Let $X$ and $Y$ be Banach spaces, and $T:X^{\ast }\to Y$ be an operator. We prove that if $X$ is Asplund and $Y$ has the approximation property, then for each Radon probability $\mu$ on $(B_{X^{\ast }},w^{\ast })$ there is a sequence of $w^{\ast }$‐to‐norm continuous operators $T_n:X^{\ast }\to Y$ such that $\parallel T_n(x^{\ast })-T(x^{\ast })\parallel \to 0$ for $\mu$‐a.e. $x^{\ast }\in B_{X^{\ast }}$; if $Y$ has the $\lambda$‐bounded approximation property for some $\lambda ⩾1$, then the sequence can be chosen in such a way that $\parallel T_n\parallel ⩽\lambda \parallel T\parallel$ for all $n\in \mathbb{N}$. The same conclusions hold if $X$ contains no subspace isomorphic to ${\ell }_1$, $Y$ has the approximation property (respectively, $\lambda$‐bounded approximation property) and $T$ has separable range. This extends to the non‐separable setting a result by Mercourakis and Stamati.

中文翻译：

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弱范数到范数连续算子的几乎所有收敛序列

让 $X$ 和 $ÿ$ 是Banach空间，并且 $Ť：X^{\ast }\to ÿ$成为运营商。我们证明$X$ 是Asplund和 $ÿ$ 具有近似属性，那么对于每个Radon概率 $\mu$ 上 $（{乙}_{X^{\ast }}，w^{\ast }）$ 有一个序列 $w^{\ast }$规范的连续运算符 ${Ť}_{ñ}：X^{\ast }\to ÿ$ 这样 $\parallel {Ť}_{ñ}（X^{\ast }）-Ť（X^{\ast }）\parallel \to 0$ 为了 $\mu$‐ae $X^{\ast }\in {乙}_{X^{\ast }}$; 如果$ÿ$ 有 $\lambda$某些的有界近似属性 $\lambda ⩾\mathrm{1个}$，那么序列可以这样选择： $\parallel {Ť}_{ñ}\parallel ⩽\lambda \parallel Ť\parallel$ 对全部 $ñ\in \mathbb{ñ}$。如果得出相同结论$X$ 不包含与的同构子空间 ${\ell }_{\mathrm{1个}}$， $ÿ$ 具有近似属性（分别是 $\lambda$有界的近似属性）和 $Ť$具有可分离的范围。这延伸到了Mercourakis和Stamati提出的不可分割的设置。