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Sharp frequency cutoffs have kernels that are not in L1

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The sharp frequency cutoffs $χ_{j} (ξ) : = 1_{(1, 2)} (2^{- j} ξ) (j \in Z)$ have uniformly bounded inverse Fourier transforms, $\sup_{j} ∥ χ_{j}^{\lor} ∥_{L^{1}} < \infty$ . Consequently the smoothness of the Littlewood-Paley partition is cosmetic: the convolution bounds of Dyadic pieces have annular Fourier support and uniformly bounded rescaled kernels would hold verbatim for the sharp cutoffs $χ_{j}$ in place of the smooth pieces $φ_{j}$ .

Facts & Assumptions

Given: Countable Choice and the interval indicator $χ : = 1_{(1, 2)}$ on $R$ and its dyadic dilates $χ_{j} (ξ) = χ (2^{- j} ξ)$ , with the negative-sign $2 π$ -normalized Fourier transform, $K_{j} : = F^{- 1} χ_{j}$ and $K_{0} : = F^{- 1} χ$ .

[F1]

$χ = 1_{(1, 2)} \in L^{1} (R) \cap L^{2} (R)$ , its integral transform is $\hat{χ} (x) = \int_{1}^{2} e^{- 2 π i x ξ} d ξ$ , the integral transform of an $L^{1}$ function represents its distributional transform and its Plancherel transform almost everywhere, and $F_{2}^{- 1} = R F_{2}$ with $R f (x) = f (- x)$ (Fourier transform on complex L1 classes, Agreement of the integral and L2 transforms, Fourier transform agrees with l one and plancherel transforms, L2 Fourier inversion).

[F2]

For $j \in Z$ , $\hat{χ (2^{- j} \cdot)} (x) = 2^{j} \hat{χ} (2^{j} x)$ : this is the one-dimensional case of the dilation law $\hat{f \circ A} (ξ) = ∣ \det A ∣^{- 1} \hat{f} (A^{- T} ξ)$ with $A = 2^{- j} I$ , and $\hat{f (- \cdot)} = \hat{f} (- \cdot)$ (Translation, modulation, linear dilation and reflection laws). For every nonnegative measurable $H$ and $b > 0$ , $\int_{R} b H (b x) d x = \int_{R} H (u) d u$ , with infinite values allowed: apply A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions to the $C^{1}$ diffeomorphism $x \mapsto b x$ .

[F3]

Complex exponential and Euler: $\exp (i θ) = \cos θ + i \sin θ$ , $∣ \exp (i θ) ∣ = 1$ , and for $x \neq 0$ the derivative of $ξ \mapsto e^{2 π i x ξ} / (2 π i x)$ is $e^{2 π i x ξ}$ , so Newton-Leibniz applies to the real and imaginary parts (The complex exponential by its power series, Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The complex exponential is entire and its complex derivative is itself, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

[F4]

The harmonic series diverges: each block $2^{k} \leq m < 2^{k + 1}$ contributes at least $2^{k} / 2^{k + 1} = 1 / 2$ , so its partial sums are unbounded. For pairwise disjoint measurable sets $E_{m}$ and $a_{m} \geq 0$ , the nonnegative simple function $s_{N} : = \sum_{m = 1}^{N} a_{m} 1_{E_{m}}$ has Lebesgue integral $\sum_{m = 1}^{N} a_{m} λ (E_{m})$ (The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions). If $s_{N} \leq f$ , monotonicity gives $\int f \geq \int s_{N}$ (Monotonicity and nonnegative homogeneity of the nonnegative integral). A closed interval has measure its length (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F5]

The smooth partition of The inhomogeneous dyadic frequency partition and its Littlewood-Paley operators has $∥ K_{j} ∥_{1} \leq C$ uniformly in $j$ (Dyadic pieces have annular Fourier support and uniformly bounded rescaled kernels).

[F6]

Cosine has period $2 π$ , vanishes at $π / 2$ and $3 π / 2$ , decreases from $π / 2$ to $π$ and increases from $π$ to $3 π / 2$ , so it is nonpositive on $[π / 2, 3 π / 2]$ and its $2 π$ translates (The zero sets of sine and cosine and the least positive common period 2 pi, Quarter-turn values and shifts by pi/2 and pi, Signs, monotonicity intervals, and ranges of sine and cosine).

