f_r$ is uniformly approximated on the closed disk by its Taylor polynomials. A character of an algebra is therefore determined by its value $\lambda$ at the coordinate function; boundedness gives $|\lambda |\le 1$, and polynomial approximation yields $\phi (f)=f(\lambda )$. Consequently the disk algebra is a natural uniform algebra on the closed disk.

A uniform algebra on $K$ is natural if every character of an algebra is evaluation at a point of $K$. The evaluation embedding then identifies its character space homeomorphically with $K$. A natural uniform algebra is characterized by the finite-generator criterion for a natural uniform algebra. In particular, $C(K)$ is natural; the disk algebra is natural when its underlying space is the closed disk.

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 $g_j=\overline{f_j}/{\sum }_k|f_k{|}^2$.

A commutative Banach algebra is semisimple when its Jacobson radical is zero. For a complex unital Banach algebra this is equivalent to injectivity of the Gelfand transform. This condition concerns the Jacobson radical; it does not assert a finite product decomposition as in finite-dimensional semisimple algebra theory.

For an algebra of complex-valued functions with a complete algebra norm, every point value $f(t)$ belongs to the spectrum of an element of $f$ in the artificial unitization. Indeed $(h,\lambda )\mapsto h(t)+\lambda$ is an algebraic unital character of an algebra, and it vanishes on $f-f(t)1$. The spectrum bound gives $|f(t)|\le \Vert f\Vert$ without first assuming that evaluation is continuous.

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.

Finite products of Banach algebra exponentials form a subgroup: inverses are $e^{-a}$ and conjugation sends $e^a$ to $e^{ga{g}^{-1}}$. Each product is joined to the identity by $t\mapsto \prod e^{t{a}_i}$. The logarithm of an element near the identity puts a norm neighborhood of $1$ in this subgroup, so its cosets are open and its complement is open. It is therefore exactly the identity component of a topological group, and is an open, closed, normal subgroup. A product of exponentials need not be a single exponential in a noncommutative algebra.

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.

A right inverse of an element $a$ is an element $b$ with $ab=1$. A left inverse of an algebra element and a right inverse of an algebra element, when both exist, coincide. For the unilateral shift operator $S$, $S^{\ast }S=1$ but $S{S}^{\ast }\ne 1$, exhibiting the distinction in the Banach algebra of bounded operators.

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.

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 $A$.

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 $b\to a$.

A Qp-Banach algebra is a ${\mathbb{Q}}_p$-algebra with a complete non-Archimedean submultiplicative norm satisfying $\Vert \lambda a\Vert =|\lambda {|}_p\Vert a\Vert$.

Every complex unital Banach algebra in which every nonzero element is invertible is isometrically isomorphic to $\mathbb{C}$.

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.

Let $S$ be a bounded multiplicatively closed subset of a unital Banach algebra. The displayed auxiliary norm $p$ is equivalent to the original norm, and each left multiplication $L_s$ is a contraction for $s\in S$. Taking the operator norm of the faithful left regular representation produces an equivalent algebra norm with $\Vert 1{\Vert }_S=1$ and $\Vert s{\Vert }_S\le 1$. If $M=\max (\Vert 1\Vert ,{\sup }_S\Vert s\Vert )$, then $\Vert a\Vert /M\le \Vert a{\Vert }_S\le M\Vert a\Vert$. The auxiliary norm is itself submultiplicative, since $p(ab)\le p(a)\Vert b\Vert \le p(a)p(b)$, but $p(1)=M$ can exceed one. The operator norm enforces the required unital normalization.

For a finite commuting family in a unital Banach algebra, divide each $x_j$ by $r(x_j)+\epsilon /2$. The spectral radius formula makes each normalized element's powers bounded. Commutativity lets every product be reordered into one power of each generator, so the entire generated semigroup is bounded. Renorming a bounded multiplicative semigroup gives the displayed bounds simultaneously. Applying this to two commuting elements proves submultiplicativity and subadditivity of their spectral radius.

