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Khintchine's inequality for finite Rademacher sums

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For every finite sequence $(a_{j})_{j \in J}$ of complex numbers, indexed by a finite set $J \subset {0, 1, 2, \dots}$ , and every $0 < p < \infty$ there are constants $0 < c_{p} \leq C_{p} < \infty$ , depending only on $p$ , such that $c_{p} (\sum_{j \in J} ∣ a_{j} ∣^{2})^{1 / 2} \leq (\int_{0}^{1} ∣ \sum_{j \in J} ε_{j} (t) a_{j} ∣^{p} d t)^{1 / p} \leq C_{p} (\sum_{j \in J} ∣ a_{j} ∣^{2})^{1 / 2},$ and the constants do not depend on the finite set $J$ . For $p = 2$ both inequalities hold with constants $1$ : $\int_{0}^{1} ∣ \sum_{j \in J} ε_{j} (t) a_{j} ∣^{2} d t = \sum_{j \in J} ∣ a_{j} ∣^{2} .$ The empty sum is zero and the empty case is trivial.

Facts & Assumptions

Given: Countable Choice, a nonempty finite set $J \subset {0, 1, 2, \dots}$ , complex numbers $a_{j}$ $(j \in J)$ , and $0 < p < \infty$ ; write $S (t) : = \sum_{j \in J} ε_{j} (t) a_{j} = A (t) + i B (t)$ with $A (t) = \sum_{j} ε_{j} (t) Re a_{j}$ and $B (t) = \sum_{j} ε_{j} (t) Im a_{j}$ , and $s^{2} : = \sum_{j} ∣ a_{j} ∣^{2}$ , $u^{2} : = \sum_{j} (Re a_{j})^{2}$ , $v^{2} : = \sum_{j} (Im a_{j})^{2}$ , so $u^{2} + v^{2} = s^{2}$ .

[F1]

For a nonempty finite set $J$ of nonnegative indices and any function $F : {\pm 1}^{J} \to C$ one has $\int_{0}^{1} F (ε_{j} (t) : j \in J) d t = 2^{- ∣ J ∣} \sum_{s \in {\pm 1}^{J}} F (s)$ ; consequently $\int_{0}^{1} ε_{j} d t = 0$ and $\int_{0}^{1} ε_{j} ε_{k} d t = δ_{j k}$ (Finite Rademacher blocks are equidistributed).

[F2]

$\exp (x) = \sum_{n \geq 0} x^{n} / n!$ for real $x$ and $\cosh y = (e^{y} + e^{- y}) / 2$ for real $y$ ; moreover $\cosh y = \sum_{k \geq 0} y^{2 k} / (2 k)!$ and $\cosh y \geq 0$ (The real exponential function and the number $e$ by a power series, The six hyperbolic functions and their natural domains). The addition law $e^{x + y} = e^{x} e^{y}$ , positivity $e^{x} > 0$ and strict increase follow from The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ and The exponential function is strictly increasing.

[F3]

For every real $y$ one has $0 \leq \cosh y \leq e^{y^{2} / 2}$ : since $(2 k)! = 1 \cdot 2 \dots 2 k \geq 2^{k} k!$ for every $k$ (pairing $2 j - 1, 2 j$ ), the nonnegative series $\sum_{k} y^{2 k} / (2 k)!$ is termwise dominated by $\sum_{k} y^{2 k} / (2^{k} k!) = e^{y^{2} / 2}$ .

[F4]

The nonnegative Lebesgue integral is monotone and scales constants: if $0 \leq f \leq g$ then $\int f \leq \int g$ , and $\int c 1_{E} = c λ (E)$ for $c \geq 0$ (Monotonicity and nonnegative homogeneity of the nonnegative integral); the layer-cake formula $\int ∣ h ∣^{p} = p \int_{0}^{\infty} λ (∣ h ∣ > t) t^{p - 1} d t$ holds for measurable complex $h$ and $0 < p < \infty$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[F5]

Holder's inequality for $L^{p}$ and $L^{q}$ with $1 / p + 1 / q = 1$ , in particular Cauchy-Schwarz, and the triangle inequality for integrals (Complex Holder, Minkowski, and the quotient norm, The modulus of an integral is bounded by the integral of the modulus).

[F6]

Applying real arithmetic closure to real and imaginary parts shows that sums and products of measurable complex functions are measurable (Arithmetic and lattice operations preserve measurability whenever they are defined).

[F7]

For every $s > 0$ the Euler integral $Γ (s) = \int_{0}^{\infty} t^{s - 1} e^{- t} d t$ converges, so it is a finite positive number (The real Gamma function by Euler's integral, Euler's Gamma integral converges exactly for positive real parameters). The monotone $C^{1}$ substitution $t = 2 s \sqrt{τ}$ on $(0, \infty)$ is valid on compact truncations and at both improper ends by Change of variable in an improper integral.

Proof

technique · direct

1.1F1F6algebra

Setup and second moment. The functions $S, A, B$ are finite sums of products of constants with the measurable functions $ε_{j}$ , hence measurable by [F6], and $∣ S ∣^{2} = A^{2} + B^{2}$ . By [F1] the mean of $S$ vanishes, $\int_{0}^{1} S d t = 0$ , and expanding $∣ S ∣^{2} = \sum_{j, k} ε_{j} ε_{k} a_{j} {\overset{ˉ}{a}}_{k}$ and integrating termwise with $\int ε_{j} ε_{k} = δ_{j k}$ gives $\int_{0}^{1} ∣ S ∣^{2} d t = \sum_{j} ∣ a_{j} ∣^{2} = s^{2}$ .

