= Solution
For a nonzero complex unital <Banach algebra>, define
$$
\sigma_A(x)=\{\lambda\in\mathbb C:\lambda1-x\text{ is not invertible in }A\}.
$$
This is the <spectrum of an element>; its complement carries the <resolvent of an element> $R_x(\lambda)=(\lambda1-x)^{-1}$.
If $|\lambda|>\|x\|$, the <Neumann series>
$$
R_x(\lambda)=\lambda^{-1}\sum_{j=0}^{\infty}(x/\lambda)^j
$$
converges in $A$. Multiplication of finite partial sums followed by a <norm> limit gives both inverse identities. Thus the spectrum is bounded by $\|x\|$. The invertible set is open by the allowed assumption, and $\lambda\mapsto\lambda1-x$ 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 gives
$$
R_x(\lambda_0+h)=\sum_{j=0}^{\infty}(-h)^jR_x(\lambda_0)^{j+1}
$$
for $|h|\|R_x(\lambda_0)\|<1$, proving <norm> analyticity. Every bounded complex <linear functional> $\psi\in A^*$ 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 gives
$$
\|R_x(\lambda)\|\leq\frac{\|1\|}{|\lambda|-\|x\|}\longrightarrow0.
$$
Here the factor $\|1\|\geq1$ 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 $(\lambda1-x)R_x(\lambda)=1$. Hence
$$
\boxed{\sigma_A(x)\text{ is nonempty and compact},\qquad \sigma_A(x)\subseteq\{|\lambda|\leq\|x\|\}}.
$$
This 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 functions>
$$
\|f\|=\sup_{|z|\leq1}|f(z)|.
$$
It 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 $Z-\lambda1$ 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 \b[<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. Let
$$
\chi_n(z)=\min\{1,\max\{0,2-8|z-n|\}\},\qquad g_n(z)=\max\{0,1-8|z-n|\},\qquad f(z)=\sum_{n\geq1}n\chi_n(z).
$$
The supports of $\chi_n$ lie in the disks $|z-n|\leq1/4$, 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 $fg_n=ng_n$; also $g_n(n)=1$, so no $g_n$ is zero. If an <algebra norm> existed, <submultiplicativity> would imply
$$
n\|g_n\|=\|fg_n\|\leq\|f\|\|g_n\|,
$$
so $n\leq\|f\|$ for every positive integer, impossible. Consequently \b[the algebra of all continuous functions on the plane admits no <algebra norm>], with no <completeness> hypothesis needed.
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