An algebra norm is submultiplicative when it satisfies the displayed product inequality. Induction gives $\Vert a_1\cdots a_n\Vert \le {\prod }_j\Vert a_j\Vert$, and in particular $\Vert a^n\Vert \le \Vert a{\Vert }^n$. The operator norm has this property because $\Vert STv\Vert \le \Vert S\Vert \Vert T\Vert \Vert v\Vert$. The supremum norm for bounded functions with pointwise multiplication is another example. This estimate supports the Neumann series and the spectral radius formula.

Take disjoint disks escaping to infinity and continuous cutoff functions $h_n$ supported in them and equal to one on smaller disks. The locally finite function $f={\sum }_{n}n{h}_n$ is continuous. Choose nonzero $g_n$ supported in the smaller disks; then $f{g}_n=n{g}_n$. Any algebra norm would give $n\Vert g_n\Vert =\Vert f{g}_n\Vert \le \Vert f\Vert \Vert g_n\Vert$, forcing $n\le \Vert f\Vert$ for every positive integer. This is impossible even without a completeness assumption.

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.

The displayed supremum norm is finite on every entire function, positive on every nonzero one by the identity theorem, and submultiplicative under pointwise multiplication. It is not complete: partial geometric sums ${\sum }_{j=0}^n(z/2{)}^j$ are Cauchy on the disk but their limit there extends meromorphically with a pole at two. More generally, no complete algebra norm exists on all entire functions because the coordinate function has algebraic spectrum of an element equal to the whole complex plane.

Any submultiplicative algebra norm on all of $C(K)$ dominates its supremum norm, even when the new norm is incomplete. Every evaluation character extends continuously to the completion: the restrictions of the completion's character space form a closed subset of $K$, and an omitted point would give a nonzero function annihilated by an invertible function in the completion.

The unitization adjoins an identity to a possibly nonunital algebra. For $C_0(X)$ it is naturally $C(X^{+})$ on the one-point compactification.

The resulting algebra is a unital algebra.

If $a$ belongs to a unital Banach algebra and $\Vert a\Vert <1$, thenAbsolute convergence and multiplication of partial sums prove the identity.

The spectrum of $a$ in a unital complex algebra $A$ isFor a nonunital algebra it is defined in the unitization of an algebra.

The algebra resolvent set consists of those scalars for which $\lambda 1-x$ is invertible in the algebra. It is open by the Neumann series, and the inverse-valued map is the resolvent of an element. A closed subalgebra can have a smaller algebra resolvent set because an ambient inverse need not lie in it.

For a closed unital subalgebra $B$ of a unital Banach algebra $A$ and $x\in B$, the smaller-algebra algebra resolvent set is relatively open and closed in the larger one. Openness follows from the Neumann series; closedness follows from continuity of inversion in $A$ and norm closedness of $B$. Every larger-algebra resolvent component is therefore either retained whole or absorbed into the smaller-algebra Banach algebra spectrum. The unbounded component is always retained, since inverses at large modulus are norm-convergent series in $x$.

For a closed subalgebra of a Banach algebra $B\subseteq A$ with common identity, the algebra resolvent set ${\rho }_B(x)$ is clopen in ${\rho }_A(x)$. A point in ${\rho }_A(x)$ has a small connected disc on which membership of ${\rho }_B(x)$ is constant, so it cannot lie in $\partial {\sigma }_B(x)$. Also ${\sigma }_A(x)\subseteq {\sigma }_B(x)$; a point in the interior of the former is in the interior of the latter. Together these observations prove the boundary inclusion, without assuming a spectral-boundary theorem.

The identity $(x+\alpha e)-\lambda e=x-(\lambda -\alpha )e$ proves the formula directly from invertibility. In a nonzero complex unital Banach algebra, the nonemptiness of the Banach-algebra spectrum implies that if $x$ is a quasinilpotent element, then $\rho (x+e)=1$. Consequently every nonscalar element cannot simultaneously be quasinilpotent: adding the identity preserves nonscalarity while translating its spectrum.

