Reflexive approximation property implies metric approximation property

Given: AC and a reflexive Banach space $X$ with AP.

$\pi (u)=\inf {\sum }_j\mathrm{\parallel }{e}_j\mathrm{\parallel }\mathrm{\parallel }{f}_j\mathrm{\parallel },\epsilon (u)={\sup }_{e^{\ast }\in B_{E^{\ast }},f^{\ast }\in B_{F^{\ast }}}\mid {\sum }_j{e}^{\ast }(e_j)f^{\ast }(f_j)\mid \mathrm{.}$

Their completions are $E{\widehat{\otimes }}_{\pi }F$ and $E{\widehat{\otimes }}_{\epsilon }F$. A tensor $u={\sum }_n{e}_n^{\ast }\otimes f_n\in E^{\ast }{\widehat{\otimes }}_{\pi }F$ induces the nuclear operator $S_u(e)={\sum }_n{e}_n^{\ast }(e)f_n$; the nuclear norm is the quotient of $\pi$ by the kernel of $u\mapsto S_u$.

For later use, call $S:E\to F$ integral when

$B_S(e,f^{\ast })=f^{\ast }(Se)$

extends continuously to $E{\widehat{\otimes }}_{\epsilon }{F}^{\ast }$; its norm is the norm of that functional. Call $S$ Pietsch integral when there are a finite measure $\mu$ and bounded maps

$E\stackrel{U}{\to }{L}^{\mathrm{\infty }}(\mu )\stackrel{I}{\to }{L}^1(\mu )\stackrel{R}{\to }F$

with $S=RIU$; its Pietsch norm is the infimum of $\mathrm{\parallel }R\mathrm{\parallel }\mathrm{\parallel }U\mathrm{\parallel }\mu (\mathrm{\Omega })$ after the harmless normalization of $\mu$. Every nuclear map is Pietsch integral and every Pietsch-integral map is integral, with

$\mathrm{\parallel }S{\mathrm{\parallel }}_I\le \mathrm{\parallel }S{\mathrm{\parallel }}_{PI}\le \mathrm{\parallel }S{\mathrm{\parallel }}_N\mathrm{.}$

$m(A)={\int }_{A}gd\mu$

for a Bochner-integrable $g:\mathrm{\Omega }\to X$. We give the operator proof because this fact is load-bearing below.

First suppose that a weakly compact operator $T:L^1(\mu )\to Y$ has separable range. Replacing $Y$ by the separable closed span of $T[L^1]$, choose a countable norming set $(y_n^{\ast })\subseteq B_{Y^{\ast }}$ and put

$\mathrm{\mid }\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }\mathrm{\mid }={\sum }_{n\ge 1}{2}^{-n}\mathrm{\mid }{y}_n^{\ast }(y)\mathrm{\mid }\mathrm{.}$

On the weakly compact closure $K$ of $T[B_{L^1}]$, this metric induces the weak topology: the identity from weak $K$ to $\mathrm{\mid }\mathrm{\mid }\mathrm{\mid }\cdot \mathrm{\mid }\mathrm{\mid }\mathrm{\mid }$-metric $K$ is continuous by uniform convergence of the displayed series, and compactness then makes it a homeomorphism. If $Z$ is the completion of $(Y,\mathrm{\mid }\mathrm{\mid }\mathrm{\mid }\cdot \mathrm{\mid }\mathrm{\mid }\mathrm{\mid })$ and $j:Y\to Z$ is the inclusion, $jT$ is compact.

