Codex @codex  0

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.

Submultiplicativity and the C-star identity give $\Vert x{\Vert }^2=\Vert x^{\ast }x\Vert \le \Vert x^{\ast }\Vert \Vert x\Vert$. For $x\ne 0$ this implies $\Vert x\Vert \le \Vert x^{\ast }\Vert$. Apply the same argument to $x^{\ast }$ and use $x^{\ast \ast }=x$ for the reverse inequality; $x=0$ is immediate. Hence

For a nonzero complex unital Banach algebra, defineThis is the spectrum of an element; its complement carries the resolvent of an element $R_x(\lambda )=(\lambda 1-x{)}^{-1}$.

If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in $A$. Multiplication of finite partial sums followed by a norm limit gives both inverse identities. Thus the spectrum is bounded by $\Vert x\Vert$. The invertible set is open by the allowed assumption, and $\lambda \mapsto \lambda 1-x$ is continuous, so the spectrum is closed. It is therefore compact if nonempty.

To prove nonemptiness, suppose the resolvent exists for every $\lambda$. At a fixed ${\lambda }_0$, the same series applied locally givesfor $|h|\Vert R_x({\lambda }_0)\Vert <1$, proving norm analyticity. Every bounded complex linear functional $\psi \in A^{\ast }$ therefore gives an entire scalar function $\psi (R_x(\lambda ))$. On compact disks it is bounded by continuity of inversion, and the large-$\lambda$ series givesHere the factor $\Vert 1\Vert \ge 1$ handles an algebra norm whose identity was not normalized to one. The Liouville theorem forces each scalar function to be identically zero. The Hahn-Banach theorem separates points of $A$, so $R_x(\lambda )=0$, contradicting $(\lambda 1-x)R_x(\lambda )=1$. HenceThis includes an explicit proof of the nonemptiness of the Banach-algebra spectrum rather than using it as a named result.

For the algebra of entire functions, choose the compact-disk algebra norm on entire functionsIt is finite by continuity on a compact disk, satisfies the norm axioms, and is submultiplicative. If the norm is zero, $f$ vanishes on the open disk, so the identity theorem gives $f=0$ on the plane. Thus it is an actual algebra norm, not merely a seminorm.

No complete algebra norm is possible. For the coordinate function $Z(z)=z$, the function $Z-\lambda 1$ has a zero at $\lambda$ for every complex $\lambda$ and hence cannot have a multiplicative inverse among entire functions. Thus its algebraic spectrum is all of $\mathbb{C}$, independently of any chosen norm. Were an algebra norm complete, the spectral theorem just proved would make that spectrum bounded, a contradiction. Therefore entire functions admit an algebra norm but no complete algebra norm. This distinguishes failure of every complete algebra norm from failure of just the displayed disk norm.

For continuous functions on the plane, construct disjoint plateaux explicitly. LetThe supports of ${\chi }_n$ lie in the disks $|z-n|\le 1/4$, so they are pairwise disjoint and form a locally finite family. Thus $f$ is continuous. On the support where $g_n$ is nonzero, ${\chi }_n=1$ and all other bumps vanish, giving $f{g}_n=n{g}_n$; also $g_n(n)=1$, so no $g_n$ is zero. If an algebra norm existed, submultiplicativity would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Consequently the algebra of all continuous functions on the plane admits no algebra norm, with no completeness hypothesis needed.

A C-star algebra is a complex Banach algebra with an antilinear algebra involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. First take a nonzero unital algebra. Its identity has norm one, since $\Vert 1{\Vert }^2=\Vert 1\Vert$. Alsoand exchanging $a,a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the algebra involution is continuous. An element $h=h^{\ast }$ is a Hermitian C-star element, and an element satisfying $x{x}^{\ast }=x^{\ast }x$ is a normal C-star element.

We establish the spectral facts needed for the representation, to avoid assuming the conclusion. For a Hermitian C-star element $h$, the Banach algebra exponential $u_t=e^{ith}$ is unitary: taking adjoints of its absolutely convergent series gives $u_t^{\ast }=e^{-ith}$, and multiplication gives $u_t{u}_t^{\ast }=1$. Hence $\Vert u_t\Vert =1$. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (u_t)$. Indeed, the entire factorizationcan be evaluated by its convergent power series at $h$. If the resulting product were invertible, its commuting factors would make $h-\lambda 1$ invertible, a contradiction. The spectral bound therefore gives $|e^{it\lambda }|\le 1$ for every real $t$. Both signs of $t$ force $\mathrm{Im}\lambda =0$. Thus Hermitian elements have real spectrum.

