Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 (C4.1, DMAC 2464, DMPC 2464)

A subset is relatively compact in $M$ when its closure in $M$ is compact.

A subset is totally bounded if for every $\epsilon >0$ it has a finite cover by balls of radius $\epsilon$. If its closure is compact, the cover by all such balls has a finite subcover, proving total boundedness.

Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as ${\sigma }_r$. These sets are infinite and nested, and two terms indexed by ${\sigma }_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in {\sigma }_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the closure as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the closure has the same limit. Sequential compactness therefore proves total boundedness implies relative compactness in a complete ambient space.

The Arzelà-Ascoli theorem says a subset of $C[0,1]$ is relatively compact exactly when it is uniformly bounded and equicontinuous. Pointwise boundedness together with equicontinuity is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\le |t-s|$, uniformly over the family, while $\Vert f{\Vert }_{\infty }\le 1$. Thus this family is relatively compact.

For $V$, the image of the unit ball has sup norm at most one and derivative $f$ of sup norm at most one; it lies in that family, so $V$ is compact.

For $W$, take $f_n=n{1}_{[0,1/n]}$, each of $L^1$ norm one. Then $W{f}_n(t)=\min (nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus $W$ is not compact. The same integration formula has different compactness properties because the domain unit balls impose different controls.

A sublinear functional on a real vector space satisfies $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for every $t\ge 0$. The dominated real Hahn-Banach theorem says that a linear functional on a subspace satisfying $f\le p$ extends to a linear functional on the whole space still dominated by $p$.

The limit-superior functional is finite on bounded sequences, positively homogeneous, and subadditive, since for every tail $\sup ({\alpha }_n+{\beta }_n)\le \sup {\alpha }_n+\sup {\beta }_n$. Passing to the tail limits proves sublinearity. On the subspace $c$ of convergent real sequences, the ordinary limit is linear and equals $p$. Hahn-Banach therefore extends it to $\phi$ on ${\ell }^{\infty }$ with

The extension already agrees with the usual limit on $c$. Alternatively, applying the domination inequality to both $x$ and $-x$ givesWhen the sequence converges, both endpoints are its limit, proving $\phi (x)=\lim {\alpha }_n$. These bounds also give $\Vert \phi \Vert =1$ and positivity on nonnegative sequences, but do not by themselves assert shift invariance.

In the infimum defining $q$, choose $k=1,j_1=1$. A finite shift leaves the limit superior unchanged, so

For $x-Sx$, choose the consecutive shifts $j_r=r$, $1\le r\le k$. The average telescopes:Its absolute value is at most $2\Vert x{\Vert }_{\infty }/k$. Taking the limit superior and then the infimum over $k$ proves $q(x-Sx)\le 0$.

For the requested functional, use the allowed sublinearity of $q$. On convergent sequences, every finite shifted average has the same limiting value, so $q$ equals the ordinary limit there. Hahn-Banach extends this limit to a linear $\psi$ satisfying $\psi \le q\le p$. Apply domination to $x-Sx$ and to $(-x)-S(-x)$ to obtain respectively $\psi (x-Sx)\le 0$ and $-\psi (x-Sx)\le 0$. HenceThis is a Banach limit; the two-sided limit bounds also show positivity, norm one, and agreement with ordinary limits.

The weak-star topology on $X^{\ast }$ is the coarsest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous. Finite intersections of evaluation inequalities form a neighborhood basis. Distinct functionals differ on some $x$, so disjoint disks around their evaluation values give disjoint neighborhoods; thus the topology is Hausdorff. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this topology.

If $f\in X^{\ast \ast }$ is weak-star continuous, some $x_1,\dots ,x_n$ and $\epsilon >0$ make $|f(\varphi )|<1$ whenever $|\varphi (x_i)|<\epsilon$ for every $i$. If all evaluations vanish, every scalar multiple of $\varphi$ lies in this neighborhood, forcing $f(\varphi )=0$. Thus $f$ vanishes on the kernel of the map $E(\varphi )=(\varphi (x_1),\dots ,\varphi (x_n))$. It factors linearly through the finite-dimensional range of $E$. Extending that factor algebraically to ${\mathbb{K}}^n$ gives scalars $a_i$ withConsequently $f=J_{X}x$ for $x={\sum }_i{a}_i{x}_i$.

For the two induced topologies, let $B$ be the common norm-closed unit ball and write ${\tau }_j=\sigma (Y,X_j)$. Under either topology $B$ is compact by Banach-Alaoglu, and the ambient space is Hausdorff. If the two topologies are comparable, the identity map from $B$ with the finer topology to $B$ with the coarser one is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism, since it carries closed compact sets to closed sets. Thus in the comparable case the topologies agree on $B$, giving alternative (a).

