Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 6 (Linear Analysis)

Start with the approximate surjectivity with geometric correction argument. For a fixed $y$, put $r_0=y$. Choose $x_j$ successively so thatInduction gives $\Vert r_j\Vert \le k^j\Vert y\Vert$ and $\Vert x_j\Vert \le R{k}^{j-1}\Vert y\Vert$. The series ${\sum }_j{x}_j$ is absolutely convergent in the Banach space $V$, because the sum of its norms is bounded by the convergent geometric series $R\Vert y\Vert /(1-k)$. Write its sum as $x$. The bounded linear operator $T$ commutes with this limit, while $T{\sum }_{j=1}^m{x}_j=y-r_m\to y$. ConsequentlyThe choices need not depend linearly on $y$; no bounded linear right inverse is being asserted.

For the bounded extension, use the restriction bounded linear operator $S:C_b(X;\mathbb{R})\to C_b(Y;\mathbb{R})$ on the Banach spaces of bounded continuous functions, each with its supremum norm. We will construct an approximation to any $g$ with $\Vert g{\Vert }_{\infty }=M>0$. The two setsare disjoint closed sets in $X$, since $Y$ is closed. If both are nonempty, the distance to a set construction defines a continuous functionThe denominator is positive at every point: zero distance to a closed set means membership in that set. Thus $\Vert u{\Vert }_{\infty }\le M/3$, with $u=-M/3$ on $A$ and $u=M/3$ on $B$. On $A$ and $B$, and also on the intervening range $|g|<M/3$, this impliesIf only $A$ is empty take $u=M/3$, if only $B$ is empty take $u=-M/3$, and if both are empty take $u=0$; the same bounds hold. For $M=0$ take $u=0$. This proves the approximation hypothesis with $R=1/3$ and $k=2/3$. Applying the argument above to $S$ yields an element $h$ of the bounded continuous functions with $h{|}_Y=g$ and $\Vert h{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }$. Restriction gives the reverse inequality, so the extension preserves the exact supremum norm:This proves the bounded metric-space case of the Tietze extension theorem rather than presupposing it.

For an arbitrary real continuous function $f$ on $Y$, let $g=\tanh f$. Its values lie strictly between $-1$ and $1$, although its supremum norm can be $1$. The bounded result supplies an extension $h$ with $|h|\le 1$. Let $D=\{x:|h(x)|=1\}$, a closed set disjoint from $Y$. If $D$ is nonempty and $Y$ is nonempty, putBoth factors are continuous functions; $\eta =1$ on $Y$ and $\eta =0$ on $D$. Off $D$ we have $|h|<1$, and on $D$ we have $q=0$, so $|q(x)|<1$ everywhere. Therefore a real continuous extension isOn $Y$ it equals $f$. If $D$ is empty use $q=h$; if $Y$ is empty simply take $F=0$. This handles unbounded $f$ without requiring a uniform margin between $|g|$ and $1$.

Here $\varphi$ is a finite real convex function on the whole real vector space; no topology or continuity is required. We prove the convex domination form of the Hahn-Banach theorem by extending across one vector. Suppose a linear functional $f$ is already dominated on a vector subspace $W$ and choose $v\notin W$. For $w\in W$ and positive $t,s$, the proposed extension $g(w+\tau v)=f(w)+\tau c$ must satisfy the boundsTo prove compatibility, take $w_1,w_2\in W$ and $t,s>0$. The convex combination of $w_1-tv$ and $w_2+sv$ with weights $s/(s+t)$ and $t/(s+t)$ lies in $W$. Domination there and convexity giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ showThus both endpoints are finite, and some $c\in [L,U]$ exists. For $\tau >0$ the upper bound gives domination of $f(w)+\tau c$; for $\tau <0$ the lower bound gives it; and for $\tau =0$ domination was assumed. This defines a dominated linear functional on $W+\mathbb{R}v$ extending $f$.

Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By Zorn's lemma there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is $V$. Hence there is a full linear extension with $\boxed{g{|}_W=f,g\le \varphi }$. A convex function here need not be positively homogeneous; this is why the bounds included all positive $s,t$.

Now fix $y\in V$. The translated convex function ${\phi }_y(z)=\varphi (y+z)-\varphi (y)$ vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on $\{0\}$ to obtain a linear functional ${\ell }_y$ with ${\ell }_y(z)\le {\phi }_y(z)$ for every $z$. SetThis is an affine function, satisfies $a_y(x)\le \varphi (x)$ for every $x$, and has $a_y(y)=\varphi (y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\varphi (x)$ at $x$. Thus the exact representation isThis is the finite convex function as supremum of affine minorants property. The linear functionals are algebraic; without a topology no assertion of continuous affine minorants is intended.

Let $\mathcal{F}=C(X,[0,1])$. Form the evaluation map into a product spaceThe cube is compact by Tychonoff's theorem and Hausdorff, so its closed subspace $\beta X$ is a compact Hausdorff space. A completely regular Hausdorff space has enough continuous functions to separate points and closed sets. Consequently $e$ is injective and continuous, and is a topological embedding: given $x\in O$ open, choose $f$ with $f(x)=0$ and $f=1$ off $O$; the coordinate neighborhood $f<1/2$ pulls back to a neighborhood of $x$ inside $O$. Identify $X$ with this dense embedded subspace. This constructs the Stone-Čech compactification.

