Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 106 / 2

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

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A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$. The character space of an algebra isIf $A$ is unital, multiplicativity and nonzeroness force $\phi (1)=1$.

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For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

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For a compact Hausdorff space $K$, every character of the Banach algebra $C(K)$ is an evaluation characterat a unique $x\in K$. The map $x\mapsto {\delta }_x$ is a homeomorphism from $K$ onto ${\Phi }_{C(K)}$ with its Gelfand topology.

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The unitization of $C_0(\mathbb{R})$ is $C({\mathbb{R}}^{+})$ on the one-point compactification of $\mathbb{R}$. The supplied homeomorphism identifies ${\mathbb{R}}^{+}$ with the circle $\mathbb{T}$. Every character $\phi$ of $C_0(\mathbb{R})$ extends to the unital characterBy part d, $\tilde{\phi }$ is evaluation at a point of ${\mathbb{R}}^{+}$. Evaluation at the point at infinity vanishes on $C_0(\mathbb{R})$ and cannot restrict to the nonzero character $\phi$. The point is therefore some $x\in \mathbb{R}$, and $\phi ={\delta }_x$. Conversely each ${\delta }_x$ is plainly a character, so

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Let $\phi$ be a character of $C(\mathbb{R})$. Put $q(t)=1/(t^2+1)$. Sincewe have $(\phi (t{)}^2+1)\phi (q)=1$, so the restriction of $\phi$ to $C_0(\mathbb{R})$ is nonzero. Part e gives a point $x$ such that this restriction is ${\delta }_x$. For $g(t)=t/(t^2+1)\in C_0(\mathbb{R})$, multiplicativity givesand hence $\phi (t)=x$.

If $h\ge 0$ is continuous, thenbelongs to $C_0(\mathbb{R})$. Applying $\phi$ to $k(h+t^2+1)=1$ yieldsso $\phi (h)=h(x)$. Every complex-valued continuous function is a complex linear combination of nonnegative continuous functions, obtained from the positive and negative parts of its real and imaginary components. Therefore $\phi ={\delta }_x$, and

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Suppose an algebra norm made $C(\mathbb{R})$ a Banach algebra. Its unital character space of an algebra would be compact in the Gelfand topology. By part i it consists of the evaluations ${\delta }_x$. The mapis continuous from the usual topology because every $h\in C(\mathbb{R})$ is continuous, and its inverse is the continuous map $\phi \mapsto \phi (t)$ defined by the coordinate function $t$. Thus the character space is homeomorphic to the noncompact space $\mathbb{R}$, a contradiction. No complete algebra norm exists.

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Let $B$ be the completion under the assumed algebra norm. Restriction sends every $\psi \in {\Phi }_B$ to a character of the dense subalgebra $C(\mathbb{R})$, hence by part i to ${\delta }_x$ for some $x$. Continuity of the restriction says exactly that $x\in K$. Conversely, every ${\delta }_x$ continuous for the algebra norm extends uniquely through the completion, and continuity of multiplication makes the extension a character of $B$. Restriction and extension therefore give a bijection

For $b\in B$, choose $h_n\in C(\mathbb{R})$ converging to $b$ in the algebra norm. Characters on the unital Banach algebra $B$ are uniformly norm bounded, so $h_n{|}_K$ converges uniformly to the function $x\mapsto {\tilde{\delta }}_x(b)$. This function is continuous. Hence $x\mapsto {\tilde{\delta }}_x$ is continuous into the Gelfand topology. Its inverse is again $\psi \mapsto \psi (t)$, so the bijection is a homeomorphism. In particular, $K$ is compact.

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Assume for contradiction that any algebra norm exists. Part iii makes $K$ compact, so choose an interval disjoint from $K$ and a nonzero $a\in C(\mathbb{R})$ supported in that interval. Choose $b\in C(\mathbb{R})$ with $b=1$ on $K$ and $b=0$ on the support of $a$. Then $a,b$ are nonzero and $ab=0$.

In the completion $B$, every character is evaluation at a point of $K$, so ${\sigma }_B(b)=\{1\}$. Thus $q=1-b$ has spectral radius zero. The equality $ab=0$ givesfor every $n$. The spectral radius formula supplies an $n$ with $\Vert q^n\Vert <1$, and submultiplicativity then givesan impossibility because $a\ne 0$. Hence $C(\mathbb{R})$ admits no algebra norm at all.

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