Past exam of the functional analysis course of the University of Oxford / 2011 (C4.1a, DMAC 2A64)

Source release: Oxford examination archive. The cover dates this examination to 17 June 2011, 9.30–11.00am. The 13 answer parts are solved in source order. The two lettered conditions inside 3(iv) are the two sides of one equivalence.

A linear operator $T:X\to Y$ is a compact operator if the image of the closed unit ball has compact closure in $Y$. Equivalently, every bounded sequence $(x_n)$ in $X$ has a subsequence whose images converge in the norm of $Y$. This implies boundedness of $T$.

The Schauder theorem for compact operators says that for a bounded linear operator between Banach spaces,where the dual map of bounded linear operator is $T^{\ast }f=f\circ T$. The forward implication also holds for normed spaces; completeness of the target gives the usual converse. The compactness definition requires the image subsequence's limit to belong to $Y$, so that completeness qualification matters when stating the theorem beyond Banach spaces.

Let $A_n$ be finite-rank bounded operators with $\Vert T-A_n\Vert \to 0$ in operator norm. Fix $ϵ>0$ and choose $n$ with $\Vert T-A_n\Vert <ϵ/3$. The set $A_n(B_X)$ is bounded in a finite-dimensional subspace, hence has a finite $ϵ/3$-net $y_1,\dots ,y_m$.

For each $x\in B_X$, some $j$ satisfiesThus $T(B_X)$ is totally bounded. Its closure is totally bounded and, as a closed subset of the Banach space $Y$, complete. A complete totally bounded metric space is compact. Consequently $T$ is compact.

The operator-norm limit is bounded, since $\Vert T\Vert \le \Vert A_n\Vert +\Vert T-A_n\Vert$ for any sufficiently good approximation. This proves directly that an approximable operator into a Banach space is compact; no assertion that every compact operator on an arbitrary Banach space is approximable is needed.

Let $Q_N$ be truncation to the first $N$ coordinates in ${\ell }^1$. For $x\in {\ell }^1$, boundedness of $T$ and absolute convergence giveApplying the difference to each $e_n$ with $n>N$ gives the reverse bound on the operator norm. HenceThe stated condition makes this tend to zero. Each $T{Q}_N$ has finite rank, so part (a)(ii) proves compactness. Thus the condition is sufficient on ${\ell }^1$.

For failure of necessity, set $Tx=({\sum }_{n\ge 1}{x}_n)e_1$. The sum defines a bounded linear functional on ${\ell }^1$, so $T$ is a bounded rank-one operator and therefore compact. But $T{e}_n=e_1$ for every $n$, so $\Vert T{e}_n{\Vert }_1=1$. Thus compactness need not make the basis images tend to zero on ${\ell }^1$.

Let $1<p<\infty$ and $q=p/(p-1)$. The dual of ${\ell }^p$ is ${\ell }^q$. For every $a\in {\ell }^q$, $⟨a,e_n⟩=a_n\to 0$, so $e_n\rightharpoonup 0$ weakly. A compact operator sends a weakly convergent sequence to a norm-convergent sequence with the corresponding limit. Indeed, any failure of $\Vert T{e}_n\Vert \to 0$ would give a subsequence bounded away from zero, with a further norm-convergent subsequence by compactness; weak continuity of $T$ forces that norm limit to be zero, a contradiction. The condition is therefore necessary.

For failure of sufficiency, partition the positive integers into consecutive blocksand define the block averaging compactness counterexampleBy Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$. Summing over disjoint blocks shows $\Vert Tx{\Vert }_p\le \Vert x{\Vert }_p$, so $T$ is bounded. If $j\in B_k$, then $T{e}_j=k^{-1/q}{e}_k$; as $j\to \infty$, also $k\to \infty$, so $\Vert T{e}_j{\Vert }_p\to 0$.

However, the vectors $x^{(k)}=k^{-1/p}{1}_{B_k}$ have norm one and satisfy $T{x}^{(k)}=e_k$. Distinct output vectors have distance $2^{1/p}$, so this bounded sequence of inputs has no norm-convergent image subsequence. Thus $T$ is not compact.