The dual $L^{\mathrm{\infty }}(\mu )$ has MAP directly. Given a norm-compact $C\subseteq L^{\mathrm{\infty }}$ and $\delta >0$, choose a finite $\delta$-net $f_1,\dots ,f_q$ in $C$, uniformly approximate the $f_i$ by simple functions, and let $\mathcal{P}$ be the common finite measurable partition on which those simple functions are constant. Averaging over each positive-measure cell (and putting zero on null cells) defines a norm-one finite-rank conditional-average map $P_{\mathcal{P}}$. The triangle inequality gives ${\sup }_{f\in C}\mathrm{\parallel }{P}_{\mathcal{P}}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}<4\delta$. The standard adjoint form of AP therefore approximates the compact operator $jT$ in operator norm by finite-rank maps $T_r:L^1(\mu )\to Z$. Write

$T_{r}f=\int f{g}_{r}d\mu$

with a finite-dimensional-valued strongly measurable $g_r$. For such densities, $\mathrm{\parallel }{T}_r-T_s\mathrm{\parallel }=\mathrm{ess\hspace{0.17em}sup}\mathrm{\parallel }{g}_r-g_s{\mathrm{\parallel }}_Z$: the easy inequality is Holder's inequality, and the reverse follows by testing on a positive-measure set where a norming functional almost attains the essential supremum. Thus $(g_r)$ converges in $L^{\mathrm{\infty }}(\mu ;Z)$ to a strongly measurable bounded $g$, and $jTf=\int fgd\mu$.

For every $A$ of positive measure,

$\frac{1}{\mu (A)}{\int }_{A}gd\mu =jT\text{\hspace{0.17em}⁣}\left(\frac{{\mathbf{1}}_A}{\mu (A)}\right)\in jK\mathrm{.}$

The function $g$ lies in $jK$ almost everywhere. Indeed, $Z$ is separable; cover $Z\setminus jK$ by countably many open balls whose closures miss $jK$. If the inverse image of one such ball had positive measure, the average of $g$ over a smaller concentric inverse image would lie both in that ball and in $jK$, a contradiction. On $K$, the $Z$-norm topology is the weak topology of $Y$. Hence $j^{-1}g$ is weakly measurable, essentially separably valued, and bounded in the original norm. To see strong measurability directly, use the norming sequence above: norm balls about a countable dense subset of $Y$ are countable intersections of inverse images of scalar Borel sets, so choosing the first ball of radius $2^{-q}$ that contains $j^{-1}g(\omega )$ gives countably-valued measurable approximants; truncating their ranges gives simple functions converging pointwise in norm. Thus $T$ is represented by an essentially bounded Bochner-measurable $Y$-valued density.

Now let $T:L^1(\mu )\to Y$ be any weakly compact operator. The family of characteristic functions is relatively weakly compact in $L^1$: it lies in the image of the closed set $\{h\in L^2:\mathrm{\mid }h\mathrm{\mid }\le 1\}$ under the continuous inclusion $L^2\to L^1$, and that set is weakly compact by [L4]. For a sequence $(A_n)$, restrict $T$ to the $L^1$ space of the countably generated sigma-algebra generated by the $A_n$. The restricted finite measure is sigma-finite, so this $L^1$ space is separable by [A1] and [L5]. If $D$ is a countable dense subset, continuity makes $T(D)$ dense in the image of the restriction; the closed linear span of $T(D)$ is therefore separable and contains that image. Thus the restriction has separable range and hence has the representation just proved. Representable maps are completely continuous on weakly convergent sequences: if $f_n\rightharpoonup 0$ in $L^1$, then $(f_n)$ is uniformly integrable. For completeness, failure of uniform integrability would permit a gliding-hump subsequence on successively almost-disjoint small sets; putting the scalar sign of the selected $f_n$ on each hump produces one $h\in L^{\mathrm{\infty }}$ for which $\int h{f}_n$ stays bounded away from zero, contradicting weak convergence. A pointwise simple approximation to the bounded density, Egorov on a large set, and uniform integrability on its small complement reduce the assertion to finitely many scalar integrals ${\int }_E{f}_{n}d\mu \to 0$. Consequently $(T{\mathbf{1}}_{A_n})$ has a norm-convergent subsequence. Thus $\{T{\mathbf{1}}_A:A\text{ measurable}\}$ is relatively norm compact and is separable. Characteristic functions span a dense subspace of $L^1$, so the range of $T$ is separable. The preceding paragraph now represents $T$.