An element of a complex unital Banach algebra is quasinilpotent if its spectral radius is zero, equivalently its spectrum of an element is $\{0\}$. A nilpotent element is quasinilpotent, but the converse need not hold. The Volterra convolution algebra on a finite interval gives powers which never vanish, yet have factorially decreasing norms, making the inverse series for $\lambda e-x$ converge for every $\lambda \ne 0$.

In a unital Banach algebra, for $\lambda \ne 0$ the existence of $R=(\lambda 1-ab{)}^{-1}$ gives $(\lambda 1-ba{)}^{-1}={\lambda }^{-1}(1+bRa)$, as verified by multiplying on either side. Interchanging $a,b$ proves the converse. Zero may differ: for the unilateral shift operator $S$ on ${\ell }^2({\mathbb{N}}_0)$, $S^{\ast }S=1$ has spectrum $\{1\}$ and $S{S}^{\ast }=1-P_0$ has spectrum $\{0,1\}$.

Every element of a nonzero complex unital Banach algebra has a nonempty compact spectrum of an element. If the resolvent of an element existed everywhere, composing it with any bounded linear functional would give a bounded entire function tending to zero at infinity. The Liouville theorem and point separation by the Hahn-Banach theorem would force the resolvent to be zero, contradicting the inverse equation. Nonunital algebras are treated by unitization.

The approximate point spectrum consists of the scalars $\lambda$ for which there are unit vectors $x_k$ satisfying $\Vert (T-\lambda I)x_k\Vert \to 0$. Equivalently, $T-\lambda I$ is not bounded below.

This is a part of the spectrum of a bounded operator.

The bilateral shift $(Tx{)}_n=x_{n+1}$ is unitary. It has empty point spectrum, while its approximate point spectrum is the unit circle. For $|\lambda |=1$, normalized vectors with entries ${\lambda }^n$ on an interval of length $N$ fail to satisfy $Tx=\lambda x$ only at the two endpoints, giving residual norm $\sqrt{2/N}$.

It is the two-sided sequence case of a shift operator.

The resolvent is $R(\lambda ,a)=(\lambda 1-a{)}^{-1}$ on the complement of the spectrum of an element. In a Banach algebra it is analytic there.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For two points $z,w$ in the resolvent set of an operator $A$,In particular, the resolvent operator is differentiable and ${\partial }_{z}R(z,A)=-R(z,A{)}^2$.

For a closed operator, the resolvent is operator-valued holomorphic on its resolvent set. On a Hilbert space its norm is subharmonic and therefore cannot have a strict interior maximum.

The $ϵ$-pseudospectrum consists of spectral points together with resolvent points where $\Vert (A-zI{)}^{-1}\Vert >{ϵ}^{-1}$. Equivalently it is described by a small lower norm.

For a spectral subset isolated by a contour $\Gamma$ in the resolvent set,projects onto its spectral subspace. Norm-small perturbations preserve the rank of this projection.

For a bounded operator on a complex Banach space, partition a disconnected spectrum into two nonempty compact pieces. The holomorphic function equal to one near the first piece and zero near the second defines a Riesz projection. The holomorphic spectral mapping theorem makes it neither zero nor identity. Its closed range is a proper nonzero invariant vector subspace. This need not be an orthogonal projection, and the argument does not require a normal operator.

If a closed unital subalgebra $A\subseteq B$ contains $x$, then ${\sigma }_A(x)$ is obtained from ${\sigma }_B(x)$ by adjoining some bounded connected components of its complement. Membership of $(x-\lambda 1{)}^{-1}$ in $A$ is constant on every connected component of the resolvent set, and it always holds on the unbounded component.

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 $A$.

For $x$ in a unital Banach algebra and $f$ holomorphic near $\sigma (x)$, the holomorphic functional calculus defineswhere $\Gamma$ winds once around the spectrum. It is a continuous unital algebra homomorphism and satisfies the spectral mapping theorem $\sigma (f(x))=f(\sigma (x))$.

