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# Alexander Shamov

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 Name Alexander Shamov Member for 1 year Seen 22 hours ago Website Location Ukraine Age 22
 Apr23 comment Preimage of zero measure setsSee, e.g., Janson "Gaussian Hilbert spaces", Chapter 15.4, where you will find a discussion of such results for "ray absolutely continuous" functions. Such conditions, generalized to the setting of infinite-dimensional Gaussian measures, are the driving force behind (the most celebrated application of) Malliavin calculus. Apr23 accepted Preimage of zero measure sets Apr23 comment Preimage of zero measure sets@Noah,Emil: Sorry, I never knew that notation could be confusing. Apr23 comment Preimage of zero measure sets@Emil: I mean that the gradient is zero. Apr23 revised Preimage of zero measure setsadded 419 characters in body; edited body Apr23 comment Preimage of zero measure sets@Noah: I guess you didn't read carefully :) Apr23 answered Preimage of zero measure sets Apr9 awarded ● Yearling Mar10 awarded ● Nice Answer Feb10 revised Interpretation of Riemann tensor antisymmetryadded 2 characters in body Feb10 revised Interpretation of Riemann tensor antisymmetrydeleted 2 characters in body Feb10 answered Interpretation of Riemann tensor antisymmetry Feb6 comment Maximal inequalities for certain functions of a martingale difference sequence1) Because it is nonnegative. 2) Right, and $\sum_{t=1}^T \xi_t$ is a martingale, so this inequality does apply to it, with no connection to 1) at all. Probably you should at least check what you wrote against the definitions to eliminate trivialities or nonsense. Feb5 answered Maximal inequalities for certain functions of a martingale difference sequence Feb4 comment Riesz representation for an infinite-dimensional spaceThe set of compact subsets is naturally ordered by inclusion. To each compact $K$ we have to associate a space - namely, $C(K)$ - and for each pair $K_1 \subset K_2$ we need a map $C(K_2) \to C(K_1)$ - in this case it is the restriction map. Feb4 comment Riesz representation for an infinite-dimensional space@Banach: Of course not. A projective limit may be defined for any directed family, which may be neither countable nor linearly ordered. Intuitively, this means that a continuous function "is" a collection of functions on compact subsets that fit together nicely, and the topology on $C(X)$ is generated by taking these restrictions. For precise definitions, see, e.g., en.wikipedia.org/wiki/Inverse_limit and en.wikipedia.org/wiki/Initial_topology. Feb3 answered Riesz representation for an infinite-dimensional space Jan9 comment Usage of set theory in undergraduate studiesI think a little experience with programming is enough to eliminate the whole free vs. bound thing :) Dec31 answered Old books still used Dec31 comment Old books still used+1 for Polya & Szego!