Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 10 / 2

Give $\mathcal{O}(U)$ the compact-open topology, that is, uniform convergence on every compact subset of $U$. Choose a bounded finite union of polygonal domains $D$ whose closure lies in $U$, whose interior contains $\sigma (x)$ and whose boundary avoids the spectrum. Orient its boundary $\Gamma$ positively, with holes oriented negatively. It has winding number one on the spectrum and zero outside $U$. For the resolvent of an element $R(\zeta ,x)=(\zeta 1-x{)}^{-1}$ define the holomorphic functional calculus byThis is an integral of a continuous Banach space-valued function. Its value is independent of the admissible contour: the integrand is holomorphic away from the spectrum, and the difference of two such cycles has winding number zero there, so the Cauchy integral theorem applies. The vector-valued theorem follows by applying bounded linear functionals and then the Hahn-Banach theorem.

Linearity is immediate. To prove multiplicativity, choose two admissible contours with the inner domain compactly contained in the outer one, and use the resolvent identityIn the double integral for ${\Theta }_x(f){\Theta }_x(g)$, integrate the first term in the outer variable $\eta$. The Cauchy integral formula gives $g(\zeta )$. For the second term, integration of $f(\zeta )/(\eta -\zeta )$ over the inner contour is zero, since $\eta$ lies outside that inner domain. Therefore the double integral equals ${\Theta }_x(fg)$. The continuous integrands on disjoint compact contours justify exchanging the integrals.

For $f=1$ and $f=Z$, deform to a large circle and use the Neumann series $R(\zeta ,x)={\sum }_{m\ge 0}{x}^m{\zeta }^{-m-1}$. Its relevant coefficients give ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. The estimateproves continuity in the stated topology.

For uniqueness, the Runge approximation theorem in the required form says that rational functions with poles outside $U$, allowing a pole at infinity, are dense in $\mathcal{O}(U)$ for uniform convergence on compact subsets. Every complex-linear unital homomorphism taking $Z$ to $x$ has prescribed values on polynomials and on $(Z-a{)}^{-1}$ for $a\notin U$: the latter value must be $(x-a1{)}^{-1}$. Hence its rational-function values are prescribed. Continuity and Runge approximation force its values on every holomorphic function. This proves continuity and uniqueness of holomorphic functional calculus, including disconnected $U$.

To prove the spectral mapping theorem, first suppose $\mu \notin f(\sigma (x))$. On a smaller open neighborhood $U^{\prime }$ of the spectrum, $f-\mu$ has no zeros, so $h=1/(f-\mu )$ is holomorphic there. The contour definition is compatible with this restriction. Multiplicativity gives ${\Theta }_x(f)-\mu 1$ the inverse ${\Theta }_x(h)$, excluding $\mu$ from its spectrum.

Conversely, if $\mu =f(\lambda )$ for $\lambda \in \sigma (x)$, write $f(z)-\mu =(z-\lambda )g(z)$, defining $g(\lambda )=f^{\prime }(\lambda )$ at its removable singularity. ThenThese two factors commute. An invertible product of commuting elements makes each factor invertible, by multiplying the inverse of the product by the other factor. That would contradict $\lambda \in \sigma (x)$. Thus

The sequence of norms is bounded. Choose $M>{\sup }_n\Vert x_n\Vert$ and $M>\Vert x\Vert$. By the spectral radius formula, every relevant spectrum lies in $|\lambda |<M$. The compact set $K=\{\lambda :|\lambda |\le M\}\setminus U$ is disjoint from $\sigma (x)$. If nonempty, continuity of the resolvent of an element gives $C={\sup }_K\Vert (\lambda 1-x{)}^{-1}\Vert <\infty$. For large $n$, $C\Vert x_n-x\Vert <1$, andis invertible for every $\lambda \in K$ by the Neumann series. If $K$ is empty there is nothing to exclude. No spectral points can lie outside the radius-$M$ disc either. Consequently $\sigma (x_n)\subset U$ for all sufficiently large $n$. Taking $U$ to be the disc of radius $r(x)+\epsilon$ also proves upper semicontinuity of the spectral radius.

Fix an admissible contour $\Gamma$ with interior $D$ compactly contained in $U$ and containing $\sigma (x)$. Apply the preceding spectral inclusion to $D$, so that it also contains $\sigma (x_n)$ for large $n$. Uniformly on the compact contour, the Neumann series makes $R(\zeta ,x_n)$ bounded, and the resolvent identity givesThe resolvents therefore converge uniformly on $\Gamma$. Both holomorphic functional calculus values use that same contour, soHence ${\Theta }_{x_n}(f)\to {\Theta }_x(f)$ for each fixed $f\in \mathcal{O}(U)$.

Define the holomorphic function $e$ on the disconnected open set $U$ to be one on $V$ and zero on $W$. Since $e^2=e$, its holomorphic functional calculus value $p={\Theta }_x(e)$ is an idempotent. The spectral mapping theorem and the nonempty spectral intersection with $V$ show $1\in \sigma (p)$; thus $p\ne 0$. These are the Riesz projection values selecting the indicated spectral component.

Part (b) gives $p_n={\Theta }_{x_n}(e)\to p$. If $\sigma (x_n)$ avoided $V$, spectral mapping would give $\sigma (p_n)=\{0\}$. A nonzero idempotent always has $1$ in its spectrum: if $1-p_n$ were invertible, $p_n(1-p_n)=0$ would force $p_n=0$. Thus avoiding $V$ forces $p_n=0$. That is impossible eventually, since $p\ne 0$ and $p_n\to p$. Therefore $\sigma (x_n)\cap V\ne \varnothing$ eventually. This is persistence of a separated spectral component, not a claim that arbitrary individual spectral points vary continuously.