If the two ambient topologies are not comparable, neither is contained in the other. By the definition of inclusion of topologies, there is a ${\tau }_1$-open set absent from ${\tau }_2$, and a ${\tau }_2$-open set absent from ${\tau }_1$. These are the requested $U_1,U_2$, giving alternative (b). Together with the comparable case, this proves that at least one of the two alternatives always holds; it does not assume that the alternatives are mutually exclusive.

TakeIt is evaluation against the first coordinate vector of $c_0$, hence continuous for $\sigma ({\ell }^1,c_0)$. The preceding continuous-dual result says that a functional continuous for $\sigma ({\ell }^1,c)$ must have coefficient sequence $({\gamma }_{\infty },{\gamma }_1,{\gamma }_2,\dots )$ for a convergent sequence $\gamma$. To represent ${\alpha }_1$, it would require ${\gamma }_{\infty }=1$ but every ${\gamma }_j=0$, impossible.

Directly, $e_1-e_n\to 0$ in $\sigma ({\ell }^1,c)$ because its pairing is ${\gamma }_{\infty }-{\gamma }_{n-1}\to 0$, whereas ${\psi }_1(e_1-e_n)=1$ for $n\ge 2$.

TakeThe series defines a bounded linear functional of norm one. It corresponds under ${\Phi }_2$ to the constant sequence ${\gamma }_j=1$, so is $\sigma ({\ell }^1,c)$-continuous. Its coefficient sequence is constantly one, which is not in $c_0$, so the continuous-dual result makes it noncontinuous for $\sigma ({\ell }^1,c_0)$. Directly $e_n\to 0$ in that topology but ${\psi }_2(e_n)=1$.

The open mapping theorem says that a bounded surjective linear map $T:E\to F$ between Banach spaces is open. Here is a proof using only Baire and completeness. Surjectivity gives $F={\bigcup }_{n\ge 1}T(n{\overline{B}}_E)$, hence also a cover by their closures. Baire makes one such closure contain a ball $B(y_0,r)$. Taking differences gives $B(0,r)\subseteq \overline{T(2n{\overline{B}}_E)}$. Scaling yields some $\delta >0$ withIf $\Vert y\Vert <\delta /2$, recursively use scaled versions of this inclusion to choose $x_k$ with $\Vert x_k\Vert \le 2^{-k}$ and residual $\Vert y-T{\sum }_{j=1}^k{x}_j\Vert <\delta 2^{-k-1}$. The sum converges in $E$ to an element of norm at most one, whose image is $y$. Thus $B(0,\delta /2)\subseteq T({\overline{B}}_E)$, and scaling gives $B(0,\delta /4)\subseteq T(\frac{1}{2}{\overline{B}}_E)\subseteq T(B_E)$. Translating and scaling proves openness on every open set. This proof does not use the closed graph or inverse mapping theorem.

Apply it to the evaluation map $E:X\to {\ell }^{\infty }$, $Ef=(f(z_n){)}_n$. It is bounded with norm at most one and onto by hypothesis. Hence its image of the unit ball contains a ball about zero, yielding a constant $c$ and, for every $\alpha$, an interpolant $g$ withFor completeness, scale a ball of radius $\eta$ inside $E(B_X)$ by $2\Vert \alpha {\Vert }_{\infty }/\eta$ to obtain $c=2/\eta$; zero data use $g=0$.

For each $n$, interpolate the data which are one at index $n$ and zero elsewhere, obtaining $g_n$ of norm at most $c$. The supplied factorization property gives $g_n=k_n{h}_n$ with $\Vert k_n{\Vert }_{\infty }=\Vert g_n{\Vert }_{\infty }\le c$. Evaluation at $z_n$ gives $1=|k_n(z_n)h_n(z_n)|\le c|h_n(z_n)|$, soThe three enumerated properties of $h_n$ are supplied hypotheses for this argument, rather than additional requests to prove them.

For the real lattice version of the Stone-Weierstrass theorem, let $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. For any $x,y\in \Omega$, separation of points and constants give $h_{xy}\in L$ matching $f$ at both $x$ and $y$: take an affine combination of a separating function, and for $x=y$ take a constant. Fix $x$. The open sets where $h_{xy}>f-\epsilon$ cover $\Omega$ as $y$ varies. Compactness gives finitely many such functions; their maximum $g_x$ belongs to the lattice $L$, satisfies $g_x>f-\epsilon$ everywhere, and has $g_x(x)=f(x)$. The open sets where $g_x<f+\epsilon$ now cover $\Omega$ as $x$ varies. Choose finitely many and take their minimum $h\in L$. Then $f-\epsilon <h<f+\epsilon$ everywhere. Thus $L$ is uniformly dense in the real continuous functions.

