Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 7 (Banach Algebras)

Use the common nonzero identity $1$. If $\Vert h\Vert <1$, completeness and submultiplicativity give the Neumann seriesFor $a\in G(A)$ and $\Vert a^{-1}\Vert \Vert b-a\Vert <1$, write $b=a(1+a^{-1}(b-a))$ and apply that series. This proves that the group of invertible elements of a Banach algebra is open. Locally the same series bounds $\Vert b^{-1}\Vert$, andproves continuity of inversion. Inversion is its own inverse, hence it is a homeomorphism of $G(A)$ onto itself.

The spectrum of an element is ${\sigma }_A(x)=\{\lambda :\lambda 1-x\notin G(A)\}$, and its complement is the algebra resolvent set ${\rho }_A(x)$. Openness of $G(A)$ makes the algebra spectrum closed. For $|\lambda |>\Vert x\Vert$,so it is bounded and thus compact. For nonemptiness, suppose $R$ existed everywhere. The local Neumann expansion around each point makes $R$ a Banach-space-valued entire function. For every bounded linear functional $\ell$ on $A$, $\ell (R(\lambda ))$ is entire, bounded on compact discs, and tends to zero at infinity by the displayed series. The Liouville theorem makes it identically zero. The Hahn-Banach theorem separates points, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the algebra spectrum is nonempty and compact.

If $B$ is a closed unital subalgebra with the same identity, then ${\rho }_B(x)\subseteq {\rho }_A(x)$. It is relatively open by the Neumann series in $B$. If ${\lambda }_j\in {\rho }_B(x)$ tends to $\lambda \in {\rho }_A(x)$, continuity of inversion in $A$ gives $({\lambda }_{j}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Closedness puts this last inverse in $B$, so $\lambda \in {\rho }_B(x)$. Therefore ${\rho }_B(x)$ is relatively open and closed in ${\rho }_A(x)$. This proves the required resolvent components in a closed unital subalgebra directly, with no spectral-boundary assumption.

Intersect a connected component $U$ of ${\rho }_A(x)$ with the clopen subset ${\rho }_B(x)$. Connectedness makes the intersection either empty or all of $U$. In the empty case $U\subseteq {\sigma }_B(x)$. In the other case $U\subseteq {\rho }_B(x)$; a connected set in ${\rho }_B(x)$ containing $U$ is also connected in ${\rho }_A(x)$, so maximality prevents it from being larger. Thus $U$ is a component of ${\rho }_B(x)$, or $U\subseteq {\sigma }_B(x)$.

The complement of a compact planar set has a unique unbounded component: outside a large disc lies a connected exterior, and every unbounded component meets that exterior. The Neumann series for $|\lambda |>\Vert x\Vert$ converges in both algebras, so this exterior belongs to ${\rho }_B(x)$. Part i retains the whole component of ${\rho }_A(x)$ containing it. That retained component is unbounded and therefore is also the unique unbounded component of ${\rho }_B(x)$. The two unbounded components coincide.

Suppose $\lambda \in \partial {\sigma }_B(x)$. If $\lambda \in {\rho }_A(x)$, choose a connected open disc about it contained in ${\rho }_A(x)$. On that disc the clopen subset ${\rho }_B(x)$ is either the whole disc or empty. Neither possibility permits a boundary point of ${\sigma }_B(x)$, so $\lambda \in {\sigma }_A(x)$. But ${\sigma }_A(x)\subseteq {\sigma }_B(x)$, so a point in the interior of ${\sigma }_A(x)$ is in the interior of ${\sigma }_B(x)$. Our point cannot be such an interior point. ConsequentlyThis is boundary inclusion of spectra in a closed unital subalgebra, established from the preceding arguments rather than quoted.

The exponential series is absolutely norm-convergent, since ${\sum }_{n\ge 1}\Vert x^n\Vert /n!\le e^{\Vert x\Vert }-1$. Completeness defines the Banach algebra exponential $e^x\in A$. If $xy=yx$, absolute convergence permits rearrangement of the product series, and the binomial identity givesTaking $y=-x$ yields $(e^x{)}^{-1}=e^{-x}$.

