= Solution
Work with a nonzero complex unital <Banach algebra>, so $1\ne0$. The <spectrum of an element> is
$$
\sigma_A(x)=\{\lambda\in\mathbb C:\lambda1-x\text{ is not invertible in }A\}.
$$
For $|\lambda|>\|x\|$, the <Neumann series>
$$
(\lambda1-x)^{-1}=\sum_{n=0}^{\infty}\frac{x^n}{\lambda^{n+1}}
$$
converges 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 $|\lambda|\le\|x\|$. 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 $R(\lambda)=(\lambda1-x)^{-1}$. Locally,
$$
R(\lambda+h)=R(\lambda)(1+hR(\lambda))^{-1}
=\sum_{j\ge0}(-h)^jR(\lambda)^{j+1},
$$
so the resolvent is holomorphic. At infinity its <Neumann series> gives $\|R(\lambda)\|=O(|\lambda|^{-1})$. For any <bounded linear functional> $\ell\in A^*$, 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 $(\lambda1-x)R(\lambda)=1$. Hence \b[the <spectrum> is nonempty and compact].
On the algebra of <entire functions>, set
$$
\boxed{\|f\|_D=\sup_{|z|\le1}|f(z)|.}
$$
It 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 cutoffs
$$
h_n(z)=\begin{cases}
1,&|z-3n|\le1/2,\\
2(1-|z-3n|),&1/2<|z-3n|<1,\\
0,&|z-3n|\ge1.
\end{cases}
$$
Their supports are disjoint and escape every compact set. Thus $f(z)=\sum_{n\ge1}nh_n(z)$ 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 $fg_n=ng_n$. Any <algebra norm> would imply
$$
n\|g_n\|=\|fg_n\|\le\|f\|\|g_n\|,
$$
so $n\le\|f\|$ for every positive integer, impossible. Therefore \b[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.
Back to article page
