CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generated

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A BMO function need not be globally integrable

Statement refuted

The claim refuted is the inclusion $B M O (R^{n}) \subseteq L^{1} (R^{n})$ , equivalently the claim that every BMO function is globally integrable. The witness is the function $b (x) = \log ∣ x ∣$ ( $b (0) = 0$ ) of The logarithm is in BMO but not in L-infinity, which satisfies $\int_{R^{n}} ∣ b (x) ∣ d x = + \infty$ ; hence $B M O (R^{n}) \subseteq̸ L^{1} (R^{n})$ . Thus BMO membership alone does not guarantee a global Lebesgue pairing $\int f b$ : absolute integrability of the product $f b$ must be checked separately. Such a pairing can exist even without cancellation of $f$ .

Facts & Assumptions

Given: The function $b (x) = \log ∣ x ∣$ for $x \neq 0$ , $b (0) = 0$ , on $R^{n}$ ; the cubes of Axis-parallel rectangles in $R^{m}$ and their volume.

[F1]

$b \in B M O (R^{n})$ and $b$ is unbounded near the origin (The logarithm is in BMO but not in L-infinity).

[F2]

For $R_{m} : = 4 \sqrt{n} + m$ and the increasing closed balls $E_{m} : = {x : ∣ x ∣ \leq R_{m}}$ , one has $E_{m} ↑ R^{n}$ and $\int_{R^{n}} ∣ b ∣ = \lim_{m \to \infty} \int_{E_{m}} ∣ b ∣$ by monotone convergence (Monotone convergence for the integral).

Counterexample

1.1F1given

For every $m \geq 1$ the cube $C_{m} : = [3, R_{m} / \sqrt{n}]^{n}$ is contained in $E_{m}$ : every $x \in C_{m}$ has $∣ x ∣ \geq 3 \sqrt{n} > 2$ and $∣ x ∣ \leq \sqrt{n} (R_{m} / \sqrt{n}) = R_{m}$ , and its volume is $∣ C_{m} ∣ = (R_{m} / \sqrt{n} - 3)^{n}$ . On $C_{m}$ one has $b (x) = \log ∣ x ∣ \geq \log 3 > 0$ , so $∣ b ∣ = b \geq \log 3$ there.

2.1step 1.1algebra

Therefore $\int_{E_{m}} ∣ b ∣ \geq \int_{C_{m}} ∣ b ∣ \geq (\log 3) ∣ C_{m} ∣ = (\log 3) (R_{m} / \sqrt{n} - 3)^{n}$ , which tends to $+ \infty$ as $m \to \infty$ .

3.1step 2.1F1F2∎

By [F2] and step 2.1, $\int_{R^{n}} ∣ b ∣ = \lim_{m \to \infty} \int_{E_{m}} ∣ b ∣ = + \infty$ ; since $b \in B M O (R^{n})$ by [F1], the inclusion $B M O (R^{n}) \subseteq L^{1} (R^{n})$ is refuted.

Depends on

Used by

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Dependency tree · two levels

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Sources

Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)
Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)