To connect with the hint, define $(Uy{)}_j=k^{-1/q}{y}_k$ for $j\in B_k$. Then $U:{\ell }^q\to {\ell }^q$ is an isometry, and the dual pairing calculation gives $T=U^{\ast }$. The vectors $U{e}_k$ have disjoint supports and norm one. The direct estimates above establish the counterexample without relying only on the hint.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

The weak topology is coarser than the norm topology, so norm convergence implies weak convergence, andFor the reverse inclusion, let $x\notin K={\overline{C}}^{\Vert \cdot \Vert }$. The set $K$ is closed and convex. The Hahn-Banach separation theorem supplies $f\in X^{\ast }$ and $a\in \mathbb{R}$ such that $f(x)>a\ge {\sup }_{y\in K}f(y)$. The set $\{z:f(z)>a\}$ is weakly open, contains $x$ and misses $C$. Hence $x$ is outside the weak closure as well. If $C$ is empty the equality is immediate. ThereforeThe separating functional explains the role of convexity: an arbitrary norm-closed subset need not be weakly closed.

A reflexive Banach space is one for which $J_X:X\to X^{\ast \ast }$ is onto. If $X$ is reflexive, every $\theta \in X^{\ast \ast }$ has the form $\theta (f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma (X^{\ast },X^{\ast \ast })$ and $\sigma (X^{\ast },X)$ are precisely the same, proving equality of the topologies.

Conversely, assume those topologies coincide. For any $\theta \in X^{\ast \ast }$, the linear functional $f\mapsto \theta (f)$ on $X^{\ast }$ is then weak-star continuous. Its continuity at zero gives finitely many $x_1,\dots ,x_m\in X$ and $ϵ>0$ such thatIf all these evaluations are zero, arbitrary scalar multiples of $f$ still satisfy the inequalities, forcing $\theta (f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto (f(x_1),\dots ,f(x_m))$. A linear functional on its image extends to ${\mathbb{R}}^m$, so for suitable scalars $a_j$,Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved

The domain carries $\sigma (Z^{\ast \ast },Z^{\ast })$, and the target carries $\sigma (X^{\ast },X)$. For each $x\in X\subseteq Z^{\ast }$,which is one of the defining weak-star continuous evaluations on $Z^{\ast \ast }$. Thus $R$ is weak-star continuous.

Restriction cannot increase a functional's norm, so $R(B_{Z^{\ast \ast }})\subseteq B_{X^{\ast }}$. Conversely, any $f\in B_{X^{\ast }}$ is a bounded linear functional on the subspace $X$ of $Z^{\ast }$. The Hahn-Banach theorem extends it to a functional $\theta \in (Z^{\ast }{)}^{\ast }=Z^{\ast \ast }$ with the same norm. Then $R\theta =f$ and $\Vert \theta \Vert \le 1$. ConsequentlyThis proves exact surjectivity on closed balls, not just density of the image.

Put $T=R\circ J_Z:Z\to X^{\ast }$. Then $(Tz)(x)=x(z)$ and $\Vert Tz\Vert \le \Vert z\Vert$. We must prove $T$ onto.

Take $f\in B_{X^{\ast }}$. Part (iv) supplies $\theta \in B_{Z^{\ast \ast }}$ with $R\theta =f$. Goldstine's theorem gives a net $z_{\alpha }\in B_Z$ with $J_Z{z}_{\alpha }\to \theta$ weak-star. By continuity of $R$, $T{z}_{\alpha }\to f$ in $\sigma (X^{\ast },X)$. Reflexivity of $X$ makes this the weak topology on $X^{\ast }$ by part (iii). Thus $B_{X^{\ast }}$ is contained in the weak closure of $T(B_Z)$.

The reverse containment holds because $T(B_Z)\subseteq B_{X^{\ast }}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\le 1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the Banach space $X^{\ast }$, givesDensity of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^{\ast }$, choose $z_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and $\Vert f-T{z}_1\Vert \le \Vert f\Vert /2$. Inductively approximate the current residual $r_{n-1}$ by $T{z}_n$ with $\Vert z_n\Vert \le \Vert r_{n-1}\Vert$ and $\Vert r_{n-1}-T{z}_n\Vert \le \Vert r_{n-1}\Vert /2$. If a residual is zero, stop.

