Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 10 (Banach Algebras)

Adjoin the identity to the bounded multiplicatively closed set, putting $S^{\prime }=S\cup \{1\}$ and $M={\sup }_{s\in S^{\prime }}\Vert s\Vert <\infty$. Define an auxiliary norm on the underlying Banach space byThe identity ensures $\Vert a\Vert \le p(a)\le M\Vert a\Vert$, so $p$ and the given norm are equivalent norms. For $s\in S$, closure under multiplication gives $p(sa)\le p(a)$. Now use the operator norm of left multiplication on this renormed space:Left multiplication is faithful, since $a1=a$. It follows that this is a norm, its triangle and scalar laws follow from the operator norm, and $L_{ab}=L_a{L}_b$ proves submultiplicativity. Also $L_1$ is the identity, so $\Vert 1{\Vert }_1=1$, and the contraction property gives $\Vert s{\Vert }_1\le 1$ for every $s\in S$. Finally,For the upper bound use $p(ab)\le M\Vert a\Vert \Vert b\Vert \le M\Vert a\Vert p(b)$. These bounds prove equivalence and completeness, so this is the required unital algebra norm. This proves renorming a bounded multiplicative semigroup; the auxiliary norm is submultiplicative but need not give the identity norm one, which is why the operator norm is used.

The spectrum of an element is $\sigma (x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is not invertible in }A\}$. The spectral radius formula isIf $r(x)<1$, choose $r(x)<q<1$. Eventually $\Vert x^m{\Vert }^{1/m}<q$, hence $\Vert x^m\Vert <q^m$. The finitely many earlier powers are bounded too. In particular the power set is bounded; indeed $x^m\to 0$ in norm.

For the commuting family, put $d_j=r(x_j)+\epsilon /2$ and $y_j=x_j/d_j$. Each $r(y_j)<1$, so $M_j={\sup }_{m\ge 0}\Vert y_j^m\Vert <\infty$. Every element of the generated multiplicative semigroup can be reordered as $y_1^{m_1}\cdots y_n^{m_n}$, and its norm is at most ${\prod }_j{M}_j$. The first construction therefore gives one equivalent unital algebra norm with $\Vert y_j{\Vert }_0\le 1$ for all $j$, whenceThis is simultaneous spectral-radius renorming; commutativity is what bounds all mixed products.

For commuting $a,b$, apply this construction to both with error $\epsilon$. Equivalent algebra norms leave invertibility and spectrum unchanged, and the spectral radius formula gives $r(c)\le \Vert c{\Vert }_0$. ThusLet $\epsilon \downarrow 0$ to obtain $r(ab)\le r(a)r(b)$ and $r(a+b)\le r(a)+r(b)$.

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.

A primitive ideal is the kernel of an algebraically irreducible unital representation, equivalently the annihilator of a module for a nonzero simple module. Irreducibility here excludes all proper nonzero invariant linear subspaces, not just closed ones. In a Banach algebra the simple modules can be represented as $A/L$ with $L$ a maximal left ideal. Such an ideal is closed: if its closure were all of $A$, it would contain an element close enough to $1$ to be invertible by the Neumann series, contradicting properness. Maximality then makes its closure itself. Thus the quotient carries a Banach-space norm and a bounded left action.

Let $P$ be a maximal proper two-sided ideal. The quotient ring $R=A/P$ is nonzero, unital and simple as a ring. By the Zorn lemma, choose a maximal left ideal $L$ of $R$. The simple module $R/L$ has an annihilator which is a proper two-sided ideal of $R$. Simplicity forces that annihilator to be zero. Its pullback to $A$ is exactly $P$, proving every maximal proper two-sided ideal is primitive. If one wants the Banach representation explicitly, realize the same module as the quotient of $A$ by the pullback maximal left ideal, which is closed by the preceding argument.

Put $K={\bigcap }_{L}L$, where $L$ ranges over all maximal left ideals. For each such $L$, the action on the simple module $A/L$ has a primitive ideal $P_L$ as its kernel. An element of that kernel annihilates $1+L$, so $P_L\subseteq L$. The intersection $J$ of all primitive ideals is consequently contained in every $L$, hence $J\subseteq K$.

