Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 6 / 3 / Solution

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.