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The Banach-Sachs property and Haar null sets. (English) Zbl 0937.46011

The author modifies the definition of Haar null set in an equivalent way and uses it to prove the following fact.
Let \(X\) be a Banach space such that \(X^*\) has the Banach-Sachs property (i.e. every bounded sequence \((x_n^*)\subset X^*\) has a subsequence \((x_{n_k}^*)\) with the norm convergent Cesaro means \(\frac 1k \sum _i^kx^*_{n_i}\)) and let \(K\subset X\) be a convex closed set with empty interior. Then there is a probability measure \(\mu \) on \(X\) such that \(\mu (K+x)=0\) for all \(x\in X\).
This may not be true if the Haar nullness is replaced by Aronszajn nullness as is noticed at the end of the paper. Namely, any closed convex set \(K\) in a separable Banach space \(X\) satisfying \(\overline {\text{span}} K=X\) cannot be Aronszajn null set.

MSC:

46B20 Geometry and structure of normed linear spaces
28A75 Length, area, volume, other geometric measure theory
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