The residual norms are bounded by $2^{-n}\Vert f\Vert$, while ${\sum }_n\Vert z_n\Vert \le 2\Vert f\Vert$. Completeness of $Z$ gives $z={\sum }_n{z}_n\in Z$, and continuity of $T$ gives $Tz=f$. For $f=0$ use $z=0$. ThereforeThis is evaluation onto the dual of a reflexive dual subspace. The correction argument is needed: a dense image alone would not imply surjectivity.

A biorthogonal system is a family $(x_i,f_i{)}_{i\in I}$ with $x_i\in X$, $f_i\in X^{\ast }$, satisfyingThus each continuous coordinate functional takes value one on its own vector and zero on all others. The vectors need not be orthogonal in any inner product, and totality of their linear span is an additional condition.

Fix $i$ and let $M_i=\overline{\mathrm{span}}\{x_j:j\ne i\}$. The hypothesis gives $M_i\ne X$. Moreover $x_i\notin M_i$, since otherwise the closed span of all the vectors would be contained in $M_i$. Hence $d_i=\mathrm{dist}(x_i,M_i)>0$.

On $M_i+\mathbb{R}{x}_i$, define ${\varphi }_i(m+t{x}_i)=t$. It is well-defined because $x_i\notin M_i$. For $t\ne 0$,so $|{\varphi }_i(m+t{x}_i)|\le \Vert m+t{x}_i\Vert /d_i$; the inequality also holds for $t=0$. The Hahn-Banach theorem extends ${\varphi }_i$ without increasing its norm to $x_i^{\ast }\in X^{\ast }$. This gives $x_i^{\ast }(x_j)={\delta }_{ij}$.

If two continuous functionals have these values, their difference vanishes on the linear span of every $x_j$, which is dense in $X$. Continuity then makes the difference zero everywhere. Thus each functional is unique, proving the required unique biorthogonal system. In fact its norm is $\Vert x_i^{\ast }\Vert =1/d_i$, by taking vectors in $M_i$ approaching the distance infimum. This is the coordinate-functional construction for a minimal total family in a normed space.

Define the finite-rank operatorIt is linear and bounded, with $\Vert P_n\Vert \le {\sum }_{i=1}^n\Vert x_i^{\ast }\Vert \Vert x_i\Vert$. Biorthogonality gives exactly the two required values on $x_j$. It also givesThus $P_n$ is a bounded projection onto $\mathrm{span}\{x_1,\dots ,x_n\}$, with kernel ${\bigcap }_{i=1}^n\ker x_i^{\ast }$. Its action on the dense span determines it uniquely among bounded linear operators. These are the partial-sum projections; no uniform bound in $n$ has yet been assumed.

Assume first that $\Vert P_n\Vert \le M$ for all $n$. Given $x\in X$ and $ϵ>0$, choose a finite linear combination $y$ of the $x_i$ with $\Vert x-y\Vert <ϵ/(M+1)$. For all sufficiently large $n$, $P_{n}y=y$. ThereforeThus $P_{n}x\to x$, and the formula in part (iii) yields the norm-convergent expansionThis proves the stated existence of coefficients.

Conversely, suppose every $x$ has some norm-convergent expansion $x={\sum }_i{\alpha }_i{x}_i$. Applying the continuous functional $x_j^{\ast }$ to its partial sums and passing to the limit gives ${\alpha }_j=x_j^{\ast }(x)$. Hence those partial sums are precisely $P_{n}x$, so $P_{n}x\to x$ for every $x$. In particular ${\sup }_n\Vert P_{n}x\Vert <\infty$ for each fixed $x$. The Uniform boundedness principle, using completeness of $X$, now givesThis proves the reverse implication. The coefficients are automatically unique, so the equivalence is the uniform partial-sum criterion for a Schauder basis for the given minimal total sequence. Its projection bound is the basis constant.