Conversely, take $x\in K$ and any nonzero simple module $E$. For every nonzero $v\in E$, simplicity makes the evaluation map $A\to E$, $a\mapsto av$, surjective. Its kernel is a maximal left ideal, so it contains $x$. Thus $xv=0$ for every nonzero $v$, and also for zero. The element $x$ therefore belongs to the annihilator of every simple module, hence to every primitive ideal. This provesSince the primitive ideals are two-sided ideals, this argument also makes $J$ a two-sided ideal, despite the one-sided ideals in this alternative description.

Suppose $x\in J$ and put $u=1+yx$. If the left ideal $Au$ were proper, it would lie in a maximal left ideal $L$. But $yx\in J\subseteq L$, so $u\in L$ would imply $1\in L$. Thus $Au=A$, and $u$ has a left inverse $v$ with $vu=1$.

A left inverse alone is insufficient in a general operator algebra. Here $v=1-vyx$ differs from the identity by an element of $J$. The same argument supplies a left inverse $w$ of $v$. Then $w=w(vu)=(wv)u=u$, hence $uv=wv=1$. So $u$ is genuinely invertible.

Conversely, if $x\notin J$, part (i) gives a maximal left ideal $L$ with $x\notin L$. Maximality implies $L+Ax=A$, so $1=l+ax$ for some $l\in L,a\in A$. Therefore $1-ax=l\in L$ cannot be invertible, since a proper left ideal contains no invertible element. Taking $y=-a$ disproves the required universal invertibility. We have proved the unit criterion for the Jacobson radical in the requested form:

If $x\in J$, part (ii) makes $1-x/\lambda$ invertible for every nonzero scalar $\lambda$. Thus $\sigma (x)\subseteq \{0\}$ and $r(x)=0$: it is a quasinilpotent element. Conversely, let $I$ be a two-sided ideal consisting of quasinilpotent elements. For $x\in I$ and $y\in A$, $yx\in I$ has spectrum $\{0\}$, so $1+yx$ is invertible. Part (ii) implies $x\in J$. This proves $J$ is the greatest ideal contained in the quasinilpotent set, the Jacobson radical as greatest quasinilpotent ideal characterization.

A semisimple Banach algebra is one with $J=\{0\}$. For $B=\mathcal{B}(X)$, its natural action on $X$ is faithful and algebraically irreducible. Indeed, given $v\ne 0$ and any $w\in X$, the Hahn-Banach theorem gives a bounded functional $g$ with $g(v)=1$. The rank-one operator $z\mapsto g(z)w$ then maps $v$ to $w$. Any invariant subspace containing $v$ consequently contains every $w$. The faithful representation has kernel zero, so zero is primitive and $B$ is semisimple.

Nevertheless, choose a nonzero bounded functional $g$ and $0\ne x_0\in \ker g$, possible when $\dim X>1$. The nonzero rank-one operator $Tz=g(z)x_0$ satisfies $T^{2}z=g(z)g(x_0)x_0=0$. Hence $\sigma (T)=\{0\}$ and $r(T)=0$, although $T\ne 0$. This proves the quasinilpotent set of $B$ is not $\{0\}$; it is not the same as the radical.

For infinite-dimensional $X$, the finite-rank operators form a nonzero proper two-sided ideal: products with bounded operators are finite-rank, and the identity has infinite rank. Thus zero is not maximal. Even if maximal ideals are required to be closed, the compact operators give a nonzero proper closed ideal containing the finite-rank operators. The identity is not compact on an infinite-dimensional Banach space, since the Riesz lemma supplies a bounded sequence with pairwise distances bounded below. Therefore zero is primitive but not maximal in infinite dimension.

Assume the compact space is nonempty. Write $M_n={\sup }_K{f}_n$. Each $M_n$ is finite and attained, and the sequence decreases to a number $L\ge 0$. Since $f\le f_n$, certainly ${\sup }_{K}f\le L$.

For any $t<L$, the sets $E_n=\{x\in K:f_n(x)\ge t\}$ are nonempty closed sets, and $E_{n+1}\subseteq E_n$. The finite intersection property and compactness give a point in their intersection. At that point $f\ge t$, so ${\sup }_{K}f\ge t$. Let $t\uparrow L$; together with the reverse bound this provesThis is supremum convergence for decreasing continuous functions. The limit need not be continuous, so the conclusion does not assert the uniform convergence of the Dini theorem. It is precisely the interchange of these decreasing suprema and pointwise limit needed in the next part.