Each $f\in \mathcal{F}$ extends as its coordinate function. Rescaling therefore extends every bounded real continuous function; real and imaginary parts extend bounded complex continuous functions. The extension is unique because $X$ is dense and the target is Hausdorff. Its supremum norm is unchanged by density, so restriction is an isometric isomorphism of normed spaces of Banach spacesThese are the space of continuous functions on a compact space and the space of bounded continuous functions, respectively.

More generally, embed a compact Hausdorff space $K$ in its own evaluation cube $[0,1{]}^{C(K,[0,1])}$. This is an embedding by complete regularity, and its image is closed by compactness. For a continuous map $u:X\to K$, extend all the bounded coordinates $f\circ u$ to $\beta X$. The resulting continuous map into the cube takes its values in the closed copy of $K$: it does so on the dense subset $X$, and the inverse image of that closed copy is closed. ThusUniqueness again follows from density. This is the universal property of the Stone-Čech compactification. It implies uniqueness of the compactification up to a homeomorphism fixing $X$, by extending the identity in both directions. Every Hausdorff compactification of $X$ is the image of $\beta X$ under a continuous surjection fixing $X$, since that image is compact and contains the dense copy of $X$. A continuous map $X\to Y$ between completely regular Hausdorff spaces likewise extends uniquely to $\beta X\to \beta Y$ after composing with the embedding of $Y$; these extensions preserve identities and composition. If $X$ is already compact, its dense copy is closed, so $\beta X=X$.

For discrete $X$, every indicator $1_D$, $D\subseteq X$, is continuous. Its extension to $\beta X$ takes only values $0,1$, since this closed condition holds on the dense subset $X$. Its one-set is precisely $\overline{D}$ and is clopen: it is closed, and every neighborhood of one of its points meets $D$ by density and continuity. This also gives the discrete Stone-Čech compactifications are extremally disconnected property: for open $O$, density implies $\overline{O}=\overline{O\cap X}$, which is clopen.

For the requested separation, a compact Hausdorff space is normal, so Urysohn's lemma supplies $u:\beta X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$. Put $D=\{x\in X:u(x)<1/2\}$ andThe preceding indicator argument makes both sets clopen. At a point of $A$, the neighborhood $u<1/2$ has its dense $X$-part inside $D$, so the point lies in $U$. At a point of $B$, the neighborhood $u>1/2$ misses $D$, so the point lies outside $U$. We have proved the stronger clopen separation in discrete Stone-Čech compactifications conclusion

Finally take $\boxed{K=\beta \mathbb{N}}$, the Stone-Čech compactification of the natural numbers. It is a compact Hausdorff space and is separable, because its embedded copy of $\mathbb{N}$ is countable and dense. For every $D\subseteq \mathbb{N}$, the indicator extends to an element $h_D\in C(K)$. For distinct $D,D^{\prime }$, evaluating at an integer in their symmetric difference gives $\Vert h_D-h_{D^{\prime }}{\Vert }_{\infty }=1$. There are uncountably many such indicators. A separable metric space cannot contain an uncountable family at pairwise distance one: disjoint balls of radius $1/3$ would require distinct members of a countable dense set. Hence $C(K)$ is not separable, despite separability of $K$.

The weak topology $\sigma (E,E^{\ast })$ is the coarsest topology making every continuous linear functional in $E^{\ast }$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\epsilon$, and a sequence converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The Hahn-Banach theorem makes the dual separate points, so the weak topology is Hausdorff.

Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the Eberlein-Šmulian theorem. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge 1\}$. It is a separable Banach space. The Hahn-Banach theorem separates points outside a closed vector subspace from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.

Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose norming functionals $f_j\in Y^{\ast }$ with $\Vert f_j\Vert =1$ and $f_j(u_j)=1$. They separate points: for $y\ne 0$, a sufficiently close $u_j$ to $y/\Vert y\Vert$ gives $|f_j(y)|>0$. This is a countable norming family. Extend the functionals to $E$ by the Hahn-Banach theorem. The evaluation mapis weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a homeomorphism onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by ${\sum }_j{2}^{-j}\min (1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact metric spaces are sequentially compact, giving a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$. If $Y=\{0\}$ the result is immediate. We did not assert that a separable Banach space has a separable dual or that its weak topology is generally metrizable.

Assume weakly null sequences have norm-null images. Let $(y_n)$ be any sequence in $T(K)$, where $K$ is weakly compact, and choose $x_n\in K$ with $y_n=T{x}_n$. By the weak subsequence result, some $x_{n_j}\rightharpoonup x\in K$. Then $x_{n_j}-x\rightharpoonup 0$, and the assumed property givesThus every sequence in $T(K)$ has a norm-convergent subsequence with limit in $T(K)$. In a metric space, sequential compactness is equivalent to compactness. Therefore $\boxed{T(K)\text{ is norm compact}}$. This is one characterization of a completely continuous operator.

