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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 $\boxed{\mathbb{C}\setminus {\sigma }_D(a)\text{ is connected}.}$

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.

Although $ab$ need not be self-adjoint, it is similar to a Positive element of a C-star algebra. Namely,By congruence preserves positivity in a C-star algebra, $d\ge 0$. In fact $b\ge 1$ givesso $d\ge a\ge 1$. Similar elements have the same spectrum of an element, because $s^{-1}(x-\lambda 1)s$ is invertible exactly when $x-\lambda 1$ is. Therefore the stronger conclusion isPositivity of the spectrum is being asserted, not positivity of the generally non-self-adjoint product itself.

Since $a-1$ and $b-1=(b-a)+(a-1)$ are positive, their spectra are nonnegative. Translation of the spectrum of an element gives $\sigma (a),\sigma (b)\subseteq [1,\infty )$, so both elements are invertible. Their inverses and inverse square roots are positive by the continuous functional calculus. Also $1-a^{-1}$ is positive because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (a)$.

For the order reversal, setThe inequality uses congruence preserves positivity in a C-star algebra. continuous functional calculus gives $k^{-1}\le 1$. Sinceanother congruence showsConsequentlyHere $b^{-1}$ is strictly positive in the stronger invertible sense: $b^{-1}\ge \Vert b{\Vert }^{-1}1>0$, because its spectrum is the set of reciprocals of the positive spectral values of $b$. This is inversion reverses the order of strictly positive elements; no commutativity of $a,b$ is required.

For an element $a$ of a complex unital Banach algebra, its spectrum of an element isThe nonzero spectra of products in opposite orders agree:For $\lambda \ne 0$, if $R=(\lambda 1-ab{)}^{-1}$ exists, direct multiplication on both sides verifiesInterchanging $a,b$ proves the converse. Zero may differ. On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, let $S$ be the unilateral shift operator $S{e}_n=e_{n+1}$ and take $a=S^{\ast }$, $b=S$. Then $ab=S^{\ast }S=I$, whereas $ba=S{S}^{\ast }=I-P_0$, where $P_0$ is projection onto the first coordinate. HenceBoth values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.

We use the spectral definition of a Positive element of a C-star algebra: $h$ is positive when $h=h^{\ast }$ and $\sigma (h)\subseteq [0,\infty )$. The continuous functional calculus, supplied by the Commutative Gelfand--Naimark theorem for the algebra generated by a Normal element of a C-star algebra and its adjoint, sends $F\in C(\sigma (h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma (F(h))=F(\sigma (h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.

For example, $F(t)=\sqrt{t}$ gives the positive square root in a C-star algebra of a Positive element of a C-star algebra. It satisfies $(h^{1/2}{)}^2=h$. Its uniqueness follows from the composition rule: if $d\ge 0$ and $d^2=h$, then the continuous functional calculus in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\le h$ and $r\ge \Vert h\Vert$, the function $r-t$ gives $r1-h\ge 0$; the same calculus givesThese transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.

To prove closure under addition without assuming an order theorem, let $a,b\ge 0$, set $\alpha =\Vert a\Vert$, $\beta =\Vert b\Vert$ and $s=\alpha +\beta$. If $s=0$, both elements are zero. Otherwise,The element $a+b$ is self-adjoint, so any $\lambda \in \sigma (a+b)$ is real and satisfies $|s-\lambda |\le s$. Thus $\lambda \ge 0$, provingWe will also use that the positive cone is proper: if $h\ge 0$ and $-h\ge 0$, then $\sigma (h)\subseteq \{0\}$, and the norm formula for self-adjoint elements gives $h=0$.

We next prove $a^{\ast }a\ge 0$ without presupposing that positivity is defined by such products. Put $h=a^{\ast }a$, which is self-adjoint, and form its negative part by the continuous functional calculus:This is positive. Let $c=a{h}_{-}^{1/2}$. Since $h_{-}$ commutes with $h$,The nonzero product-spectrum identity proved above gives $\sigma (c{c}^{\ast })\setminus \{0\}=\sigma (c^{\ast }c)\setminus \{0\}$. As $c{c}^{\ast }$ is self-adjoint, it too has spectrum in $(-\infty ,0]$, so $c{c}^{\ast }\le 0$.