If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

If a spectrum of an element splits into disjoint nonempty compact pieces $X,Y$, choose disjoint open neighborhoods and a holomorphic function equal to one on the first and zero on the second. The holomorphic functional calculus sends it to an idempotent with the corresponding zero/one values under every algebra character. It is neither zero nor identity: if one piece were killed, a reciprocal function on the other piece would make an originally spectral coordinate difference invertible.

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.

If the spectrum of an element $x$ avoids $(-\infty ,0]$, its holomorphic functional calculus value for the principal cube root is the unique $y$ with $y^3=x$ and spectrum in $\{z\ne 0:|\arg z|<\pi /3\}$. For uniqueness, any other such $v$ defines a continuous unital algebra homomorphism $f\mapsto {\Theta }_v(f\circ (z\mapsto z^3))$ from the slit-plane holomorphic functions to the algebra. It sends the coordinate to $x$, so continuity and uniqueness of holomorphic functional calculus identify it with ${\Theta }_x$. Applying it to the principal cube root gives $v=y$. This argument does not assume commutativity of the ambient algebra or separation by algebra characters.

If the spectrum of an element $x$ lies in the open right half-plane, holomorphic functional calculus gives the displayed square root with spectrum in that half-plane. It is unique there. Two such roots $y,v$ have character values equal to the unique scalar square root with positive real part. Every character of an algebra is therefore nonzero on $y+v$, so the character description of the spectrum makes $y+v$ invertible. From $(y-v)(y+v)=0$ one obtains $y=v$. Equal character values alone do not prove equality in a possibly nonsemisimple algebra.

The closed unital algebra generated by a bounded operator and all its resolvents is commutative, because the generators commute. It has exactly the original operator spectrum at that operator: all required inverses outside that spectrum were included, and an inverse in the smaller algebra is also an operator inverse. This permits commutative-algebra linear functional calculus without replacing the spectrum by the possibly larger spectrum of a polynomial-generated algebra.

Equip holomorphic functions on an open spectral neighborhood with the compact-open topology. A fixed admissible contour gives a bound on $\Vert {\Theta }_x(f)\Vert$ by a constant times the supremum of $|f|$ on the contour, proving continuity. The Runge theorem makes rational functions with poles outside the open set dense. A unital homomorphism sending the coordinate function to $x$ must send each inverse coordinate difference to the corresponding resolvent, so its rational values and then all its holomorphic values are forced.

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.

Choose a cycle $\Gamma$ surrounding $\sigma (x)$ and a cycle $\Delta$ surrounding $f(\sigma (x))$, with $f(\Gamma )$ inside the region of winding number one for $\Delta$. The algebra-homomorphism property givesSubstitute this into the contour formula for $g(f(x))$, interchange integrals, and apply the Cauchy integral formula to the inner $\zeta$ integral. This proves composition as an equality of algebra elements, not merely an equality on characters, which need not separate a nonsemisimple algebra.

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.

The full spectrum of an element is its spectrum together with every bounded component of its complement. Its complement is the unbounded resolvent component, and the maximum modulus principle controls a holomorphic function on each filled hole by its values on the spectral boundary.

A character on a complex algebra is a nonzero multiplicative linear functional $\phi :A\to \mathbb{C}$.

Algebra characters form the domain of the Gelfand representation; they are distinct from traces of group representations.

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.

For a self-adjoint element $h$, the exponentials $\exp (ith)$ are unitary. A continuous algebra character therefore satisfies $|\exp (it\varphi (h))|\le 1$ for every real $t$, forcing $\varphi (h)$ to be real. Decomposing an arbitrary element into real and imaginary self-adjoint parts proves the displayed identity. Thus every algebra character is a star-preserving scalar homomorphism; commutativity of the whole algebra is not needed for this fact.

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 character space ${\Phi }_A$ is the set of all characters on $A$.

For a commutative complex unital Banach algebra, maximal ideals correspond bijectively to characters of an algebra by taking kernels. Give the maximal ideal space the Gelfand topology through this correspondence. It is a compact Hausdorff space, since the characters form a weak-star closed subset of the dual unit ball, which is compact by the Banach-Alaoglu theorem.

