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The Carleson maximal operator is not strong type (1,1)
Statement refuted
There exists a finite constant bounding by that constant times for every .
In fact, for every there is a nonnegative trigonometric polynomial on the period-one torus, with Haar mass one, such that
As a further consequence under DC, there exists a real with .
Facts & Assumptions
Given: and normalized Haar measure on . DC is assumed only for the further norm-divergence consequence.
The measurable maximal operator is for (Carleson maximal partial-sum operator).
The Fejer kernel satisfies and for every (The Fejer kernel is a positive approximate identity).
Fejer means of a continuous one-periodic complex function converge uniformly to that function (Fejer means converge uniformly for continuous periodic functions).
For every one-period integrable and every , (Fourier partial sums are Dirichlet convolutions).
The quantities equal the norms of the partial-sum operators on continuous functions and are at least for (Fourier partial-sum operator norm equals the Lebesgue constant).
For every measure space and , is complete in its norm (Riesz-Fischer completeness of for ).
Under DC, pointwise bounded families of bounded linear maps from a Banach space to a normed space have uniformly bounded norms (Uniform boundedness principle).
For two sigma-finite measure spaces and a nonnegative product-measurable function, the product integral equals both iterated integrals (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).
Counterexample
Choose an integer with , possible from the logarithmic lower bound. For each , the function is a real nonnegative trigonometric polynomial of norm one.
Expanding the square formula for gives coefficient for and zero otherwise. Orthogonality of finite characters therefore gives : either side has coefficient for and zero elsewhere. Since is continuous, as , with fixed.
For the additional consequence, let be a real integrable periodic representative. The convolution formula, after a periodic change of variables, gives . The integrand is product-measurable: depends on one coordinate and is continuous. Both measure spaces are finite, so Tonelli and translation invariance give . Thus is a bounded real linear operator on real .
Choose large enough that this uniform error is less than one. The measure has mass one, so . Since , the polynomial is the required witness. Its maximal function is finite and bounded: when the partial-sum index is at least , the partial sum equals , so only finitely many distinct continuous polynomials enter its supremum. Thus the strict norm inequality is an ordinary finite integral, not an artifact of an infinite value.
For each fixed , step 1.2 and the unit-norm tests imply . These operator norms are consequently unbounded.
Now assume DC. Real is Banach by completeness. If were finite for each , uniform boundedness would contradict step 2.2. Therefore some real integrable has unbounded norms of its partial sums. This is a norm-divergence conclusion and does not assert almost-everywhere divergence.
Depends on
- Carleson maximal partial-sum operator
- The Fejer kernel is a positive approximate identity
- Fejer means converge uniformly for continuous periodic functions
- Fourier partial sums are Dirichlet convolutions
- Fourier partial-sum operator norm equals the Lebesgue constant
- Riesz-Fischer completeness of $L^p$ for $1 \le p \le \infty$
- Uniform boundedness principle
- Tonelli's theorem for nonnegative measurable functions on a sigma-finite product
Used by
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Sources
- Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)