Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 6 / 3

For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.

A complex Banach algebra is an associative complex algebra with a complete norm satisfying submultiplicativity, $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. A field has a nonzero identity $e$, and every nonzero element is invertible. We derive the Gelfand-Mazur theorem from these facts.

If $\Vert z\Vert <1$, 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 $|\lambda |>\Vert x\Vert$,This resolvent of an element tends to zero in norm at infinity; the estimate follows from the geometric bound on $\Vert x^n\Vert$.

We next prove that the spectrum of an element cannot be empty. Suppose every $\lambda e-x$ were invertible. At each ${\lambda }_0$, writeThe 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 $z\ne 0$, define $g(tz)=t\Vert z\Vert$ 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 $\mathrm{Re}\ell (z)=\Vert z\Vert \ne 0$. 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 $(\lambda e-x)R(\lambda )=e\ne 0$. 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 mapis therefore a bijective unital complex-algebra homomorphism. It and its inverse are bounded, since $\Vert \lambda e\Vert =|\lambda |\Vert e\Vert$. Thus it is an isomorphism of Banach algebras. If the usual normalization $\Vert e\Vert =1$ is included in the definition, it is an isometry; without that normalization, it is still the required continuous algebra isomorphism.

The first algebra exists. Let $A=L^1(0,1)$ with truncated convolutionThe Tonelli theorem givesCommutativity follows by replacing $s$ with $t-s$; associativity follows from the Fubini theorem applied to the simplex $s,u\ge 0$, $s+u\le t$. Together with completeness of $L^1$, this makes $A$ a commutative Banach algebra, not necessarily unital. Adjoin an identity using unitization:This norm is complete and submultiplicative by the preceding bound, and $e=(1,0)$ is the identity. This is the Volterra convolution algebra on a finite interval with an identity adjoined.

Take $x=(0,1)$, where $1(t)=1$. Direct induction under convolution givesThus no power vanishes. For every $\lambda \ne 0$ the seriesconverges absolutely, since its norms have factorial denominators. Multiplication by $\lambda e-x$ telescopes to $e$, so it is an inverse. On the other hand $x$ cannot be invertible: its scalar coordinate is zero, and scalar coordinates multiply, so $xy$ can never equal $e$. HenceThis explicitly exhibits a nonnilpotent quasinilpotent element without assuming a spectral-radius formula.

The second algebra does not exist. Suppose it did, and choose a nonscalar $y$. Nonemptiness of its spectrum of an element, together with $\rho (y)=0$, gives ${\sigma }_B(y)=\{0\}$. But $y+e$ is still nonscalar: if $y+e=\alpha e$, then $y=(\alpha -1)e$. By translation of the spectrum by a scalar,This contradicts the required zero spectral radius for every nonscalar element. The contradiction already uses the spectral condition alone; the nonvanishing-power condition cannot remedy it.