Write $c=u+iv$ with $u=u^{\ast }$ and $v=v^{\ast }$. The continuous functional calculus makes $u^2$ and $v^2$ positive, and the addition result givesBut both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^{\ast }c+c{c}^{\ast }=0$. Now $c^{\ast }c=-c{c}^{\ast }$ is positive as well as nonpositive, so $c^{\ast }c=0$. The C-star identity gives $c=0$, whence $h_{-}^2=0$ and $h_{-}=0$. Thus $h$ has no negative spectral part:This proves the positivity of adjoint products from spectral positivity and the equivalence with the usual definition by elements of the form $d^{\ast }d$: the reverse direction is the factorization $h=(h^{1/2}{)}^{\ast }{h}^{1/2}$ for $h\ge 0$.

Finally, if $a\ge 0$ and $c\in A$, writeThe preceding result therefore provesIn particular, congruence preserves positivity in a C-star algebra, and $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. These facts will justify the noncommutative order manipulations in the numbered parts.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

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.

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving

In a unital C-star algebra $A$, an element is Hermitian when $x=x^{\ast }$, unitary when $x^{\ast }x=x{x}^{\ast }=1$, and normal when $x^{\ast }x=x{x}^{\ast }$. It is a Positive element of a C-star algebra when $x=y^{\ast }y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty )$.

Suppose first that $\Vert \tau \Vert =\tau (1)$, and normalize so that both equal one. If $x=x^{\ast }$, the C-star identity givesWriting $\tau (x)=a+ib$, we obtainfor every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives

Let $x=x^{\ast }$. Every character of an algebra on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the spectrum of an element therefore gives $\sigma (x)\subset \mathbb{R}$. Iterating the C-star identity,so $\Vert x\Vert =r(x)$. If $c=\Vert x\Vert$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; henceis a difference of positive elements.

If $0\le x$ and $\Vert x\Vert \le 1$, then spectral translation givesThus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\Vert 1-x\Vert \le 1$.

Return to a norm-one $\tau$ with $\tau (1)=1$. For $0\le x\le 1$, the preceding paragraph and reality on Hermitian elements giveScaling proves that $\tau$ is a positive functional on a C-star algebra. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functionalis positive but is not multiplicative, and hence is not a character.

Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity offor every $\lambda \in \mathbb{C}$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequalityTaking $y=1$ yields $|\tau (x){|}^2\le \tau (x^{\ast }x)\tau (1)$. Since $0\le x^{\ast }x\le \Vert x{\Vert }^{2}1$, positivity givesTogether with $\tau (1)\le \Vert \tau \Vert$, this proves $\Vert \tau \Vert =\tau (1)$.

Convex combinations preserve positivity and value one at the identity, so the state space $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^{\ast }(1,x)$ is commutative and its Gelfand transform identifies it with $C(\sigma (x))$. Choose $\lambda \in \sigma (x)$ with $|\lambda |=\Vert x\Vert$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving Hahn--Banach extension to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau (x)|=\Vert x\Vert$.

The state space is nonempty, convex, and weak-star compact by the Banach-Alaoglu theorem. The Krein-Milman theorem gives an extreme point, so a pure state on a C-star algebra exists. For positive $x$, the preceding norm-attainment result makesa nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

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.

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 $\lambda \ne 0$.

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.

In a complex unital Banach algebra, right multiplication by $a$ induces a bounded module endomorphism $T$ of $A/L$ when $La\subseteq L$. Every nonzero module endomorphism of this simple module is algebraically bijective, since its kernel and range are submodules. Its inverse is bounded by the bounded inverse theorem. Choosing $\lambda$ in the nonempty spectrum of an element of $T$ forces $T-\lambda I=0$. Applying this identity to $1+L$ gives $a-\lambda 1\in L$, with uniqueness because $1\notin L$.