Alexander Shamov
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Registered User
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Apr 23 |
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Preimage of zero measure sets See, 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. |
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Apr 23 |
accepted | Preimage of zero measure sets |
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Apr 23 |
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Preimage of zero measure sets @Noah,Emil: Sorry, I never knew that notation could be confusing. |
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Apr 23 |
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Preimage of zero measure sets @Emil: I mean that the gradient is zero. |
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Apr 23 |
revised |
Preimage of zero measure sets added 419 characters in body; edited body |
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Apr 23 |
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Preimage of zero measure sets @Noah: I guess you didn't read carefully :) |
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Apr 23 |
answered | Preimage of zero measure sets |
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Apr 9 |
awarded | ● Yearling |
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Mar 10 |
awarded | ● Nice Answer |
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Feb 10 |
revised |
Interpretation of Riemann tensor antisymmetry added 2 characters in body |
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Feb 10 |
revised |
Interpretation of Riemann tensor antisymmetry deleted 2 characters in body |
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Feb 10 |
answered | Interpretation of Riemann tensor antisymmetry |
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Feb 6 |
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Maximal inequalities for certain functions of a martingale difference sequence 1) 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. |
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Feb 5 |
answered | Maximal inequalities for certain functions of a martingale difference sequence |
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Feb 4 |
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Riesz representation for an infinite-dimensional space The 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. |
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Feb 4 |
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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. |
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Feb 3 |
answered | Riesz representation for an infinite-dimensional space |
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Jan 9 |
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Usage of set theory in undergraduate studies I think a little experience with programming is enough to eliminate the whole free vs. bound thing :) |
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Dec 31 |
answered | Old books still used |
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Dec 31 |
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Old books still used +1 for Polya & Szego! |