Assume instead that $T$ sends weakly compact sets to norm-compact sets. If $x_n\rightharpoonup 0$, then $K=\{0,x_1,x_2,\dots \}$ is weakly compact: an open cover has a member containing zero, which contains all but finitely many terms of the convergent sequence. Hence $T(K)$ is norm compact. A bounded linear operator is weak-to-weak continuous, because each $g\in F^{\ast }$ pulls back to $g\circ T\in E^{\ast }$. Thus $T{x}_n\rightharpoonup 0$. If $\Vert T{x}_n\Vert$ did not tend to zero, a subsequence with norms bounded below by some $\epsilon >0$ would have a norm-convergent further subsequence with limit $y$. That further subsequence also converges weakly to both $y$ and zero; uniqueness of limits in the Hausdorff weak topology forces $y=0$, contradicting the lower bound. Therefore $\boxed{\Vert T{x}_n\Vert \to 0}$, proving the equivalence and the characterization of completely continuous operators.

Now consider the inclusion of the space of continuous functions on a compact space $C(\mathbb{T})$ into L2 space for normalized Lebesgue measure. It is a bounded linear operator, since $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$. For a weakly null sequence $(f_n)$ in $C(\mathbb{T})$, point evaluations are continuous linear functionals, so $f_n(t)\to 0$ for every $t$. The Uniform boundedness principle, applied to the canonical embeddings $f_n\in C(\mathbb{T}{)}^{\ast \ast }$, gives a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$: each dual functional is bounded on the weakly convergent sequence. The dominated convergence theorem now yieldsThus $j_2$ satisfies both equivalent conditions.

It is nevertheless not a compact operator. Over complex scalars the functions $f_n(t)=e^{int}$ lie in the unit ball of $C(\mathbb{T})$ and form an orthonormal sequence in $L^2(\mu )$, soThere is no norm-convergent subsequence. Over real scalars use $\cos (nt)$, whose distinct images have distance $1$. Complete continuity controls weakly compact sets, rather than all bounded sets.

For a bounded linear operator $T:E\to F$, write the weak square norm of a finite family asIt is a 2-summing operator if $({\sum }_j\Vert T{x}_j{\Vert }^2{)}^{1/2}\le C{w}_2(x_1,\dots ,x_m)$ for all finite families, and ${\pi }_2(T)$ is the least such $C$. This specializes the definition of an absolutely p-summing operator.

First let $E=H_1$, $F=H_2$ be Hilbert spaces. For the column operator $A:{\ell }_2^m\to H_1$, $A{e}_j=x_j$, the Riesz representation theorem gives $w_2(x_1,\dots ,x_m)=\Vert A^{\ast }\Vert =\Vert A\Vert$. If $T$ is a Hilbert-Schmidt operator, the Parseval identity applied to orthonormal bases givesThus ${\pi }_2(T)\le \Vert T{\Vert }_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the Bessel inequality, with equality by choosing one of its vectors. Hence ${\sum }_j\Vert T{e}_j{\Vert }^2\le {\pi }_2(T{)}^2$. Taking the supremum over finite subsets of an orthonormal basis proves the Hilbert-Schmidt operator condition and the reverse norm inequality. These basis sums work even for nonseparable Hilbert spaces, being understood as suprema of finite subsums. Therefore

For an $n$-dimensional normed vector space $E$, again associate a family $(x_j)$ with $A:{\ell }_2^m\to E$. By the Hahn-Banach theorem and scalar Euclidean duality,Let $Q$ be the orthogonal projection onto $(\ker A{)}^{\perp }$, whose dimension is at most $n$. Since $A=AQ$, the trace of $Q$ givesThis proves ${\pi }_2(I_E)\le \sqrt{n}$ for every norm, without comparing that norm to an arbitrary Euclidean norm.

To obtain equality and the projection bound, choose a maximum-volume centered ellipsoid inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the John ellipsoid; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in \partial B_E\cap \partial B_2$ and positive weights $c_r$ with the John contact decompositionFor completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\mathrm{tr}D>0$ and $u^{\ast }Du\le 0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\mathrm{tr}D+O(t^2)$ in the real case and $|{\det }_{\mathbb{C}}(I+tD){|}^2=1+2t\mathrm{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.

At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is ${\varphi }_r(x)=u_r^{\ast }x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|{\varphi }_r(x)|\le \Vert x{\Vert }_E$. Also $B_2\subseteq B_E$ gives $\Vert x{\Vert }_E\le |x{|}_2$. For any dual-unit functional $\varphi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}{u}_r$:The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence ${\pi }_2(I_E)\ge \sqrt{n}$. Together with the upper estimate, the norm isThis proves the 2-summing norm of a finite-dimensional identity formula.

Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each ${\varphi }_r$ to a dual-unit functional ${\tilde{\varphi }}_r$ on $X$ using the real or complex Hahn-Banach theorem. DefineThe contact identity gives $P{|}_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto {\sum }_r\sqrt{c_r}{a}_r{u}_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. ThereforeThus the linear projection satisfies $\boxed{\Vert P\Vert \le \sqrt{n}}$, the Kadec-Snobar projection bound. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.