For the spectral hypothesis, put $K={\sigma }_A(u)$. Its complement has an open, hence path-connected, unbounded component containing $0$. Join $0$ to an exterior point by a simple polygonal arc in that component, and continue it to infinity, avoiding $K$. Removing this polygonal slit gives a simply connected open neighborhood $\Omega$ of $K$ on which $z$ has a branch of the complex logarithm $L(z)$. The holomorphic functional calculus therefore defines $x=L(u)$ and its composition rule givesThe relevant slit is chosen through the resolvent, not assumed to be a fixed negative-real-axis cut. This is the logarithm from a spectral slit in a Banach algebra.

Let $H$ be the subgroup of finite products of exponentials. It is a subgroup because $e^a{e}^{-a}=1$ and reversing a product gives its inverse. Each product is joined to $1$ by $t\mapsto e^{t{a}_1}\cdots e^{t{a}_m}$, so $H\subseteq G_0$. Conversely, if $\Vert v-1\Vert <1$, the convergent logarithm seriessatisfies $e^{\log v}=v$, by the scalar analytic identity and functional calculus. Thus $H$ contains a neighborhood of $1$. Its translates make $H$ open, and all other cosets are open too, making $H$ closed. Connectedness forces $G_0\subseteq H$. Finally $g{e}^a{g}^{-1}=e^{ga{g}^{-1}}$, so $H$ is normal. Thereforean open-and-closed normal subgroup. This is the identity component of Banach-algebra invertibles; finite products are essential and are not asserted to be single exponentials.

An algebra involution is a conjugate-linear map $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $(a^{\ast }{)}^{\ast }=a$; on a unital algebra it fixes $1$. In the Banach-star convention it is continuous, although the following spectral identity uses only its algebraic properties. An element $a$ is invertible precisely when $a^{\ast }$ is: its inverse is $(a^{-1}{)}^{\ast }$. Applying this to $a=\lambda 1-x$ provesFor a nonunital algebra, extend the involution to its unitization of an algebra by $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$ and use the spectrum in that unitization; the same argument proves the identity.

A unital C-star algebra is a complex Banach algebra with an involution satisfying the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A Normal element of a C-star algebra satisfies $x^{\ast }x=x{x}^{\ast }$. The norm identity for the nonzero identity implies $\Vert 1\Vert =1$.

Submultiplicativity and the C-star identity give $\Vert x{\Vert }^2=\Vert x^{\ast }x\Vert \le \Vert x^{\ast }\Vert \Vert x\Vert$. For $x\ne 0$ this implies $\Vert x\Vert \le \Vert x^{\ast }\Vert$. Apply the same argument to $x^{\ast }$ and use $x^{\ast \ast }=x$ for the reverse inequality; $x=0$ is immediate. Hence

For a self-adjoint C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $x$ is a normal C-star element, thenAll powers are normal C-star elements, so induction gives $\Vert x^{2^j}\Vert =\Vert x{\Vert }^{2^j}$. The general spectral radius formula now gives $r(x)=\Vert x\Vert$.

One can also see the needed growth estimate directly. For $R>r(x)$ the resolvent Cauchy coefficient formula givesThe formula follows by the Neumann series on a larger circle and contour deformation through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to \infty$ to get $\Vert x\Vert \le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the norm bound on the algebra spectrum. ThusThis proves spectral radius norm equality for normal elements from the norm identity.

A algebra character $\phi$ is nonzero, linear and multiplicative, with $\phi (1)=1$. Since $\phi (a)\in {\sigma }_A(a)$, it satisfies $|\phi (a)|\le \Vert a\Vert$ and is automatically continuous. If $h=h^{\ast }$ and $t\in \mathbb{R}$, then $(e^{ith}{)}^{\ast }=e^{-ith}$, so $e^{ith}$ is a Unitary element of a C-star algebra and has norm one. HenceUsing both signs of $t$ forces $\mathrm{Im}\phi (h)=0$. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ both self-adjoint C-star elements. Their algebra character values are real, and thereforeThis is the conjugation in the original PDF; the converted TeX loses the bar.

