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

Statement refuted

Under Countable Choice, there exists a finite constant bounding $∥ C f ∥_{1}$ by that constant times $∥ f ∥_{1}$ for every $f \in L^{1} (T)$ .

Assume the Axiom of Countable Choice. In fact, for every $A > 0$ there is a nonnegative trigonometric polynomial $f$ on the period-one torus, with Haar mass one, such that

$∥ f ∥_{1} = 1, ∥ C f ∥_{1} > A .$

As a further consequence under DC, there exists a real $h \in L^{1} (T)$ with $\sup_{N} ∥ S_{N} h ∥_{1} = \infty$ .

Facts & Assumptions

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

[F1]

The measurable maximal operator is $C f = \sup_{N \geq 0} ∣ S_{N} f ∣$ for $f \in L^{1}$ (Carleson maximal partial-sum operator).

[F2]

The Fejer kernel satisfies $F_{K} (t) = (K + 1)^{- 1} ∣ \sum_{j = 0}^{K} e^{2 π i j t} ∣^{2} \geq 0$ and $\int_{0}^{1} F_{K} (t) d t = 1$ for every $K \geq 0$ (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 $x$ , $S_{N} f (x) = (f * D_{N}) (x)$ (Fourier partial sums are Dirichlet convolutions).

[F5]

The quantities $∥ D_{N} ∥_{1}$ equal the norms of the partial-sum operators on continuous functions and are at least $(3 π)^{- 1} \log (N + 1)$ for $N \geq 1$ (Fourier partial-sum operator norm equals the Lebesgue constant).

[F6]

For every measure space and $1 \leq p \leq \infty$ , $L^{p}$ is complete in its norm (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ).

[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

1.1F2F5given

Choose an integer $N \geq 1$ with $∥ D_{N} ∥_{1} > A + 1$ , possible from the logarithmic lower bound. For each $K \geq 0$ , the function $F_{K}$ is a real nonnegative trigonometric polynomial of $L^{1}$ norm one.

1.2F2F3F4

Expanding the square formula for $F_{K}$ gives coefficient $1 - ∣ k ∣ / (K + 1)$ for $∣ k ∣ \leq K$ and zero otherwise. Orthogonality of finite characters therefore gives $S_{N} F_{K} = F_{K} * D_{N} = σ_{K} D_{N}$ : either side has coefficient $1 - ∣ k ∣ / (K + 1)$ for $∣ k ∣ \leq \min {K, N}$ and zero elsewhere. Since $D_{N}$ is continuous, $∥ S_{N} F_{K} - D_{N} ∥_{\infty} \to 0$ as $K \to \infty$ , with $N$ fixed.

1.3F4F8

For the additional consequence, let $h$ be a real integrable periodic representative. The convolution formula, after a periodic change of variables, gives $∣ S_{N} h (x) ∣ \leq \int_{0}^{1} ∣ h (u) ∣ ∣ D_{N} (x - u) ∣ d u$ . The integrand is product-measurable: $h (u)$ depends on one coordinate and $D_{N} (x - u)$ is continuous. Both measure spaces are finite, so Tonelli and translation invariance give $∥ S_{N} h ∥_{1} \leq \int_{0}^{1} ∣ h (u) ∣ \int_{0}^{1} ∣ D_{N} (x - u) ∣ d x d u = ∥ D_{N} ∥_{1} ∥ h ∥_{1}$ . Thus $S_{N}$ is a bounded real linear operator on real $L^{1}$ .

2.1F1step 1.1step 1.2

Choose $K$ large enough that this uniform error is less than one. The measure has mass one, so $∥ S_{N} F_{K} ∥_{1} \geq ∥ D_{N} ∥_{1} - ∥ S_{N} F_{K} - D_{N} ∥_{1} > A$ . Since $C F_{K} \geq ∣ S_{N} F_{K} ∣$ , the polynomial $f = F_{K}$ is the required witness. Its maximal function is finite and bounded: when the partial-sum index is at least $K$ , the partial sum equals $F_{K}$ , 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.

2.2F2F5step 1.2step 1.3

For each fixed $N$ , step 1.2 and the unit-norm tests $F_{K}$ imply $∥ S_{N} : L^{1} (T, R) \to L^{1} (T, R) ∥ \geq \lim_{K \to \infty} ∥ S_{N} F_{K} ∥_{1} = ∥ D_{N} ∥_{1}$ . These operator norms are consequently unbounded.

3.1F6F7step 1.3step 2.2given∎

Now assume DC. Real $L^{1}$ is Banach by completeness. If $\sup_{N} ∥ S_{N} h ∥_{1}$ were finite for each $h$ , uniform boundedness would contradict step 2.2. Therefore some real integrable $h$ has unbounded $L^{1}$ norms of its partial sums. This is a norm-divergence conclusion and does not assert almost-everywhere divergence.

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Sources

Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)