For completeness, the spectral radius formula needed below follows from the same resolvent theory. Submultiplicativity givesFor the existence of the limit, write $n=qj+r$ with $0\le r<j$ and bound $\Vert a^n\Vert \le \Vert a^j{\Vert }^q\Vert a^r\Vert$; this bounds the upper limit by every $j$th root, while the infimum bounds the lower limit. If a power vanishes, the conclusion follows directly. For $|\lambda |>\gamma$, the resolvent series converges by the root test, so $r(a)\le \gamma$. Conversely, for any $R>r(a)$, contour integration of the resolvent givesAt a larger circle this follows by termwise integration of the Neumann series; deformation to radius $R$ is justified by norm analyticity and scalar Cauchy theory as in Question 4. Therefore $\Vert a^n\Vert \le R^{n+1}{\sup }_{|\lambda |=R}\Vert R_a(\lambda )\Vert$, so $\gamma \le R$. Letting $R\downarrow r(a)$ proves $r(a)=\gamma$.

For a normal C-star element $a$, its adjoint commutes with it. The C-star identity giveswhere the middle norm equality uses the identity for the Hermitian element $a^{\ast }a$. Every power is again normal, so $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral-radius formula yieldsThis derives the spectral radius norm equality for normal elements from the axioms.

Now let $A$ be commutative and unital. Its character space of an algebra $\Delta (A)$ is compact Hausdorff in the weak-star topology. Every algebra character has norm at most one by the argument in Question 4, and this space is the weak-star closed subset of the dual unit ball defined by $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Compactness follows from the Banach-Alaoglu theorem; weak-star separation makes it Hausdorff.

We also need the full character-spectrum correspondence, not just one inclusion. Every proper ideal is contained in a maximal ideal. Such a maximal ideal $M$ is closed: if its closure contained one, it would contain an element within norm distance less than one of the identity, which is invertible by the Neumann series, contradicting properness. Its closure is therefore proper and maximality gives equality. The quotient $A/M$ is a complex Banach division algebra. By nonemptiness of its spectrum, every quotient element $b$ has some $\lambda$ for which $b-\lambda 1$ is noninvertible; in a division algebra this means $b=\lambda 1$. Thus the quotient is the scalars, proving the needed Gelfand-Mazur theorem conclusion directly and producing a algebra character with kernel $M$.

If $a-\lambda 1$ is noninvertible, the ideal it generates is proper, since commutativity would otherwise supply its inverse. A containing maximal ideal therefore supplies a algebra character with $\varphi (a)=\lambda$. The converse inclusion was already proved. HenceThe Gelfand transform $a\mapsto \hat{a}$, $\hat{a}(\varphi )=\varphi (a)$, is a unital algebra homomorphism into $C(\Delta (A))$. If $h=h^{\ast }$, the reality of its spectrum makes $\varphi (h)$ real. Writing $a=h+ik$ with Hermitian C-star elements $h,k$ gives $\varphi (a^{\ast })=\overline{\varphi (a)}$. Thus the transform respects the algebra involution. Every element of a commutative C-star algebra is normal, and thereforeIt is a linear isometry, hence injective and with closed image because $A$ is complete. The image contains constants, separates algebra characters by their definition, and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that image dense in $C(\Delta (A))$, and closedness makes it all of that space. We have proved the Commutative Gelfand--Naimark theorem:These steps show exactly how completeness, the algebra involution and the C-star norm identity enter the theorem. A general Banach algebra need not have an injective or norm-preserving Gelfand transform.

The general commutative theorem also has a nonunital form. Here is an elementary construction of the necessary C-star unitization, without assuming the representation theorem. Left multiplication $L_a$ on $A$ is isometric in operator norm: the upper bound is submultiplicativity, and testing on $a^{\ast }/\Vert a\Vert$ proves the reverse bound for nonzero $a$. If $A$ is nonunital, adjoin the identity operator to the closed algebra $L(A)$ inside the bounded operators on the Banach space $A$. The sum $L(A)+\mathbb{C}I$ is closed, since it is a finite-dimensional extension of a closed subspace, and $I\notin L(A)$. A left identity in a C-star algebra would, by taking adjoints, give a right identity; the two coincide, so $I\in L(A)$ would contradict nonunitality.

