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Why is any norm-bounded family $T \subseteq L(X)$ on a reflexive Banach space $X$ relatively weakly compact?

Rasmus
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Youssef
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  • Why do you assume that this might be true? – Rasmus Sep 15 '13 at 21:15
  • Did you mean a norm-bounded family in the dual? – Daniel Fischer Sep 15 '13 at 21:16
  • @DanielFischer no – Youssef Sep 15 '13 at 21:40
  • @Rasmus Lemma 2.7 page 312 One parameter Semigroups For Linear Evolution Equation. – Youssef Sep 15 '13 at 21:55
  • Then what do you mean by "relatively weakly compact"? Relatively compact in the topology of pointwise convergence as maps to $\sigma(X,X')$? – Daniel Fischer Sep 15 '13 at 22:10
  • For a set of operators T ⊂ L(X), X a Banach space, the following assertions are equivalent.

    (a) T is relatively compact in Lσ(X).

    (b) {T x : T ∈ T } is relatively weakly compact in X for all x ∈ X.

    (c) T is bounded, and {T x : T ∈ T } is relatively weakly compact in X for all x in some dense subset of X.

    Lσ(X) the space L(X) endowed with the weak operator topology

     – Youssef Sep 15 '13 at 22:59

1 Answers1

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Recall $L(X,Y')\cong_1 (Y\otimes_{\pi} X)'$ via cononical isometric isomorphism $I_{X,Y}$ well defined by $I_{X,Y}(T)(y\otimes_\pi x)=(T(x))(y)$. Since $X=X''$ we conclude $$ L(X)\cong_1 (X'\otimes_{\pi} X)' $$ As $I_{X,X'}$ is isometric isomorphism, then the family $\mathcal{S}:=I_{X,X'}(\mathcal{T})\subset (X'\otimes_{\pi} X)'$ is norm bounded, hence $\operatorname{cl}_{w^*}(\mathcal{S})$ is $w^*$-compact. Since $L^\sigma(X)\cong((X'\otimes_{\pi} X)',w^*)$ via $I_{X,X'}$, then $\operatorname{cl}_{L^\sigma(X)}(\mathcal{T})=I_{X,X'}^{-1}(\operatorname{cl}_{w^*}(\mathcal{S}))$ is compact in $L^\sigma(X)$. Thus $\mathcal{T}$ is a relatively weakly compact family.

Norbert
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