For a compact set $K\subset \mathbb{C}$, let $\mathcal{R}(K)$ be the uniform closure on $K$ of the rational functions with no poles on $K$. Every character of $\mathcal{R}(K)$ is evaluation at a unique point of $K$, so ${\Phi }_{\mathcal{R}(K)}$ is naturally homeomorphic to $K$.

If $X=Y\oplus Z$ and the Banach spaces $Y$ and $Z$ are isomorphic, then the operator algebra $\mathcal{B}(X)$ has no characters. The two complementary projections have unequal character values, while mutually inverse off-diagonal maps force those values to be equal.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$, equivalently the coarsest topology making every map $\phi \mapsto \phi (a)$ continuous.

This is the topology on the character space used in the Gelfand representation.

For an algebra of functions on a space, evaluation at $x$ is the character ${\delta }_x(f)=f(x)$.

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 $C({\Phi }_A)$.

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.

For a commutative complex unital Banach algebra, a character of an algebra annihilates $a-\varphi (a)1$, so this element cannot be invertible. Conversely a noninvertible $a-\lambda 1$ generates a proper ideal, which lies in a maximal ideal. The corresponding character then has value $\lambda$ at $a$. Thus the Gelfand transform determines every element's spectrum even when the algebra is not semisimple.

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 closed unital subalgebra generated by $x$ is the norm closure of the polynomials in $x$. Its character space is homeomorphic to $\sigma (x)$ through $\phi \mapsto \phi (x)$, and the complement of that spectrum is connected.

For $B=\overline{\mathbb{C}[1,x]}$ inside $A$, the displayed set is the full spectrum of $x$ in $A$. On a bounded complementary component, the maximum modulus principle gives $|p(\lambda )|\le {\max }_{{\sigma }_A(x)}|p|\le \Vert p(x)\Vert$. Evaluation $p(x)\mapsto p(\lambda )$ therefore extends to a character of an algebra on $B$, showing $\lambda \in {\sigma }_B(x)$. Conversely the unbounded resolvent component survives by resolvent components in a closed unital subalgebra. In particular, if zero lies in that component, $x^{-1}$ belongs to the norm closure of polynomials in $x$.

A C-star algebra is a complex Banach algebra with an involution satisfying $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

For a nonunital C-star algebra, adjoining an identity gives multiplication $(a,\lambda )(b,\mu )=(ab+\lambda b+\mu a,\lambda \mu )$ and involution $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$. Its standard C-star norm extends that of $A$, with the new identity of norm one. It can be constructed by adjoining the identity operator to a faithful nondegenerate operator representation and taking the operator norm. The original algebra is a closed ideal. The usual spectrum of an element in a nonunital algebra is taken in this unitization, permitting unital spectral arguments without requiring the original algebra to contain an identity.

The left multiplication representation of a C-star algebra is isometric: test $L_a$ on $a^{\ast }/\Vert a\Vert$ to get $\Vert L_a\Vert =\Vert a\Vert$. For a nonunital algebra, its closed image plus the identity operator is the algebraic unitization. For a formal $z=a+\lambda 1$ and $b\in A$, $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Thus $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$; applying the same estimate to $z^{\ast }$ proves the algebra involution is isometric and the C-star identity holds. This gives a unitization without first assuming an operator representation on a Hilbert space.

For an infinite-dimensional complex Hilbert space, the Calkin algebra is the unital quotient of the bounded operators by the closed two-sided ideal of compact operators. Its spectrum records invertibility up to compact errors. An operator is Fredholm exactly when its image in this quotient is invertible.

A C-star subalgebra is a norm-closed subalgebra closed under the involution, with the inherited C-star identity. A unital inclusion in this context uses the same identity as the ambient C-star algebra. The spectral permanence for C-star algebras says the inclusion preserves the spectrum of an element.

A C-star homomorphism is a complex linear multiplicative map between C-star algebras that preserves the involution. Such maps are contractive; an injective one is an isometry.