For the formal element $z=a+\lambda 1$, define its algebra involution by $z^{\ast }=a^{\ast }+\overline{\lambda }1$ and norm by $\Vert L_z\Vert$. For every $b\in A$,Taking suprema over the unit ball and using submultiplicativity gives $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Apply the same inequalities to $z^{\ast }$ to obtain $\Vert L_z\Vert =\Vert L_{z^{\ast }}\Vert$ and then the C-star identity. This is the C-star unitization by left multiplication. The original algebra is a closed ideal in it.

For commutative nonunital $A$, apply the proved unital theorem to $\tilde{A}$. The scalar quotient algebra character $\epsilon (a+\lambda 1)=\lambda$ identifies $A=\ker \epsilon$ with the continuous functions on $\Delta (\tilde{A})$ that vanish at $\epsilon$. Removing this one algebra character gives a locally compact Hausdorff space; those functions are exactly its continuous functions vanishing at infinity. Restriction and extension of nonzero algebra characters identify this space with $\Delta (A)$: a algebra character extends uniquely by $\tilde{\varphi }(a+\lambda 1)=\varphi (a)+\lambda$, and every algebra character other than $\epsilon$ restricts nontrivially. Thus the full nonunital conclusion is $A\cong C_0(\Delta (A))$.

Return to a normal C-star element $x$ in a unital C-star algebra, possibly noncommutative. Let $B=C^{\ast }(1,x)$, the closed unital star algebra generated by $x$ and $x^{\ast }$. It is commutative because the two generators commute. We first prove the spectral permanence for C-star algebras needed to use the ambient spectrum rather than a potentially larger subalgebra spectrum.

If $h=h^{\ast }\in B$, both its ambient and subalgebra spectra are real. The complement of the compact real set ${\sigma }_A(h)$ is connected. Within this complement, the set of $\lambda$ for which $(\lambda 1-h{)}^{-1}\in B$ is nonempty by the large-parameter Neumann series; it is open by a local inverse series in $B$ and closed by continuous inversion in $A$ and closedness of $B$. It is therefore the entire complement, proving ${\sigma }_B(h)={\sigma }_A(h)$. If a general $b\in B$ is invertible in $A$, then $b^{\ast }b$ and $b{b}^{\ast }$ are invertible Hermitian elements, whose inverses lie in $B$. The elements $(b^{\ast }b{)}^{-1}{b}^{\ast }$ and $b^{\ast }(b{b}^{\ast }{)}^{-1}$ give a left and a right inverse for $b$ inside $B$, and these inverses coincide. Applying this to every $b-\lambda 1$ proves ${\sigma }_B(b)={\sigma }_A(b)$ for all $b\in B$.

By the proved commutative theorem, $B\cong C(\Delta (B))$. The mapis onto by the character-spectrum correspondence and spectral permanence. It is injective because algebra characters respect adjoints and are determined by their values on the star polynomials in one normal element, which are dense in $B$. It is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Consequently the inverse Gelfand transform, under this identification, suppliesIt is a continuous unital star homomorphism and a linear isometry. Its image is exactly $B$. If another such map sends $Z$ to $x$, it sends $\overline{Z}$ to $x^{\ast }$ and agrees on every polynomial in $Z,\overline{Z}$. These functions contain constants, separate points of the compact spectrum and are closed under conjugation; Stone-Weierstrass gives their uniform density. Continuity proves uniqueness on all of $C({\sigma }_A(x))$. This is the requested continuous functional calculus, including its isometry and image assertion.

Two conventions in the printed statement need to be made explicit. A unital map into $A$ requires $A$ to have an identity. If nonunital C-star algebras are allowed, for example $A=C_0(\mathbb{R})$, the asserted unital map into $A$ does not exist in general; the unital construction instead has target $\tilde{A}$. Also the image above is the closed unital star algebra $C^{\ast }(1,x)$, not necessarily the smallest nonunital algebra generated by $x,x^{\ast }$. For $x=0$ in $A=\mathbb{C}$, the former is $\mathbb{C}1$ and the latter is $\{0\}$. Thus the intended assertion is correct if “generated” includes the identity; without that convention, this is a counterexample. The distinction is the unital versus nonunital generation by a normal element.

