Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 (Functional Analysis)

For an element $x$ of a unital complex Banach algebra $A$, its spectrum of an element isIf $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so the spectrum is bounded. If ${\lambda }_0$ is in the resolvent set, thenis invertible for $\lambda$ sufficiently close to ${\lambda }_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.

If the spectrum were empty, the resolvent $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^{\ast }$, the scalar entire function $f(R(\lambda ))$ would be bounded and therefore constant by Liouville theorem. It would be zero because of the limit at infinity. The Hahn-Banach theorem would then force $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Therefore ${\sigma }_A(x)$ is nonempty.

Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, soOn each connected component of $\mathbb{C}\setminus {\sigma }_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda |$. ConsequentlyThis is the spectrum in a closed unital subalgebra theorem.

For the bilateral shift $T$ on ${\ell }^1(\mathbb{Z})$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda |>1$ and to $T^{-1}$ for $|\lambda |<1$ givesFor $|\lambda |=1$, normalize the vector given by $x_n={\lambda }^{-n}$ on $-N\le n\le N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence

The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}{e}_0=e_{-1}$. Therefore $0\in {\sigma }_{\mathcal{A}}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb{C}\setminus \mathbb{T}$, giving

Goldstine theorem states that the canonical image $J(B_X)$ of the closed unit ball of a normed vector space $X$ is weak-star dense in $B_{X^{\ast \ast }}$.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact in the weak-star topology. To prove it, map each $f\in B_{X^{\ast }}$ to its values inEvery factor is compact, so Tychonoff theorem makes $P$ compact. The image of $B_{X^{\ast }}$ is cut out by the closed linearity conditionsand is therefore closed in $P$. The product topology restricted to this image is exactly pointwise convergence on $X$, namely the weak-star topology. Hence the ball is compact.

This statement is true. If $(x_n)$ is dense in $B_X$, then on $B_{X^{\ast }}$ the weak-star topology is induced by a metric such asThe Banach-Alaoglu theorem makes this metric space compact, and every compact metric space is sequentially compact.

This statement is also true. The preceding compact metrizable ball $B_{X^{\ast }}$ is separable. Each ball $n{B}_{X^{\ast }}$ is therefore weak-star separable, andThe union of countable dense subsets of these balls is a countable weak-star dense subset of $X^{\ast }$.

This statement is false. Let $I$ have the cardinality of the continuum and take $X={\ell }^1(I)$, which is nonseparable because its unit coordinate vectors form an uncountable discrete set. Its dual ball with the weak-star topology is the product cubeThe Hewitt–Marczewski–Pondiczery theorem says that a product of at most continuum many separable spaces is separable, so this cube is separable. Since $X^{\ast }={\bigcup }_{n\ge 1}n{B}_{X^{\ast }}$, the whole dual is weak-star separable despite $X$ being nonseparable.

For $p\ge 2$, integrate the stated scalar inequality to obtain the Clarkson inequalityIf $f,g$ belong to the unit ball and $\Vert f-g{\Vert }_p\ge \epsilon$, thenThus $L^p[0,1]$ is uniformly convex.

Now let $X$ be uniformly convex. It is enough to show that every $\Phi \in S_{X^{\ast \ast }}$ lies in the canonical image of $X$. Given $\epsilon >0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^{\ast }}$ with $\Phi (f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, thenso $\Vert x-y\Vert <\epsilon$. By Goldstine theorem, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_{\alpha })$; completeness gives $x_{\alpha }\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is reflexive.

For $p\ge 2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p{)}^{\ast }=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ for which every bounded linear functional in $X^{\ast }$ is continuous. Thus every $f\in X^{\ast }$ is weakly continuous. Conversely, if a linear functional $f$ is weakly continuous at zero, some basic weak neighbourhood gives $f_1,\dots ,f_n\in X^{\ast }$ and $\epsilon >0$ such thatIt follows that ${\bigcap }_j\ker f_j\subseteq \ker f$, and elementary linear algebra then gives $f\in \mathrm{span}\{f_1,\dots ,f_n\}\subseteq X^{\ast }$.

