Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 / A

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.