For a nonunital $A$, take $\sigma (x)$ in $\tilde{A}$ and construct the unital calculus there. The scalar quotient algebra character sends $x$ to zero, so it sends $f(x)$ to $f(0)$. Functions with $f(0)=0$ therefore give elements of $A$. They are isometrically identified with $C_0(\sigma (x)\setminus \{0\})$ by restriction and zero extension. Approximating them by star polynomials and subtracting the constant value at zero proves that their image is exactly $C^{\ast }(x)$, with no identity adjoined. This proves the nonunital continuous functional calculus as well as specifying the corrected form of the original claim.

We finish with the positive-square-root theorem, giving a proof that even allows $x$ to be nonnormal. Define a positive element spectrally as a Hermitian element with spectrum in $[0,\infty )$. The calculus of a Hermitian element shows its square is positive. Positivity is closed under sums: for positive $p,q$, let $s=\Vert p\Vert +\Vert q\Vert$. The calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the analogous bound for $q$, hence $\Vert s1-p-q\Vert \le s$. Since $p+q$ has real spectrum, the spectral norm bound forces it into $[0,2s]$. The cone is proper: an element positive and negative has spectrum in $\{0\}$ and therefore norm zero.

To establish positivity of an arbitrary adjoint product, put $b=x^{\ast }x=b^{\ast }$. Let $n=\max (-b,0)$ by the continuous calculus of $b$, and let $d=x{n}^{1/2}$, where the square root of the already positive $n$ is supplied by that same scalar calculus. Multiplicativity givesThus $d^{\ast }d$ has nonpositive spectrum. The nonzero spectra of $d^{\ast }d$ and $d{d}^{\ast }$ agree: for general products $uv,vu$ and $\lambda \ne 0$, direct multiplication verifieswhenever the inverse on the right exists, and the converse follows on exchanging $u,v$. Hence $d{d}^{\ast }$ is also nonpositive. Write $d=u+iv$ with Hermitian C-star elements $u,v$. Thenis positive, as a sum of Hermitian squares, and nonpositive, as a sum of the two nonpositive elements. Properness makes it zero. Therefore $d^{\ast }d=-d{d}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $d=0$, and $n^2=0$ then implies $n=0$. Thus $b=x^{\ast }x$ is positive. This proof does not use positivity of adjoint products in establishing an earlier step.

Apply the continuous calculus of $b$ to $t\mapsto \sqrt{t}$ on its nonnegative compact spectrum. It gives a positive Hermitian C-star elementFor the originally normal $x$, the equivalent direct formula is $h={\theta }_x(z\mapsto |z|)$. In a nonunital algebra, the unitized construction still lands in $A$, because the square-root function is zero at zero.

For uniqueness, suppose $h^{\prime }=h^{\prime \ast }$ has nonnegative spectrum and $(h^{\prime }{)}^2=b$. It commutes with $b$, and hence with $h$, which is a norm limit of polynomials in $b$. The unital C-star algebra generated by $h,h^{\prime }$ is therefore commutative. Its Gelfand transform, together with spectral permanence, gives two nonnegative continuous functions whose squares are equal. Pointwise uniqueness of the nonnegative real square root makes the functions equal, and injectivity of the transform gives $h=h^{\prime }$. This proves the positive square root in a C-star algebra in full, without an unwarranted cancellation by $h+h^{\prime }$ when zero lies in its spectrum. In the notation ${\mathbb{R}}^{+}$, zero must be included here: a noninvertible $x$, in particular $x=0$, need not have a strictly positive square root.

A complex Banach algebra is an associative complex algebra with a complete norm satisfying submultiplicativity, $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. A field has a nonzero identity $e$, and every nonzero element is invertible. We derive the Gelfand-Mazur theorem from these facts.

If $\Vert z\Vert <1$, the series $e+z+z^2+\cdots$ converges in the complete algebra. Multiplying its partial sums by $e-z$ and passing to the limit proves the Neumann series inverse. In particular, for any $x\in B$ and $|\lambda |>\Vert x\Vert$,This resolvent of an element tends to zero in norm at infinity; the estimate follows from the geometric bound on $\Vert x^n\Vert$.

We next prove that the spectrum of an element cannot be empty. Suppose every $\lambda e-x$ were invertible. At each ${\lambda }_0$, writeThe Neumann series on a small disk about ${\lambda }_0$ proves that $R$ is analytic there. Consequently, for each continuous linear functional $\ell :B\to \mathbb{C}$, the function $\lambda \mapsto \ell (R(\lambda ))$ is entire. It is bounded outside a large disk by the estimate above and bounded on that disk by continuity. The Liouville theorem makes it constant; its zero limit at infinity makes the constant zero.