For the final unheaded request, let $B=C^{\ast }(1,x)\subseteq A$. Normality makes this a commutative unital C-star subalgebra. We first justify spectral permanence, so that we do not silently use the possibly larger algebra spectrum in a mere non-star subalgebra. A self-adjoint C-star element $h$ in a C-star algebra has real algebra spectrum: $e^{ith}$ has norm one, and $\lambda \in \sigma (h)$ implies $e^{it\lambda }\in \sigma (e^{ith})$ for each real $t$. The latter assertion follows from the factorization $e^{ith}-e^{it\lambda }1=(h-\lambda 1)g_t(h)$ with commuting factors. An invertible product of commuting factors would make $h-\lambda 1$ invertible. Both signs of $t$ then force $\lambda \in \mathbb{R}$.

A compact subset of the real line has connected planar complement. Apply Question 1 to the self-adjoint C-star element $h\in B$: all of ${\rho }_A(h)$ is its unbounded component and is retained in $B$. Thus ${\sigma }_A(h)={\sigma }_B(h)$. If $a\in B$ is invertible in $A$, both $a^{\ast }a$ and $a{a}^{\ast }$ are self-adjoint C-star elements and invertible in $A$, hence invertible in $B$. The elements $(a^{\ast }a{)}^{-1}{a}^{\ast }$ and $a^{\ast }(a{a}^{\ast }{)}^{-1}$ are respectively a left and a right inverse in $B$, so coincide and give its inverse. This proves spectral permanence for C-star algebras.

In a commutative unital complex Banach algebra, algebra spectra are the ranges of the characters. Indeed a noninvertible $b-\lambda 1$ lies in a maximal ideal. Such an ideal is closed, since its proper closure cannot contain an element sufficiently near $1$. The quotient is a complex Banach division algebra; nonemptiness of the algebra spectrum in that quotient makes each element scalar, producing an algebra character with value $\lambda$ at $b$. Conversely, an algebra character value cannot correspond to an invertible difference.

Apply this to $B$. The preceding involution result gives $\phi (x^{\ast })=\overline{\phi (x)}$, and spectral permanence applies also to $p(x,x^{\ast })$. ConsequentlyThus the second argument is conjugated as in the original PDF, not a second copy of $\lambda$. This is the spectral mapping rule for a star polynomial in one normal element.

The group $G(K)$ consists of the nonvanishing continuous functions, and $E(K)$ is a subgroup because multiplication is commutative: $e^g{e}^h=e^{g+h}$. The objective is a logarithm decomposition with finitely many integer obstructions, one for each bounded complementary component. No local connectivity of $K$ is assumed.

We use two elementary finite-neighborhood facts, which make the outline independent of the general Jordan theorem. Given a neighborhood $V$ of $K$, take a finite grid neighborhood $N\subset V$ with $K\subset \mathrm{int}N$. It can be chosen to be a finite union of polygonal surfaces with boundary. Its complement has finitely many bounded holes. Holes that lie in the same component of $\mathbb{C}\setminus K$ can be joined by thin polygonal corridors disjoint from $K$, which are removed from $N$. A hole lying in the unbounded component is similarly joined to the exterior. For the remaining finitely many components, corridors may be continued to include their chosen representatives $a_i$. Small polygonal adjustments preserve $K\subset \mathrm{int}N$ and $N\subset V$. Thus every remaining bounded hole lies in a distinct bounded component of $\mathbb{C}\setminus K$ and contains its representative. One may also arrange this for any prescribed finite list of bounded components by first keeping their representative points out of the grid neighborhood. All paths used are compact and disjoint from $K$, so sufficiently thin corridors remain disjoint from $K$.