First establish the norm estimate behind the spectral-radius three-circle inequality. If $g$ is a holomorphic Banach space-valued function on a neighborhood of an annulus $r_0\le |z|\le r_1$, then for $r_0<r<r_1$ and $\theta =\log (r/r_0)/\log (r_1/r_0)$,Here is the underlying Hadamard three-circle theorem argument. For any bounded linear functional $\ell$ of norm at most one, $\ell (g)$ is scalar holomorphic. Its logarithmic modulus is subharmonic. Compare it on the annulus with the harmonic function affine in $\log |z|$ whose boundary values are the logarithms of the two boundary norm bounds. The maximum principle for subharmonic functions gives the interpolated bound inside. Using slightly larger positive bounds and then taking limits handles zeros. The Hahn-Banach theorem identifies the norm with the supremum over these scalar functionals, proving the displayed Banach-valued estimate.

Apply this estimate to $g(z)=f(z{)}^{2^n}$ and take the $2^n$-th root. On either boundary circle the nonnegative continuous functionsdecrease, because submultiplicativity gives $q_{n+1}\le q_n$. The spectral radius formula gives their pointwise limit $r(f(z))$. Part (i) passes their boundary suprema to the limit. With $M_r={\sup }_{|z|=r}r(f(z))$ we obtain the general estimateIn particular, take $r_0=R^{-1}$, $r=1$, $r_1=R$, so $\theta =1/2$. Since $r(f(1))\le M_1$,This proves the requested inequality without assuming continuity of the spectral radius on the circles.

For the final spectral-radius Liouville theorem, suppose $r(f(z))\le C$ on the whole plane. The case $C=0$ is immediate. Center circles at an arbitrary $a$, and put $M_a(t)={\sup }_{|z-a|=t}r(f(z))$. For $0<\delta <r<R$, the same argument givesIf $M_a(\delta )=0$, this forces $M_a(r)=0$. Otherwise let $R\to \infty$ to get $M_a(r)\le M_a(\delta )$. Upper semicontinuity, proved in 2(a), gives $M_a(\delta )\le r(f(a))+\epsilon$ for all sufficiently small $\delta$. Thus every $z$ satisfies $r(f(z))\le r(f(a))$. The center $a$ is arbitrary, so reversing the two points gives equality. Hence the spectral radius is constant. The algebra-valued function itself need not be constant; $f(z)=zN$ with a nonzero square-zero matrix $N$ is a useful example.

The functional's stipulated positivity initially means $f(z^{\ast }z)\ge 0$, without assuming continuity. For a Hermitian element of a C-star algebra $h$, the identityexpresses it as a real linear combination of two adjoint squares. Their functional values are real, so $f(h)$ is real. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ Hermitian. Complex linearity gives $f(x)=f(h)+if(k)$ and $f(x^{\ast })=f(h)-if(k)$. ThereforeThis is the Hermitian symmetry used in the Cauchy–Schwarz inequality for positive C-star functionals.

The inequality printed in this part is false for general positive functionals. In $A=M_2(\mathbb{C})$ take the state on a C-star algebra $f(a)=a_{22}$ and $x=y=E_{12}$. It is positive since $f(z^{\ast }z)=\Vert z{e}_2{\Vert }^2\ge 0$, and it has norm one. ButThus the literal requested inequality would say $1\le 0$.

The correct Cauchy–Schwarz inequality for positive C-star functionals keeps the orders consistent:To prove it, set $a=f(x^{\ast }x)$, $d=f(y^{\ast }y)$ and $b=f(x^{\ast }y)$. By part (i) and positivity,If $d>0$, choose $t=-\overline{b}/d$ to obtain $|b{|}^2\le ad$. If $d=0$, arbitrary choices of $t$ force $b=0$, and the inequality again follows. Replacing $x,y$ by $x^{\ast },y^{\ast }$ also yields the valid alternativeEither replacing the printed right-hand factors by $f(x^{\ast }x),f(y^{\ast }y)$, or replacing the left-hand product by $x{y}^{\ast }$, repairs the statement. No tracial hypothesis is supplied, so it cannot be silently used to exchange those products.