To prove Mazur theorem, let $K$ be norm-closed and convex and let $x\notin K$. The Hahn-Banach separation theorem strictly separates $x$ from $K$ by some member of $X^{\ast }$. The corresponding open half-space is weakly open, contains $x$, and misses $K$. Thus $K$ is weakly closed.

If $X$ is reflexive, the Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and the canonical identification transports this to weak compactness of $B_X$. Conversely, if $B_X$ is weakly compact, then $J(B_X)$ is weak-star compact and hence weak-star closed in $B_{X^{\ast \ast }}$. Goldstine theorem says it is weak-star dense there, so it equals $B_{X^{\ast \ast }}$ and $X$ is reflexive.

When $X$ is reflexive, weak and weak-star topologies coincide on $X^{\ast }$, so Banach–Alaoglu makes $B_{X^{\ast }}$ weakly compact and $X^{\ast }$ is reflexive. If $Y\subseteq X$ is closed, then $B_Y$ is a weakly closed subset of $B_X$, hence weakly compact. The quotient map sends a suitable weakly compact ball of $X$ onto the unit ball of $X/Y$, which is therefore weakly compact. Thus $Y$ and $X/Y$ are reflexive as well.

For each $n$, define $J{x}_n:X^{\ast }\to \mathbb{R}$ by $J{x}_n(f)=f(x_n)$. Weak convergence makes $(J{x}_n(f){)}_n$ bounded for every $f\in X^{\ast }$. The Uniform boundedness principle gives

Let $K$ be the closed convex hull of the $x_n$. Given a sequence in $K$, approximate its terms in norm by finite convex combinations of the $x_n$. A diagonal subsequence makes every coefficient converge. Any loss of total coefficient mass is assigned to zero, which belongs to $K$ by Mazur theorem because $x_n\rightharpoonup 0$. Since $f(x_n)\to 0$ for every $f\in X^{\ast }$, splitting each sum into a finite head and a uniformly small tail proves weak convergence of this subsequence to the corresponding convex combination. Hence $K$ is weakly sequentially compact and, by the stated theorem, weakly compact.

DefineIt is bounded because $(x_n)$ is norm bounded, and its values lie in $c_0$ because $x_n\rightharpoonup 0$. Its adjoint-on-preduals map isIf $K$ had nonempty norm interior, then $K-K$ would contain a ball about zero. The quantitative open-mapping argument applied to the convex combinations above would make $T_{\ast }$ surjective, and hence make $T$ bounded below. Its range $R$ would be a closed infinite-dimensional subspace of $c_0$ whose unit ball is compact for coordinatewise convergence, since it lies in the coordinatewise compact image of a weak-star compact ball of $X^{\ast }$.

By the stated structural theorem, $R$ contains a closed subspace isomorphic to $c_0$. The bounded partial sums of the image of the standard $c_0$ basis would then have a coordinatewise convergent subnet. Uniform boundedness turns coordinatewise convergence in $c_0$ into weak convergence, and norm-closed subspaces are weakly closed. Pulling the limit back would make the partial sums of the standard basis converge weakly in $c_0$, impossible because their coordinate values force the putative limit to be the constant-one sequence. This contradiction proves that $K$ has empty norm interior.

By the spectral radius formula, $r(T)<1$. ChooseThe formula gives a constant $C$ such that $\Vert T^n\Vert \le C{r}^n$ for every $n$. Since $f$ is holomorphic on the unit disc, the Cauchy estimate on the circle of radius $s$ gives $|a_n|\le M_s{s}^{-n}$. ThereforeThus ${\sum }_{n\ge 0}{a}_n{T}^n$ converges absolutely in operator norm.

