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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.