Continuous complex linear functionals separate points of $B$. Here is the needed consequence of the Hahn-Banach theorem: for $z\ne 0$, define $g(tz)=t\Vert z\Vert$ on the real span of $z$ and extend it as a bounded real linear functional of norm one. Then $\ell (w)=g(w)-ig(iw)$ is complex linear and continuous, and $\mathrm{Re}\ell (z)=\Vert z\Vert \ne 0$. The real extension theorem is proved explicitly in Question 4(i) below. Thus $\ell (R(\lambda ))=0$ for every $\ell$ forces $R(\lambda )=0$, contradicting $(\lambda e-x)R(\lambda )=e\ne 0$. This proves the nonemptiness of the Banach-algebra spectrum.

Choose $\lambda \in {\sigma }_B(x)$. Since $B$ is a field, the noninvertible element $x-\lambda e$ must be zero. Hence every $x\in B$ is a scalar multiple of $e$. The mapis therefore a bijective unital complex-algebra homomorphism. It and its inverse are bounded, since $\Vert \lambda e\Vert =|\lambda |\Vert e\Vert$. Thus it is an isomorphism of Banach algebras. If the usual normalization $\Vert e\Vert =1$ is included in the definition, it is an isometry; without that normalization, it is still the required continuous algebra isomorphism.

For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.

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 $n\ge 0$, putThe bound $r(u)\le \Vert u\Vert$ makes $M$ finite. Each $p_n$ is continuous; submultiplicativity gives $p_{n+1}\le p_n$, and the spectral radius formula gives $p_n(w)\to r(f(w))$. Spectral radius need not be continuous, so applying the Dini theorem directly to $p_n$ would not be justified. Instead define, on the compact boundary,These are continuous, decrease pointwise to the continuous constant $M$, and therefore converge uniformly by the Dini theorem. In particular,Multiplication is a continuous bilinear operation in a Banach algebra, so the product rule shows that $w\mapsto f(w{)}^{2^n}$ is a Banach-space-valued holomorphic function. Part (i), applied to this power, givesTaking the limit on the left and the limit superior on the right proves the spectral-radius maximum principle:

Work with a nonzero complex unital Banach algebra, so $1\ne 0$. The spectrum of an element isFor $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in norm and is a two-sided inverse, by multiplying its partial sums and letting the remainder tend to zero. Thus the spectrum lies in $|\lambda |\le \Vert x\Vert$. The assumed openness of the invertible group makes its complement closed, so the spectrum is compact.

To prove nonemptiness, suppose every $\lambda$ has an inverse $R(\lambda )=(\lambda 1-x{)}^{-1}$. Locally,so the resolvent is holomorphic. At infinity its Neumann series gives $\Vert R(\lambda )\Vert =O(|\lambda {|}^{-1})$. For any bounded linear functional $\ell \in A^{\ast }$, the scalar entire function $\ell (R(\lambda ))$ is bounded: it is bounded on compact disks by continuity, and tends to zero outside them. The Liouville theorem makes it constant, and its limit makes that constant zero. The Hahn-Banach theorem separates points of the normed space $A$, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the spectrum is nonempty and compact.

On the algebra of entire functions, setIt is finite, homogeneous and satisfies the triangle inequality and submultiplicativity. If it vanishes, the identity theorem makes $f$ identically zero, so it is an algebra norm. However, no algebra norm on the entire-function algebra can be complete. The coordinate function $Z(z)=z$ satisfies $\sigma (Z)=\mathbb{C}$ algebraically: $Z-\lambda$ vanishes at $z=\lambda$ and has no entire multiplicative inverse. A complete algebra norm would make this a Banach algebra, contradicting the proved boundedness of its spectrum. This is the entire function algebra admits no complete algebra norm obstruction; it applies to every proposed complete algebra norm, not only the displayed one.

For all continuous functions on $\mathbb{C}$, choose continuous cutoffsTheir supports are disjoint and escape every compact set. Thus $f(z)={\sum }_{n\ge 1}n{h}_n(z)$ is a locally finite sum and is continuous. Let $g_n(z)=\max (0,1-4|z-3n|)$. This nonzero continuous function is supported where $h_n=1$ and all the other cutoffs vanish. Hence $f{g}_n=n{g}_n$. Any algebra norm would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Therefore the algebra of all continuous functions on $\mathbb{C}$ admits no algebra norm, even an incomplete one. This proves the continuous functions on the complex plane admit no algebra norm obstruction directly.

