= Solution
A complex <Banach algebra> is an associative complex algebra with a complete <norm> satisfying <submultiplicativity>, $\|ab\|\leq\|a\|\|b\|$. A <field> has a nonzero identity $e$, and every nonzero element is invertible. We derive the <Gelfand-Mazur theorem> from these facts.
If $\|z\|<1$, 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 $|\lambda|>\|x\|$,
$$
R(\lambda)=(\lambda e-x)^{-1}
=\frac{e}{\lambda}+\sum_{n=1}^{\infty}\frac{x^n}{\lambda^{n+1}}.
$$
This <resolvent of an element> tends to zero in <norm> at infinity; the estimate follows from the geometric bound on $\|x^n\|$.
We next prove that the <spectrum of an element> cannot be empty. Suppose every $\lambda e-x$ were invertible. At each $\lambda_0$, write
$$
\lambda e-x=(\lambda_0e-x)\big(e+(\lambda-\lambda_0)R(\lambda_0)\big).
$$
The <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 $z\ne0$, define $g(tz)=t\|z\|$ 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 $\operatorname{Re}\ell(z)=\|z\|\ne0$. 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 $(\lambda e-x)R(\lambda)=e\ne0$. 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 map
$$
\boxed{\Phi:\mathbb C\longrightarrow B,\qquad\Phi(\lambda)=\lambda e}
$$
is therefore a bijective unital complex-algebra homomorphism. It and its inverse are bounded, since $\|\lambda e\|=|\lambda|\|e\|$. Thus it is an isomorphism of <Banach algebras>. If the usual normalization $\|e\|=1$ is included in the definition, it is an isometry; without that normalization, it is still the required continuous algebra isomorphism.
Back to article page