Apply the correct Cauchy–Schwarz inequality for positive C-star functionals to $x^{\ast }$ and $1$. Its left side is $|f(x){|}^2$, and its diagonal values are $f(x{x}^{\ast })$ and $f(1)$. Thus the printed conclusion of this part is valid:This does not depend on the false mixed-order version in the preceding part.

Put $c=f(1)\ge 0$. For a given $x$, let $M=\Vert x{\Vert }^2=\Vert x{x}^{\ast }\Vert$ by the C-star identity. For $\epsilon >0$ the binomial seriesconverges absolutely in the Banach algebra, because $\Vert x{x}^{\ast }/(M+\epsilon )\Vert <1$. Its coefficients are real and $x{x}^{\ast }$ is Hermitian, so $b_{\epsilon }^{\ast }=b_{\epsilon }$; multiplication of the absolutely convergent series gives $b_{\epsilon }^{\ast }{b}_{\epsilon }=(M+\epsilon )1-x{x}^{\ast }$. Positivity yields $f(x{x}^{\ast })\le (M+\epsilon )c$. Let $\epsilon \downarrow 0$. Part (iii) now givesThus $f$ is a bounded linear functional with $\Vert f\Vert \le c$. Since the nonzero identity in a C-star algebra has norm one, evaluating at it gives the reverse bound. HenceThis proves automatic continuity from the original algebraic positivity assumption, including the case $c=0$, when $f=0$.

For the converse, let $f$ be bounded and $\Vert f\Vert =f(1)=c$. If $c=0$, it is zero and positive. Suppose $c>0$. For any Hermitian $h$, the Banach algebra exponential $e^{ith}$ is unitary for real $t$, so its norm is one by the C-star identity. ConsequentlyWriting $f(h)=u+iv$ and squaring moduli gives $|f(e^{ith}){|}^2=c^2-2cvt+O(t^2)$. The bound for both signs of small $t$ forces $v=0$, so $f(h)$ is real.

For a Hermitian element $a$, repeated applications of the C-star identity give $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula therefore gives $\Vert a\Vert =r(a)$. Take $h=x^{\ast }x$; by the permitted spectral positivity assumption, $\sigma (h)\subseteq [0,\Vert h\Vert ]$. If $M=\Vert h\Vert >0$, affine spectral mapping and the just-proved Hermitian norm identity give $\Vert 1-h/M\Vert \le 1$. Hence $|c-f(h)/M|\le c$. Since $f(h)$ is real, this implies $f(h)\ge 0$. If $M=0$, $h=0$ and the result is immediate. Thus a bounded functional with $\Vert f\Vert =f(1)$ is positive.

For the spectral-value request, let $\lambda \in \sigma (x)$ and define a functional on the linear span of $1,x$ by $g(a1+bx)=a+b\lambda$. If $b\ne 0$, the scalar $a+b\lambda$ belongs to $\sigma (a1+bx)$: subtracting it from that element gives $b(x-\lambda 1)$, which is not invertible. The same norm bound is immediate for $b=0$. ThereforeThis also proves well-definedness if that span is one-dimensional. We have $g(1)=1$ and $\Vert g\Vert =1$. The complex Hahn-Banach theorem extends it to a functional $f$ on all of $A$ with norm one and $f(1)=1$. The converse just proved makes that extension positive. Thus there is a state with $f(x)=\lambda$, the spectral values attained by C-star states result.

Finally, for $x\ne 0$, write $x=h+ik$ with $h,k$ Hermitian. At least one of them is nonzero. Its norm equals its spectral radius, so compactness of its nonempty spectrum supplies a nonzero spectral value. Apply the preceding state construction to that Hermitian element. Part (i) makes $f(h),f(k)$ real; the selected one is nonzero, so $f(x)=f(h)+if(k)\ne 0$. This proves positive functionals separate every nonzero element, the states separate elements of a C-star algebra property. Looking only at $\sigma (x)$ would not prove this last step, since a nonzero quasinilpotent $x$ can have spectrum $\{0\}$.