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Strong type (1,1) fails for the Hilbert transform

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

The interval indicator $f = 1_{(0, 1]}$ lies in $L^{1} (R)$ , but its Hilbert transform is $H f (x) = π^{- 1} \log ∣ x / (x - 1) ∣$ for $x \notin {0, 1}$ , which is not integrable because $∣ H f (x) ∣ ≳ 1 / ∣ x ∣$ for large $x$ . Hence the Hilbert transform has no compatible strong type $(1, 1)$ extension: there is no bounded $L^{1} \to L^{1}$ extension agreeing with the $L^{2}$ Hilbert transform on $L^{1} \cap L^{2}$ . No weak $(1, 1)$ estimate is refuted here; the Hilbert transform is a Calderón–Zygmund operator and the companion page proves the weak endpoint instead.

Facts & Assumptions

Given: Countable Choice; the indicator $f = 1_{(0, 1]}$ ; the function $q (x) = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣}$ on $R ∖ {0, 1}$ ; the dominating function $G (x) : = ∣ q (x) ∣$ off ${0, 1}$ , with arbitrary values on that null set; the $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions.

[F1]

For $h \in L^{p} (R)$ , $1 \leq p < \infty$ , and $ε > 0$ , the truncation $H_{ε} h (x) = \frac{1}{π} \int_{∣ x - y ∣ > ε} \frac{h (y)}{x - y} d y$ is an absolutely convergent Lebesgue integral, is defined for every $x$ , and depends only on the almost-everywhere class of $h$ (Truncated Hilbert transform and principal value). The interval-specific domination $∣ H_{ε} f ∣ \leq G$ is proved in step 1.1 below.

[F2]

For every Schwartz function $φ$ the principal value $\lim_{ε ↓ 0} H_{ε} φ (y)$ exists at every $y$ and equals $H φ (y)$ ; the Hilbert transform extends to an isometric $L^{2}$ multiplier operator with $H^{*} = - H$ for the first-variable-linear pairing, so $⟨ H h, ψ ⟩ = ⟨ h, H^{*} ψ ⟩$ (The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, The Hilbert transform is an L2 isometry and squares to minus the identity, The Hilbert transform is skew-adjoint on L2, The Hilbert-space adjoint of a bounded operator).

[F3]

Dominated convergence holds for sequences dominated by an integrable function on a fixed measure space (Dominated convergence); Countable Choice is assumed throughout and is used only through the cited suppliers.

[F4]

Fubini interchanges absolutely integrable complex double integrals, and locally integrable functions with equal distribution pairings agree almost everywhere. (Fubini's theorem for L^1 functions on a sigma-finite product, Locally integrable functions embed in distributions)

Counterexample

technique · direct

1.1F1givenalgebra

Explicit values of the truncations. For $x \notin {0, 1}$ and $0 < ε < d (x) : = \min {∣ x ∣, ∣ x - 1 ∣}$ the truncation is a sum of ordinary integrals with no singular point in the domain, and direct antiderivatives give: for $x < 0$ or $x > 1$ , $H_{ε} f (x) = \frac{1}{π} \int_{0}^{1} \frac{d y}{x - y} = \frac{1}{π} \log \frac{∣ x ∣}{∣ x - 1 ∣} = q (x)$ ; for $0 < x < 1$ , $H_{ε} f (x) = \frac{1}{π} (\int_{0}^{x - ε} + \int_{x + ε}^{1}) \frac{d y}{x - y} = \frac{1}{π} (\log x - \log ε + \log ε - \log (1 - x)) = \frac{1}{π} \log \frac{x}{1 - x} = q (x)$ , since $∣ x / (x - 1) ∣ = x / (1 - x)$ there. Hence $H_{ε} f (x) = q (x)$ for every $ε < d (x)$ , so $\lim_{ε ↓ 0} H_{ε} f (x) = q (x)$ at every $x \notin {0, 1}$ . Moreover, for $0 < x < 1$ direct integration gives $π H_{ε} f (x) = (\log (x / ε))_{+} - (\log ((1 - x) / ε))_{+}$ . Since $u \mapsto u_{+}$ is $1$ -Lipschitz, $∣ H_{ε} f (x) ∣ \leq ∣ q (x) ∣$ . Outside $[0, 1]$ the integrand has one sign, so deleting part of the integration domain also gives $∣ H_{ε} f (x) ∣ \leq ∣ q (x) ∣$ . Thus $G$ dominates the truncations almost everywhere. It is locally integrable because $∣ q (x) ∣ \leq π^{- 1} (∣ \log ∣ x ∣ ∣ + ∣ \log ∣ x - 1 ∣ ∣)$ and $\int_{0}^{a} ∣ \log t ∣ d t < \infty$ for finite $a > 0$ .

