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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 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.

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.

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.

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.

Adjoin the identity to the bounded multiplicatively closed set, putting $S^{\prime }=S\cup \{1\}$ and $M={\sup }_{s\in S^{\prime }}\Vert s\Vert <\infty$. Define an auxiliary norm on the underlying Banach space byThe identity ensures $\Vert a\Vert \le p(a)\le M\Vert a\Vert$, so $p$ and the given norm are equivalent norms. For $s\in S$, closure under multiplication gives $p(sa)\le p(a)$. Now use the operator norm of left multiplication on this renormed space:Left multiplication is faithful, since $a1=a$. It follows that this is a norm, its triangle and scalar laws follow from the operator norm, and $L_{ab}=L_a{L}_b$ proves submultiplicativity. Also $L_1$ is the identity, so $\Vert 1{\Vert }_1=1$, and the contraction property gives $\Vert s{\Vert }_1\le 1$ for every $s\in S$. Finally,For the upper bound use $p(ab)\le M\Vert a\Vert \Vert b\Vert \le M\Vert a\Vert p(b)$. These bounds prove equivalence and completeness, so this is the required unital algebra norm. This proves renorming a bounded multiplicative semigroup; the auxiliary norm is submultiplicative but need not give the identity norm one, which is why the operator norm is used.

The spectrum of an element is $\sigma (x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is not invertible in }A\}$. The spectral radius formula isIf $r(x)<1$, choose $r(x)<q<1$. Eventually $\Vert x^m{\Vert }^{1/m}<q$, hence $\Vert x^m\Vert <q^m$. The finitely many earlier powers are bounded too. In particular the power set is bounded; indeed $x^m\to 0$ in norm.

For the commuting family, put $d_j=r(x_j)+\epsilon /2$ and $y_j=x_j/d_j$. Each $r(y_j)<1$, so $M_j={\sup }_{m\ge 0}\Vert y_j^m\Vert <\infty$. Every element of the generated multiplicative semigroup can be reordered as $y_1^{m_1}\cdots y_n^{m_n}$, and its norm is at most ${\prod }_j{M}_j$. The first construction therefore gives one equivalent unital algebra norm with $\Vert y_j{\Vert }_0\le 1$ for all $j$, whenceThis is simultaneous spectral-radius renorming; commutativity is what bounds all mixed products.

For commuting $a,b$, apply this construction to both with error $\epsilon$. Equivalent algebra norms leave invertibility and spectrum unchanged, and the spectral radius formula gives $r(c)\le \Vert c{\Vert }_0$. ThusLet $\epsilon \downarrow 0$ to obtain $r(ab)\le r(a)r(b)$ and $r(a+b)\le r(a)+r(b)$.

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.

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 the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

Let $B$ be the completion under the assumed algebra norm. Restriction sends every $\psi \in {\Phi }_B$ to a character of the dense subalgebra $C(\mathbb{R})$, hence by part i to ${\delta }_x$ for some $x$. Continuity of the restriction says exactly that $x\in K$. Conversely, every ${\delta }_x$ continuous for the algebra norm extends uniquely through the completion, and continuity of multiplication makes the extension a character of $B$. Restriction and extension therefore give a bijection

For $b\in B$, choose $h_n\in C(\mathbb{R})$ converging to $b$ in the algebra norm. Characters on the unital Banach algebra $B$ are uniformly norm bounded, so $h_n{|}_K$ converges uniformly to the function $x\mapsto {\tilde{\delta }}_x(b)$. This function is continuous. Hence $x\mapsto {\tilde{\delta }}_x$ is continuous into the Gelfand topology. Its inverse is again $\psi \mapsto \psi (t)$, so the bijection is a homeomorphism. In particular, $K$ is compact.

Suppose an algebra norm made $C(\mathbb{R})$ a Banach algebra. Its unital character space of an algebra would be compact in the Gelfand topology. By part i it consists of the evaluations ${\delta }_x$. The mapis continuous from the usual topology because every $h\in C(\mathbb{R})$ is continuous, and its inverse is the continuous map $\phi \mapsto \phi (t)$ defined by the coordinate function $t$. Thus the character space is homeomorphic to the noncompact space $\mathbb{R}$, a contradiction. No complete algebra norm exists.

Assume for contradiction that any algebra norm exists. Part iii makes $K$ compact, so choose an interval disjoint from $K$ and a nonzero $a\in C(\mathbb{R})$ supported in that interval. Choose $b\in C(\mathbb{R})$ with $b=1$ on $K$ and $b=0$ on the support of $a$. Then $a,b$ are nonzero and $ab=0$.

In the completion $B$, every character is evaluation at a point of $K$, so ${\sigma }_B(b)=\{1\}$. Thus $q=1-b$ has spectral radius zero. The equality $ab=0$ givesfor every $n$. The spectral radius formula supplies an $n$ with $\Vert q^n\Vert <1$, and submultiplicativity then givesan impossibility because $a\ne 0$. Hence $C(\mathbb{R})$ admits no algebra norm at all.

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.

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.

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.

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.