Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 10 / 3 / Solution

For Banach spaces $E,F$ and $1\le p<\infty$, define the weak $p$-norm of a finite family byA bounded linear operator $T:E\to F$ is an absolutely p-summing operator if $({\sum }_i\Vert T{x}_i{\Vert }^p{)}^{1/p}\le C{w}_p(x_1,\dots ,x_m)$ for every finite family. The least constant is ${\pi }_p(T)$.

The Pietsch factorization theorem states that such an operator admits a probability measure $\mu$ on $K=B_{E^{\ast }}$ with its weak-star topology such thatEquivalently, let $i:E\to C(K)$ be the evaluation embedding $i(x)(\varphi )=\varphi (x)$ and $j_p:C(K)\to L^p(\mu )$ the natural map. There is a bounded linear operator $V:H\to F$, where $H=\overline{j_{p}i(E)}\subset L^p(\mu )$, such that $T=V{j}_{p}i$ and $\Vert V\Vert \le {\pi }_p(T)$. Conversely, any such factorization makes $T$ an absolutely p-summing operator, with ${\pi }_p(T)\le \Vert V\Vert$. The final operator is required on this closed subspace $H$; no extension to all of $L^p(\mu )$ is assumed.

Here is the existence proof. If $T=0$ the conclusion is immediate. Otherwise write $C={\pi }_p(T)>0$. The Banach-Alaoglu theorem makes $K$ compact, and $\varphi \mapsto |\varphi (x){|}^p$ is continuous for each $x\in E$. For finitely many vectors $x_1,\dots ,x_m$, putWe claim the compact convex hull of $v(K)$ meets $a+[0,\infty {)}^m$. If it did not, finite-dimensional Hahn-Banach separation theorem would give weights ${\lambda }_i\ge 0$ withThe weights are nonnegative because the second convex set is unbounded in every positive coordinate direction. This strict inequality contradicts the defining summing inequality applied to ${\lambda }_i^{1/p}{x}_i$. Thus a finite convex combination of the vectors $v(\varphi )$ lies coordinatewise above $a$. Equivalently, a finitely supported probability measure satisfies all these finitely many domination constraints.

By the Riesz-Markov-Kakutani representation theorem, probability measures on compact $K$ are the positive norm-one functionals on $C(K)$ taking the constant function to one. They form a compact set in the weak-star topology, by the Banach-Alaoglu theorem. Each domination constraint $\int |\varphi (x){|}^{p}d\mu \ge C^{-p}\Vert Tx{\Vert }^p$ is closed. The finite feasibility just proved gives the finite intersection property, hence one measure $\mu$ satisfies every constraint.

Define $V(j_{p}i(x))=Tx$. The domination inequality makes this well-defined and bounded by $C$, and completeness of $F$ extends it to $H$. Conversely, any factorization satisfiesThis proves both directions of the Pietsch factorization theorem, including the optimal factorization norm.

For the multiplication operator, let $f\in L^p(0,1)$ and take $g_1,\dots ,g_m\in C([0,1])$. Point evaluations are norm-one linear functionals on $C([0,1])$, soThus $M_f$ is an absolutely p-summing operator and ${\pi }_p(M_f)\le \Vert f{\Vert }_p$. Taking the single function $g=1$ gives the reverse inequality, soFor $f\ne 0$, the explicit domination measure on $[0,1]$ is $|f(t){|}^{p}dt/\Vert f{\Vert }_p^p$, pushed into $K$ by $t\mapsto {\delta }_t$. For $f=0$ everything is zero; no essential boundedness of $f$ is needed.