1.2F1F2F3givenalgebraF4

Distributional convergence to the $L^{2}$ transform. Let $φ \in C_{c}^{\infty} (R)$ and put $b_{ε} (y) : = \frac{1}{π} \int_{∣ x - y ∣ > ε} \frac{\overline{φ (x)}}{x - y} d x$ ; on the domain $∣ x - y ∣ \geq ε$ the double integral $\iint_{∣ x - y ∣ > ε} \frac{∣ f (y) φ (x) ∣}{∣ x - y ∣} d y d x$ is finite because $∣ x - y ∣^{- 1} \leq ε^{- 1}$ and $f$ and $φ$ are bounded with bounded support, so Fubini applies and $⟨ H_{ε} f, φ ⟩ = \int_{0}^{1} f (y) b_{ε} (y) d y$ . As $ε ↓ 0$ one has $b_{ε} (y) = - \frac{1}{π} \int_{∣ x - y ∣ > ε} \frac{\overline{φ (x)}}{y - x} d x \to - H \overline{φ} (y) = - \overline{H φ (y)}$ , using that the kernel is real and [F2]; the convergence is dominated by a constant depending on $φ$ , because applying the estimate $∣ \int_{∣ t ∣ > ε} ψ (y + t) t^{- 1} d t ∣ \leq \int_{∣ t ∣ \leq 1} ∣ ψ (y + t) - ψ (y) ∣ ∣ t ∣^{- 1} d t + \int_{∣ t ∣ > 1} ∣ ψ (y + t) ∣ d t \leq 2 ∥ ψ^{'} ∥_{\infty} + 2 ∥ ψ ∥_{1}$ to $ψ = \overline{φ}$ bounds all $b_{ε} (y)$ uniformly. Hence dominated convergence on the finite-measure set $(0, 1]$ gives $⟨ H_{ε} f, φ ⟩ \to - \int_{0}^{1} f (y) \overline{H φ (y)} d y = - ⟨ f, H φ ⟩ = ⟨ f, H^{*} φ ⟩ = ⟨ H f, φ ⟩$ .

1.3givenalgebra

$q \notin L^{1} (R)$ : for $x \geq 2$ one has $\frac{x}{x - 1} = 1 + \frac{1}{x - 1}$ with $\frac{1}{x - 1} \leq 1$ , and $\log (1 + u) \geq \frac{u}{2}$ for $0 \leq u \leq 1$ , so $q (x) = \frac{1}{π} \log (1 + \frac{1}{x - 1}) \geq \frac{1}{2 π (x - 1)} \geq \frac{1}{2 π x}$ . Therefore $\int_{2}^{\infty} ∣ q ∣ \geq \frac{1}{2 π} \int_{2}^{\infty} \frac{d x}{x} = + \infty$ .

2.1F2step 1.1step 1.2algebraF4

Identification of the limit. On each compact $K$ the domination $∣ H_{ε} f ∣ \leq G$ of step 1.1 with $G \in L^{1} (K)$ and the pointwise convergence $H_{ε} f \to q$ off the null set ${0, 1}$ let dominated convergence pass the limit inside the pairing: $⟨ H_{ε} f, φ ⟩ \to \int q \overline{φ}$ for every test function $φ$ supported in $K$ , hence for every test function. Comparing with step 1.2, $\int (q - H f) \overline{φ} = 0$ for every $φ \in C_{c}^{\infty}$ , and both $q$ and the $L^{2}$ class $H f$ are locally integrable, so $q = H f$ almost everywhere; in particular $f \in L^{1} \cap L^{2}$ with $∥ f ∥_{1} = 1$ and $H f = q$ .

3.1step 2.1step 1.3algebra∎

Suppose $T : L^{1} (R; C) \to L^{1} (R; C)$ were bounded and agreed with the $L^{2}$ Hilbert transform on $L^{1} \cap L^{2}$ . Since $f \in L^{1} \cap L^{2}$ by step 2.1, the $L^{1}$ class $T f$ would equal the $L^{1}$ class $H f$ , which step 2.1 identifies with $q$ ; but $q \notin L^{1}$ by step 1.3, whereas $T f \in L^{1}$ by definition of $T$ . This contradiction shows that no such $T$ exists, which is exactly the failure of strong type $(1, 1)$ ; the statement says nothing about weak type $(1, 1)$ .

Depends on

Used by

Calderón–Zygmund operators need not map L∞ to L∞ Counterexample

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Sources

Loukas Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)