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Differentiation of L-one functions for a doubling weight
Statement
Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain), hence Countable Choice (Dependent choice implies countable choice). Let be a weight on whose measure is doubling, and let (Weights, their associated measures, and the spaces L^p(w)). Then for -almost every , more generally the same limit holds along any family of balls or cubes shrinking nicely to (A family shrinking nicely to a point) with -measure comparable to the corresponding ball.
Facts & Assumptions
Given: Dependent Choice, a weight with doubling, and .
is the weighted maximal function of The weighted maximal function of a doubling weight, and it obeys the weak bound for every (The weighted maximal function of a doubling weight is weak (1,1)).
is dense in because is a Radon measure (C_c(X) is dense in L^p(mu) for a Radon measure), with complex density obtained by approximating the two real components separately; a function is continuous, so for every and every family shrinking nicely to the weighted averages of converge to (the averages of are bounded by the maximum of over the shrinking sets, which tends to ).
Chebyshev's inequality: for a nonnegative measurable and , (Chebyshev-Markov inequality for the integral).
Dominated convergence gives continuity in radius for local weighted integrals, and Lebesgue-measurable functions have Borel representatives: apply A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra componentwise using is exactly the completion of the restriction of to the Borel sets, setting infinite values on null sets to zero. These representatives give jointly Borel integrands by composition with continuous maps; Tonelli gives measurable section integrals (Dominated convergence, Borel representatives make the convolution integrand Borel measurable, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).
Proof
First suppose globally and put . For and , the centred weighted oscillation is at most , since the corresponding oscillation of continuous tends to zero. These limsups are measurable: for each fixed the integrals are continuous in positive radius by dominated convergence, hence rational radii suffice; joint measurability follows after choosing Borel representatives, and changing them on a null set does not change the a.e. assertion.
For , . The weak bound and Chebyshev give . Taking the infimum over using [F2] makes this measure zero. The countable union over is null, so the centred absolute oscillation tends to zero a.e. For local , apply this result to ; on , sufficiently small centred balls see . The countable union of these exceptional sets is null and , proving the same conclusion for every local input.
Outside the null set of step 2.1, the absolute oscillation over any shrinking set with and is bounded by times the centred oscillation. It therefore tends to zero, and the modulus of the difference between the average of and is at most this oscillation. Balls and cubes shrinking nicely with the stipulated weighted comparability meet these hypotheses. This proves all the stated limits, on a common full-measure set.
Depends on
- Weights, their associated measures, and the spaces L^p(w)
- The weighted maximal function of a doubling weight
- The weighted maximal function of a doubling weight is weak (1,1)
- C_c(X) is dense in L^p(mu) for a Radon measure
- Chebyshev-Markov inequality for the integral
- A family shrinking nicely to a point
- The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain
- Dependent choice implies countable choice
- Dominated convergence
- Borel representatives make the convolution integrand Borel measurable
- Complex translation, convolution, approximate identities, and mollification
- A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra
- $\mathcal{L}(\mathbb{R}^n)$ is exactly the completion of the restriction of $\lambda_n$ to the Borel sets
- Tonelli's theorem for nonnegative measurable functions on a sigma-finite product
Used by
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Sources
- Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249, 2014) (standard reference, not scraped)
- Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)