Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 5

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is isomorphic to $\mathbb{C}$ by the scalar map $\lambda \mapsto \lambda 1$. With the usual normalization $\Vert 1\Vert =1$, that isomorphism is isometric. Commutativity need not be assumed in the theorem; it follows from the conclusion.

Every complex Banach division algebra is the scalar field. We use complex scalars throughout this question, including for $C^1[0,1]$; the complex-algebra character and spectrum conclusions depend on that convention.

The assertion requires a proper ideal. If $I=A$, no proper ideal can contain it, so the literal statement without that restriction is false.

Let $I$ be proper, and consider the partially ordered collection of proper ideals containing $I$. It is nonempty. The union of a chain is an ideal, because any two of its elements belong together to some member of the chain. It is still proper: if it contained $1$, some member of the chain would contain $1$ and equal $A$. Thus every chain has an upper bound in the collection. The Zorn lemma gives a maximal member, and any proper ideal containing it still belongs to the collection. Hence this is a maximal ideal of $A$.

Therefore every proper ideal is contained in a maximal proper ideal. This algebraic argument only needs a nonzero unital commutative ring; the Banach algebra assumptions become important for closedness of maximal ideals in part (c).

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.

For every character of an algebra $\varphi$, the element $a-\varphi (a)1$ cannot be invertible, so $\varphi (a)\in {\sigma }_A(a)$. Conversely, if $\lambda \in {\sigma }_A(a)$, the principal ideal $A(a-\lambda 1)$ is proper: otherwise $b(a-\lambda 1)=1$ for some $b$, which in this commutative algebra gives an inverse. Part (b) puts that ideal inside a maximal ideal $M$, and the corresponding algebra character from part (c) has $\varphi (a)=\lambda$. ThereforeHere $\hat{a}(\varphi )=\varphi (a)$ is the Gelfand transform. The spectrum equals character values in a commutative Banach algebra result does not require the Gelfand transform to be injective: the algebra need not be semisimple.

Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.