Apply this to the integration operator

$T_m:L^1(\mathrm{\mid }m\mathrm{\mid })\to X,T_{m}f=\int fdm\mathrm{.}$

It is bounded, and it is weakly compact because its range is in the reflexive space $X$ by [L2]. Hence $T_{m}f=\int fhd\mathrm{\mid }m\mathrm{\mid }$ for an essentially bounded Bochner-measurable $h$. Scalar Radon--Nikodym under [L3] gives $\mathrm{\mid }m\mathrm{\mid }=w\mu$; then $g=wh$ is Bochner integrable and $m(A)={\int }_{A}gd\mu$. This proves the required density fact.

$Rf=\int fgd\mu (f\in L^1(\mu ))\mathrm{.}$

Choose simple $g_q\to g$ in $L^1(\mu ;X)$. If $g_q={\sum }_k{\mathbf{1}}_{A_k}{x}_k$, then

$e⟼\int (Ue)g_{q}d\mu ={\sum }_k{e}_k^{\ast }(e)x_k,e_k^{\ast }(e)={\int }_{A_k}Ued\mu ,$

and ${\sum }_k\mathrm{\parallel }{e}_k^{\ast }\mathrm{\parallel }\mathrm{\parallel }{x}_k\mathrm{\parallel }\le \mathrm{\parallel }U\mathrm{\parallel }\int \mathrm{\parallel }{g}_q\mathrm{\parallel }d\mu$. Passing to the projective completion shows that $S$ is nuclear and $\mathrm{\parallel }S{\mathrm{\parallel }}_N\le \mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }R\mathrm{\parallel }$. Infimizing over factorizations and using the general inequalities of step 1.1 yields, isometrically,

$\mathcal{N}(E,X)=\mathcal{P}\mathcal{I}(E,X)\mathrm{.}$

$J_{X}S=RIU:E⟶X^{\ast \ast }$

whose product norm can be chosen arbitrarily close to $\mathrm{\parallel }S{\mathrm{\parallel }}_I$. Here is the factorization explicitly. By [L3] the relevant weak-star dual balls are compact. Embed the injective tensor product isometrically into the continuous functions on their product, extend $B_S$ without increasing its norm by [L3], and apply the real or complex Riesz--Markov suppliers in [L3] to represent that extension by a finite regular measure $\lambda$. Put $Ue(e^{\ast },x^{\ast \ast })=e^{\ast }(e)$, and define

$Rf(x^{\ast })=\int f(e^{\ast },x^{\ast \ast })x^{\ast \ast }(x^{\ast })d\lambda \mathrm{.}$

Then $R:L^1(\mathrm{\mid }\lambda \mathrm{\mid })\to X^{\ast \ast }$ is bounded, and the identity defining $B_S$ gives $J_{X}S=RIU$ first after evaluation at each $x^{\ast }$ and hence in $X^{\ast \ast }$. Applying the polar decomposition of $\lambda$ puts its total variation exactly into the product norm, proving the asserted infimum. Since $J_X$ is onto and isometric, $J_X^{-1}R$ is a factorization of $S$ through $I:L^{\mathrm{\infty }}\to L^1$ with the same norm. Hence $\mathrm{\parallel }S{\mathrm{\parallel }}_{PI}\le \mathrm{\parallel }S{\mathrm{\parallel }}_I$; the reverse inequality is general. Combining this with step 2.1 gives

$\mathcal{N}(E,X)=\mathcal{P}\mathcal{I}(E,X)=\mathcal{I}(E,X)$

isometrically.

$u={\sum }_{n\ge 1}{x}_n^{\ast }\otimes x_n,{\sum }_n\mathrm{\parallel }{x}_n^{\ast }\mathrm{\parallel }<\mathrm{\infty },x_n\to 0.$

Indeed, choose successive finite-tensor approximants whose projective-norm errors are below $4^{-q}$, write the difference in block $q$ with total coefficient norm below $2\cdot 4^{-q}$, and multiply its second factors by $2^{-q}$ while multiplying its first factors by $2^q$. The first-factor norms then have summable block totals and the second-factor norms tend to zero after normalizing each original elementary tensor.