For the complex *-algebra, its real subalgebra $A_{\mathbb{R}}$ of Hermitian C-star elements contains the real and imaginary parts of every member. It separates points, because at least one of those parts distinguishes any pair distinguished by a complex function. Its uniform closure is a unital real subalgebra and is a lattice by the allowed result. The real theorem makes this closure all of $C_{\mathbb{R}}(\Omega )$. Approximating the real and imaginary parts separately proves $A$ is uniformly dense in $C(\Omega )$.

On $[0,1]$, the algebra of polynomials in $t^2$ is unital, closed under conjugation, and separates points because $t\mapsto t^2$ is injective there. Choose $p(t^2)$ within $\epsilon /3$ of $a(t)$, and correct its endpoints byThis is still a polynomial in $t^2$ and has exactly the requested endpoint values. The correction has absolute value at most $\epsilon /3$, because it is a convex combination of two errors of that size. Therefore

The spectrum of an element is ${\sigma }_A(a)=\{\lambda \in \mathbb{C}:a-\lambda 1_A\text{ is not invertible in }A\}$. A Normal element of a C-star algebra satisfies $a^{\ast }a=a{a}^{\ast }$.

For a Hermitian C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. For normal $a$, commutation givesAll powers remain normal, so induction yields $\Vert a^{2^k}\Vert =\Vert a{\Vert }^{2^k}$. Taking the spectral-radius limit along these powers givesLet $\phi :A\to B$ be a C-star homomorphism, without assuming continuity or preservation of the ambient unit. Put $p=\phi (1_A)$, a projection in a C-star algebra. If $p=0$, then $\phi =0$. Otherwise its image lies in the unital corner $pBp$ and it is unital as a map into that corner. An inverse in $A$ maps to an inverse in the corner, so ${\sigma }_{pBp}(\phi (b))\subseteq {\sigma }_A(b)$. Apply this to the normal element $b=a^{\ast }a$ and use the C-star identity in both algebras:Thus every C-star homomorphism is norm non-increasing, including nonunital maps.

A Unitary element of a C-star algebra satisfies $u^{\ast }u=u{u}^{\ast }=1$. A projection in a C-star algebra satisfies $p=p^{\ast }=p^2$.

If $q=u^{\ast }u$ is a projection in a C-star algebra, the C-star identity gives $\Vert u(1-q){\Vert }^2=\Vert (1-q)q(1-q)\Vert =0$. Hence $u=uq=u{u}^{\ast }u$. Conversely $u{u}^{\ast }u=u$ implies $(u^{\ast }u{)}^2=u^{\ast }u$, and $u^{\ast }u$ is a Hermitian C-star element, so it is a projection in a C-star algebra. Taking adjoints of the triple-product equality gives the corresponding equality for $u^{\ast }$, provingFor nonzero $u$, let $p=u{u}^{\ast }$ and $q=u^{\ast }u$. These are projections, $u=pu=uq$, and $\Vert u\Vert =1$. The rectangular corner $pAq$ is closed because it is the range of the bounded idempotent map $a\mapsto paq$.

Define $\Theta :pAq\to qAq$ by $\Theta (a)=u^{\ast }a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $u{u}^{\ast }a=pa=a$ and $u^{\ast }ub=qb=b$ verify both compositions. Both maps are contractive because $\Vert u\Vert =\Vert u^{\ast }\Vert =1$, so $\Theta$ is an isometry. For the proposed multiplication and involution,Thus the operations correspond exactly to the ordinary operations in the closed C-star algebra $qAq$. This proves associativity, closure, all involution identities and the C*-norm identity at once. Moreover $\Theta (u)=q$, so the unit of $A_u$ is $u$. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^{\ast }a$ is an onto isometric *-isomorphism from $A_u$ to $A$.

First suppose $x$ is a positive linear functional, without assuming boundedness. Every Hermitian C-star element $h$ is a difference of positive elements, since $\Vert h\Vert 1\pm h\ge 0$. Thus $x(h)$ is real. For a Hermitian C-star element $h$ of norm at most one, $-1\le h\le 1$ gives $|x(h)|\le x(1)$.

For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The real part of $\zeta a$ is a Hermitian C-star element and has norm at most $\Vert a\Vert$. Since $x$ is real on Hermitian C-star elements, it respects real parts, givingThis proves boundedness and $\Vert x\Vert \le x(1)$. Evaluation at the unit gives the reverse inequality, so positivity implies $\Vert x\Vert =x(1)$.

Conversely let $x$ be bounded with $x(1)=\Vert x\Vert =c\ge 0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a Hermitian C-star element $h$ and real $t$, the C-star identity givesWriting $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\le t^2\Vert h{\Vert }^2$. Taking $t\to 0$ through positive and negative values forces $b=0$. Hence $x$ is real on Hermitian C-star elements. If $0\le a\le 1$, then $1-a$ has norm at most one, so $x(1-a)\le 1$ and $x(a)\ge 0$. Scaling proves nonnegativity on every positive element, in particular every $b^{\ast }b$. Therefore