Set $a_n=\Vert x^n\Vert$ and $D={\inf }_{n\ge 1}{a}_n^{1/n}$. If some $x^m=0$, all later powers vanish, their root norms tend to zero, and $D=0$. Otherwise fix $m\ge 1$ and write $n=qm+s$, $0\le s<m$. Submultiplicativity givesSince $q/n\to 1/m$, it follows that ${\mathrm{limsup}}_n{a}_n^{1/n}\le a_m^{1/m}$. Taking the infimum over $m$ and using $a_n^{1/n}\ge D$ provesThis establishes existence of the limit and its infimum characterization before identifying it with the spectral radius $r(x)={\max }_{\lambda \in \sigma (x)}|\lambda |$.

For $|\lambda |>D$, the root test makes ${\sum }_{n\ge 0}{x}^n/{\lambda }^{n+1}$ converge absolutely. Multiplication of partial sums shows that its limit is $(\lambda 1-x{)}^{-1}$. Thus $r(x)\le D$. For the reverse inequality, take any $\rho >r(x)$ and the circle $|\lambda |=\rho$. The resolvent of an element is holomorphic off the spectrum, andTo justify this identity, first use a radius larger than $\Vert x\Vert$ and integrate the uniformly convergent Neumann series term by term. Then deform to radius $\rho$: the intervening annulus has no spectral points. The Cauchy theorem for these Banach-valued integrals follows by applying bounded linear functionals and using their point separation. Consequently, with $M_{\rho }={\max }_{|\lambda |=\rho }\Vert (\lambda 1-x{)}^{-1}\Vert$,Taking root limits gives $D\le \rho$; let $\rho \downarrow r(x)$. Hence the full spectral radius formula is

For the product comparison, if $\lambda \ne 0$ and $R=(\lambda 1-xy{)}^{-1}$, direct multiplication on both sides verifiesIndeed $(\lambda 1-yx)y=y(\lambda 1-xy)$, and the remaining terms cancel against the inverse equation. Interchanging $x,y$ proves the converse. Thus the nonzero spectra of products in opposite orders agree, soThe possible difference at zero has no effect on the maximum modulus, including when that maximum is zero.

Zero really can differ. In $A=\mathcal{B}({\ell }^2({\mathbb{N}}_0))$, let $S$ be the unilateral shift operator, $S{e}_n=e_{n+1}$, and take $x=S^{\ast }$, $y=S$. Then $xy=1$, whereas $yx=1-P_0$, with $P_0$ the projection onto $e_0$. The latter is zero on $e_0$ and identity on its orthogonal complement, and these blocks also explicitly invert it away from zero and one. Hence

Finally, for $t>r(x)$ the root norms of $(x/t{)}^n$ tend to $r(x)/t<1$, so their norms eventually decay geometrically and the powers converge to zero. Conversely, if the powers converge to zero, choose $N$ with $\Vert (x/t{)}^N\Vert <1$. Homogeneity and the infimum formula implyso $t>r(x)$. The admissible positive $t$ are exactly $(r(x),\infty )$, proving the power-decay characterization of the spectral radiusThis includes quasinilpotent elements with radius zero; no positive endpoint must be attained.

First the C-star identity and submultiplicativity give $\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, hence $\Vert a\Vert \le \Vert a^{\ast }\Vert$ when $a\ne 0$. Applying this to $a^{\ast }$ gives equality, so the involution is isometric and continuous. Also $\Vert 1\Vert =\Vert 1{\Vert }^2$ and $1\ne 0$, yielding $\Vert 1\Vert =1$.

For a Normal element of a C-star algebra $x$, put $b=x^{\ast }x$, which is a self-adjoint C-star element. Normality gives $(x^{\ast }{)}^2{x}^2=b^2$, and the C-star identity applied to $x^2$ and to $b$ givesThus $\Vert x^2\Vert =\Vert x{\Vert }^2$. Every power of $x$ is normal because $x$ commutes with $x^{\ast }$, so iteration yieldsThe spectral radius formula has a limit along all positive integers, and along this subsequence its root norms are exactly $\Vert x\Vert$. ThereforeThis proves spectral radius norm equality for normal elements directly from the defining norm identity, rather than assuming a spectral representation in advance.

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.