If $T$ is normal, then $S=T-I$ is normal and $\sigma (S)=\{0\}$. The C-star identity and normality imply $\Vert S^n\Vert =\Vert S{\Vert }^n$. The spectral radius formula consequently givesso $T=I$.

Normality is essential. On ${\ell }^2$, let $N$ send $e_2$ to $e_1$ and every other basis vector to zero. Then $N\ne 0$ but $N^2=0$, sowhile $I+N\ne I$.

The Commutative Gelfand--Naimark theorem says that every commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C(\Delta (A))$, where $\Delta (A)$ is its compact character space and the map is the Gelfand transform

Indeed, maximal ideals give enough characters to identify $\sigma (a)$ with the range of $\hat{a}$. The C-star identity and the spectral radius formula giveso $\Gamma$ is isometric and has closed range. Characters send $a^{\ast }$ to $\overline{\chi (a)}$, so the range is self-conjugate; it contains constants and separates distinct characters. The complex Stone-Weierstrass theorem makes the range dense in $C(\Delta (A))$, and closedness makes it all of that algebra. This proves the theorem.

An element $x$ of a C-star algebra is positive when it is self-adjoint and $\sigma (x)\subseteq [0,\infty )$. Consider the commutative C-star subalgebra $C^{\ast }(1,x)$. Under its Gelfand–Naimark isomorphism, $x$ becomes a nonnegative continuous function $\hat{x}$. The function $\sqrt{\hat{x}}$ is continuous and nonnegative, so its inverse image $y$ is positive and satisfies

Let $p_C(x)=\inf \{t>0:x\in tC\}$ be the Minkowski functional of $C$. Openness, convexity, and $0\in C$ make $p_C$ sublinear, with $p_C(x)<1$ for $x\in C$ and $p_C(x_0)\ge 1$. Define a linear functional on $\mathbb{R}{x}_0$ by $f(t{x}_0)=t{p}_C(x_0)$. The real Hahn-Banach theorem extends it to $X$ with $f\le p_C$. Hence

Apply this separation to the open ball of radius $\Vert x_0\Vert$ and rescale to obtain $f\in S_{X^{\ast }}$ with $f(x_0)=\Vert x_0\Vert$. For a closed subspace $Y$, apply it toThe separator must vanish on $Y$ because $C$ contains every translate along $Y$, and normalization gives

Let $F\subseteq X^{\ast }$ be finite-dimensional and $\Phi \in B_{X^{\ast \ast }}$. ConsiderIf $\Phi {|}_F$ did not belong to $S((1+\epsilon )B_X)$, finite-dimensional strict separation would produce some $f\in F$ withcontradicting $\Vert \Phi \Vert \le 1$. Thus there is $x\in X$ with $\Vert x\Vert <1+\epsilon$ and $f(x)=\Phi (f)$ for every $f\in F$.

Given a basic weak-star neighbourhood of $\Phi$ in $B_{X^{\ast \ast }}$, apply this result to the finite-dimensional span of its defining functionals and then replace $x$ by $x/(1+\epsilon )$. As $\epsilon \downarrow 0$, the resulting points of $J(B_X)$ enter that neighbourhood. Hence $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$, proving Goldstine theorem.

Identify $X$ with its canonical image in $X^{\ast \ast }$ and putFor $x\in S_X$ and $\lambda \in \mathbb{R}$, if $|\lambda |\le 1/(d+1)$ thenIf $|\lambda |\ge 1/(d+1)$, thenTherefore $d(x,\mathrm{span}\{\Phi \})\ge c$.

The restriction of $x\in X^{\ast \ast }$ to $\ker \Phi \subseteq X^{\ast }$ has normwhere the first equality is the Hahn-Banach distance formula and $(\ker \Phi {)}^{\perp }=\mathrm{span}\{\Phi \}$. Scaling from $S_X$ givesfor every $x\in X$. Hence $\ker \Phi$ is $c$-norming for $X$.