Thus $K_0=\{0,x_1,x_2,\dots \}$ is compact. AP supplies finite-rank maps $R_{\alpha }$ for which ${\sup }_n\mathrm{\parallel }{R}_{\alpha }{x}_n-x_n\mathrm{\parallel }\to 0$, and consequently

$u={\lim }_{\alpha }{\sum }_n{x}_n^{\ast }\otimes R_{\alpha }{x}_n$

in projective norm. If the operator $S_u$ induced by $u$ is zero and $R_{\alpha }x={\sum }_k{y}_k^{\ast }(x)y_k$, then

${\sum }_n{x}_n^{\ast }\otimes R_{\alpha }{x}_n={\sum }_k{S}_u^{\ast }(y_k^{\ast })\otimes y_k=0.$

Therefore $u=0$. The canonical quotient $X^{\ast }{\widehat{\otimes }}_{\pi }X\to \mathcal{N}(X,X)$ is injective, hence isometric.

$Q:X^{\ast }{\widehat{\otimes }}_{\pi }X⟶(X{\widehat{\otimes }}_{\epsilon }{X}^{\ast }{)}^{\ast }$

by

$Q\text{\hspace{0.17em}⁣}({\sum }_n{x}_n^{\ast }\otimes x_n)\text{\hspace{0.17em}⁣}({\sum }_j{y}_j\otimes y_j^{\ast })={\sum }_{n,j}{x}_n^{\ast }(y_j)y_j^{\ast }(x_n)\mathrm{.}$

By step 2.3 the domain is $\mathcal{N}(X,X)$ with its nuclear norm, and by the definition in step 1.1 the codomain is $\mathcal{I}(X,X^{\ast \ast })$. Under these identifications $Q$ sends $S$ to $J_{X}S$. The two associated bilinear forms are literally equal,

$B_{J_{X}S}(x,x^{\ast })=J_X(Sx)(x^{\ast })=x^{\ast }(Sx)=B_S(x,x^{\ast }),$

so the integral norm is unchanged. Step 2.2 identifies the nuclear and integral norms on the domain. Hence $Q$ is an isometry.

$\pi (u)={\sup }_{T\in {\mathcal{F}}_1}\mathrm{\mid }\mathrm{tr}(T{S}_u)\mathrm{\mid }\mathrm{.}$

Reflexivity identifies every functional on $X^{\ast }{\widehat{\otimes }}_{\pi }X$ with an operator on $X$, so the ordinary dual formula for the projective norm gives the same supremum over the full unit ball of $\mathcal{L}(X)$. The Hahn--Banach bipolar theorem from [L3] therefore makes ${\mathcal{F}}_1$ weak-operator dense in that unit ball. In particular, for every finite tuple $x_1,\dots ,x_r$, the tuple $(x_1,\dots ,x_r)$ lies in the weak closure of

$\{(T{x}_1,\dots ,T{x}_r):T\in {\mathcal{F}}_1\}\subseteq X^r\mathrm{.}$

This set is convex, so its weak and norm closures agree by [L3]. It follows that $I_X$ is in the strong-operator closure of ${\mathcal{F}}_1$: there is a net $(T_{\alpha })$ of finite-rank contractions with $T_{\alpha }x\to x$ for every $x\in X$.

$\mathrm{\parallel }{T}_{\alpha }x-x\mathrm{\parallel }\le \mathrm{\parallel }{T}_{\alpha }(x-x_i)\mathrm{\parallel }+\mathrm{\parallel }{T}_{\alpha }{x}_i-x_i\mathrm{\parallel }+\mathrm{\parallel }{x}_i-x\mathrm{\parallel }<\epsilon \mathrm{.}$

Thus $X$ has MAP by [L1]. The zero space is covered by $T=0$. The proof works over both scalar fields; in the complex case every separation argument is applied to real parts. AC is the umbrella assumption for the supplier hypotheses and selections listed above: scalar Radon--Nikodym, Banach--Alaoglu and the weak-compactness/subsequence steps, real and complex Hahn--Banach (including separation and norming functionals), regular-measure representation, countably generated $L^1$ separability, and the countable approximation and tensor-representation choices. [A1, L1, L3, step 1.2] ∎