For such a finite polygonal $N$, winding indices identify the homology group $H_1(N;\mathbb{Z})$ with the free abelian group on its bounded holes. Here is a finite-chain explanation. Triangulate a large disc with $N$ as a subcomplex. Every integral one-cycle bounds a two-chain in the disc. Its filling coefficient is constant on each component of the complement of $N$, is zero on the exterior, and is the winding index there. If all bounded-hole indices vanish, the filling is supported in $N$, so the cycle already bounds in $N$. Conversely the sum of the oriented faces in one hole has its chain boundary in $N$ and has index one in that hole and zero in the others. Thus cycles ${\gamma }_j$ can be chosen with $\mathrm{wind}({\gamma }_j,a_i)={\delta }_{ij}$. These can be integral sums of boundary curves, which matters when components of $N$ are nested. This proves the needed statement using finite triangulation, not separation of an arbitrary simple closed curve. Together these are finite-hole neighborhoods of a planar compact set.

Now extend $f\in G(K)$ to a continuous $F$ near $K$, using the Tietze extension theorem on its real and imaginary parts. Since ${\min }_K|f|>0$, shrink the neighborhood until $F$ has no zeros, and choose $N$ within it as above. The winding of $F$ on the cycles ${\gamma }_j$ defines integers $k_j$. Setwhere only the finitely many represented holes are used. The winding of $F/r$ on every cycle of $N$ is zero. On each component choose a local complex logarithm at a base point and continue it along paths. Zero winding number around every loop makes the continued value independent of the path; local logarithm charts make it continuous. The continuous logarithm lifting criterion therefore gives $F/r=e^g$ on $N$, and restriction gives $f=e^{g{|}_K}r{|}_K$. Thus $G(K)=E(K)G(\mathbf{a})$, even when $K$ has infinitely many holes: each individual function uses finitely many factors.

To prove uniqueness, suppose a finite product $r=\prod (z-a_i{)}^{k_i}$ equals $e^g$ on $K$. Extend $g$ continuously to a neighborhood. There $r{e}^{-g}$ is close to $1$ after shrinking, so it has its small-branch logarithm. Hence $r$ itself has a continuous function giving a complex logarithm on that neighborhood. Choose $N$ there detecting all the finitely many $a_i$ as separate holes. Its dual cycles give $\mathrm{wind}(r\circ {\gamma }_j,0)=k_j$, while a function with a logarithm has zero winding number. Therefore every $k_j=0$. This proves both freeness of the listed rational functions and $E(K)\cap G(\mathbf{a})=\{1\}$. All subgroups commute, soThis is planar logarithm factorization. If there are no bounded components the second factor is trivial.

For the Jordan curve deduction, let $J$ be homeomorphic to the circle. Parametrize it by $\gamma :[0,1]\to J$ with just the two endpoints identified. For $f\in G(J)$, lift a logarithm of $f\circ \gamma$ along the interval. Its endpoint increment is $2\pi in$. The integer $n$ is additive, every integer is attained by a function with $f(\gamma (t))=e^{2\pi int}$, and $n=0$ precisely when the lift descends to a logarithm on $J$. Hence $G(J)/E(J)\cong \mathbb{Z}$. The factorization says this is free abelian on the bounded complementary components; comparison of ranks, or tensoring with $\mathbb{Q}$, shows that exactly one bounded component exists. There is also the unique unbounded component, so the complement has exactly two components.

We still need the common-boundary assertion. A proper closed arc has trivial exponential quotient because every nonvanishing function on an interval admits a logarithm. Factorization consequently makes the complement of such an arc connected, and since it is open it is path-connected. Fix $p\in J$ and any neighborhood $V$ of $p$, and choose a small open arc $A\subset J\cap V$ containing $p$. The remainder $J\setminus A$ is a proper closed arc. Join a point in each of the two complementary components by a path avoiding $J\setminus A$. This path must meet $J$, and every such meeting lies in $A$. The segment before its first meeting lies in the first region, and the segment after its last meeting lies in the second; continuity makes both regions meet $V$. Thus every point of $J$ lies on both boundaries. Conversely either boundary is contained in the closed set $J$. There are exactly two regions, one bounded and one unbounded, with common boundary $J$, the Jordan curve theorem. This is Jordan separation from the exponential quotient; it does not assume the theorem in the factorization proof.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.