Counterexample

1.1F1F3algebra

The inverse transform of the sharp cutoff. Since $χ \in L^{1} \cap L^{2}$ , [F1] gives $F^{- 1} χ = R F_{2} χ$ as an $L^{2}$ class, and $F_{2} χ = \hat{χ}$ almost everywhere, so $K_{0} (x) = \hat{χ} (- x) = \int_{1}^{2} e^{2 π i x ξ} d ξ$ for almost every $x$ , the last function being continuous and hence the correct representative. For $x \neq 0$ , [F3] gives $\int_{1}^{2} e^{2 π i x ξ} d ξ = (e^{4 π i x} - e^{2 π i x}) / (2 π i x) = e^{3 π i x} (e^{π i x} - e^{- π i x}) / (2 π i x) = e^{3 π i x} \sin (π x) / (π x)$ , using $e^{π i x} - e^{- π i x} = 2 i \sin (π x)$ from Euler's formula, and $K_{0} (0) = \int_{1}^{2} d ξ = 1$ ; consequently $∣ K_{0} (x) ∣ = ∣ \sin (π x) ∣ / (π ∣ x ∣)$ for $x \neq 0$ .

2.1F4F6step 1.1algebra

The kernel is not in $L^{1}$ . For $m \geq 1$ let $I_{m} : = [m + \frac{1}{4}, m + \frac{3}{4}]$ . On $I_{m}$ one has $2 π x \in [2 π m + \frac{π}{2}, 2 π m + \frac{3 π}{2}]$ , so $\cos (2 π x) \leq 0$ ; since $∣ K_{0} (x) ∣^{2} = ∣ e^{4 π i x} - e^{2 π i x} ∣^{2} / (4 π^{2} x^{2}) = (2 - 2 \cos (2 π x)) / (4 π^{2} x^{2}) \geq 1 / (2 π^{2} x^{2})$ on $I_{m}$ , and $∣ x ∣ \leq m + \frac{3}{4}$ , there holds $\int_{I_{m}} ∣ K_{0} ∣ \geq \frac{1}{π \sqrt{2}} \cdot \frac{1}{m + 3 / 4} \cdot \frac{1}{2} \geq \frac{c}{m + 1}$ with $c : = 1 / (4 π \sqrt{2})$ . The intervals $I_{m}$ are pairwise disjoint. For every $N \geq 1$ set $a_{m} : = 1 / (π \sqrt{2} (m + 3 / 4))$ and $s_{N} : = \sum_{m = 1}^{N} a_{m} 1_{I_{m}}$ . The pointwise bound just proved gives $s_{N} \leq ∣ K_{0} ∣ 1_{[1, \infty)}$ , and [F4] yields $\int_{1}^{\infty} ∣ K_{0} ∣ \geq \int s_{N} = \sum_{m = 1}^{N} a_{m} / 2 \geq c \sum_{m = 1}^{N} 1 / (m + 1)$ . These finite lower bounds are unbounded by the harmonic-series argument in [F4], so $\int_{1}^{\infty} ∣ K_{0} ∣ = + \infty$ and hence $∥ K_{0} ∥_{L^{1}} = + \infty$ .

3.1F2F5step 1.1step 2.1algebra∎

No uniform bound and the failure of the sharp replacement. By [F2] and step 1.1, $\hat{χ_{j}} (x) = 2^{j} \hat{χ} (2^{j} x)$ and hence $K_{j} (x) = \hat{χ_{j}} (- x) = 2^{j} \hat{χ} (- 2^{j} x) = 2^{j} K_{0} (2^{j} x)$ for every $j \in Z$ ; the nonnegative change of variables in [F2], with $b = 2^{j} > 0$ , gives $∥ K_{j} ∥_{L^{1}} = ∥ K_{0} ∥_{L^{1}} = + \infty$ for every $j$ , so in particular $\sup_{j} ∥ χ_{j}^{\lor} ∥_{L^{1}} = \infty$ . This contradicts the uniform bound $∥ K_{j} ∥_{1} \leq C$ of the smooth partition [F5]: the sharp-cutoff family cannot replace the smooth annular cutoffs in the convolution estimates, and smoothness of the partition is used essentially, not cosmetically.

Remarks

Recorded orientation, not proved here. Nonintegrability of these kernels does not rule out strict-range $L^{p}$ multiplier bounds. Grafakos, §6.1.3, Theorem 6.1.5 and the discussion preceding it (printed p. 427), proves that the one-dimensional sharp dyadic square function does characterise $L^{p}$ for $1 < p < \infty$ . The same discussion records that in $R^{n}$ , $n \geq 2$ , the sharp-annulus square function fails to characterise $L^{p}$ when $1 < p < \infty$ and $p \neq 2$ , because the ball indicator is not an $L^{p}$ multiplier. These source records are not used in the kernel computation above.

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Sources

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249) (standard reference, not scraped)
Terence Tao, Math 247A Lecture Notes 4 (UCLA, Fall 2006) (standard reference, not scraped)