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The Carleson maximal operator is not strong type (1,1)

Statement refuted

There exists a finite constant bounding Cf1 by that constant times f1 for every fL1(T).

In fact, for every A>0 there is a nonnegative trigonometric polynomial f on the period-one torus, with Haar mass one, such that

f1=1,Cf1>A.

As a further consequence under DC, there exists a real hL1(T) with supNSNh1=.

Facts & Assumptions

Given: A>0 and normalized Haar measure on T=R/Z. DC is assumed only for the further norm-divergence consequence.

[F1]

The measurable maximal operator is Cf=supN0SNf for fL1 (Carleson maximal partial-sum operator).

[F2]

The Fejer kernel satisfies FK(t)=(K+1)1j=0Ke2πijt20 and 01FK(t)dt=1 for every K0 (The Fejer kernel is a positive approximate identity).

[F3]

Fejer means of a continuous one-periodic complex function converge uniformly to that function (Fejer means converge uniformly for continuous periodic functions).

[F4]

For every one-period integrable f and every x, SNf(x)=(fDN)(x) (Fourier partial sums are Dirichlet convolutions).

[F5]

The quantities DN1 equal the norms of the partial-sum operators on continuous functions and are at least (3π)1log(N+1) for N1 (Fourier partial-sum operator norm equals the Lebesgue constant).

[F6]

For every measure space and 1p, Lp is complete in its norm (Riesz-Fischer completeness of Lp for 1p).

[F7]

Under DC, pointwise bounded families of bounded linear maps from a Banach space to a normed space have uniformly bounded norms (Uniform boundedness principle).

[F8]

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

technique · Fejer-kernel tests, with an additional uniform-boundedness consequence
1.1

Choose an integer N1 with DN1>A+1, possible from the logarithmic lower bound. For each K0, the function FK is a real nonnegative trigonometric polynomial of L1 norm one.

F2F5given
1.2

Expanding the square formula for FK gives coefficient 1k/(K+1) for kK and zero otherwise. Orthogonality of finite characters therefore gives SNFK=FKDN=σKDN: either side has coefficient 1k/(K+1) for kmin{K,N} and zero elsewhere. Since DN is continuous, SNFKDN0 as K, with N fixed.

F2F3F4
1.3

For the additional consequence, let h be a real integrable periodic representative. The convolution formula, after a periodic change of variables, gives SNh(x)01h(u)DN(xu)du. The integrand is product-measurable: h(u) depends on one coordinate and DN(xu) is continuous. Both measure spaces are finite, so Tonelli and translation invariance give SNh101h(u)01DN(xu)dxdu=DN1h1. Thus SN is a bounded real linear operator on real L1.

F4F8
2.1

Choose K large enough that this uniform error is less than one. The measure has mass one, so SNFK1DN1SNFKDN1>A. Since CFKSNFK, the polynomial f=FK is the required witness. Its maximal function is finite and bounded: when the partial-sum index is at least K, the partial sum equals FK, 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.

F1step 1.1step 1.2
2.2

For each fixed N, step 1.2 and the unit-norm tests FK imply SN:L1(T,R)L1(T,R)limKSNFK1=DN1. These operator norms are consequently unbounded.

F2F5step 1.2step 1.3
3.1

Now assume DC. Real L1 is Banach by completeness. If supNSNh1 were finite for each h, uniform boundedness would contradict step 2.2. Therefore some real integrable h has unbounded L1 norms of its partial sums. This is a norm-divergence conclusion and does not assert almost-everywhere divergence.

F6F7step 1.3step 2.2given

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