2.1F1F2F3step 1.1algebra

Exponential moments. For real $ρ$ the function $t \mapsto e^{ρ A (t)}$ is a finite product $\prod_{j \in J} e^{ρ ε_{j} (t) Re a_{j}}$ of functions of the individual signs, so [F1] applied to $F (s) = \prod_{j} e^{ρ s_{j} Re a_{j}}$ gives $\int_{0}^{1} e^{ρ A} d t = \prod_{j \in J} \frac{e^{ρ Re a_{j}} + e^{- ρ Re a_{j}}}{2} = \prod_{j} \cosh (ρ Re a_{j}) \leq \prod_{j} e^{(ρ Re a_{j})^{2} / 2} = e^{ρ^{2} u^{2} / 2}$ , where [F2] and [F3] were used termwise and the product of exponentials was combined.

3.1F2F4step 2.1algebra

Tail bounds. Let $λ > 0$ . If $u > 0$ , take $ρ = λ / u^{2}$ in step 2.1 and use monotonicity of the integral on ${A > λ}$ to get $λ ({A > λ}) \leq e^{- ρ λ} \int e^{ρ A} \leq e^{- λ^{2} / (2 u^{2})}$ ; applying the same argument to $- A$ gives $λ ({∣ A ∣ > λ}) \leq 2 e^{- λ^{2} / (2 u^{2})}$ . If $u = 0$ , then $A = 0$ and this tail measure is $0$ . The same bounds hold for $B$ with $v$ . If $s = 0$ , then $S = 0$ and the tail measure is $0$ . Otherwise $s > 0$ , and ${∣ S ∣ > λ} \subset {∣ A ∣ > λ / \sqrt{2}} \cup {∣ B ∣ > λ / \sqrt{2}}$ because if both component moduli are at most $λ / \sqrt{2}$ , then $∣ S ∣^{2} = A^{2} + B^{2} \leq λ^{2}$ . Applying the component bound at threshold $λ / \sqrt{2}$ separately to the labeled $A$ - and $B$ -tails, and omitting a contribution when its variance is zero, gives $λ ({∣ S ∣ > λ}) \leq {\begin{cases} 2 e^{- λ^{2} / (4 u^{2})}, & u > 0, \\ 0, & u = 0 \end{cases} + {\begin{cases} 2 e^{- λ^{2} / (4 v^{2})}, & v > 0, \\ 0, & v = 0 \end{cases} \leq 4 e^{- λ^{2} / (4 s^{2})},$ because each positive variance among $u, v$ is at most $s$ and the two labeled contributions are each at most $2 e^{- λ^{2} / (4 s^{2})}$ . Equal positive variances still contribute twice, as required by the union bound.

4.1F4F7step 3.1algebra

Upper bound. For $s > 0$ the layer-cake formula [F4] applied to $S$ and step 3.1 give $\int_{0}^{1} ∣ S ∣^{p} d t = p \int_{0}^{\infty} λ ({∣ S ∣ > t}) t^{p - 1} d t \leq 4 p \int_{0}^{\infty} e^{- t^{2} / (4 s^{2})} t^{p - 1} d t$ ; substituting $t = 2 s \sqrt{τ}$ , $d t = s τ^{- 1 / 2} d τ$ , turns the last integral into $4 p 2^{p - 1} s^{p} \int_{0}^{\infty} τ^{p / 2 - 1} e^{- τ} d τ = 4 p 2^{p - 1} Γ (p / 2) s^{p}$ , which is finite by [F7]. Hence $∥ S ∥_{p} \leq C_{p} s$ with $C_{p} : = (4 p 2^{p - 1} Γ (p / 2))^{1 / p} < \infty$ and, for $s = 0$ , $∥ S ∥_{p} = 0$ .

5.1F5step 1.1step 4.1algebra

Lower bound. Let $s > 0$ . Step 4.1 with $p = 4$ gives $\int_{0}^{1} ∣ S ∣^{4} d t \leq C_{4}^{4} s^{4}$ ; splitting the integral of $∣ S ∣^{2}$ over ${∣ S ∣ \leq s / 2}$ and ${∣ S ∣ > s / 2}$ and applying Cauchy-Schwarz [F5] to the second piece, $\int_{0}^{1} ∣ S ∣^{2} d t \leq (s / 2)^{2} + (\int ∣ S ∣^{4})^{1 / 2} λ ({∣ S ∣ > s / 2})^{1 / 2} \leq s^{2} / 4 + C_{4}^{2} s^{2} λ ({∣ S ∣ > s / 2})^{1 / 2}$ ; with $\int ∣ S ∣^{2} = s^{2}$ from step 1.1 this gives $λ ({∣ S ∣ > s / 2}) \geq (3 / (4 C_{4}^{2}))^{2} = 9 / (16 C_{4}^{4})$ . Consequently, for every $p > 0$ , $\int_{0}^{1} ∣ S ∣^{p} d t \geq \int_{{∣ S ∣ > s / 2}} ∣ S ∣^{p} d t \geq (s / 2)^{p} \cdot 9 / (16 C_{4}^{4}) = c_{p}^{p} s^{p}$ with $c_{p} : = (9 / (16 C_{4}^{4}))^{1 / p} / 2 > 0$ .

6.1step 1.1step 4.1step 5.1algebra∎

Conclusion. For a nonempty $J$ and $s > 0$ , steps 4.1 and 5.1 give the two-sided inequality with constants $c_{p}, C_{p}$ that are explicit functions of $p$ alone, in particular independent of $J$ and of the coefficients; for $s = 0$ all quantities vanish and the inequality is trivial, and for the empty set $J = \emptyset$ both sides are $0$ . The case $p = 2$ is step 1.1, where the identity $\int ∣ S ∣^{2} = s^{2}$ gives both inequalities with constants $1$ .

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Sources

Terence Tao, Math 247A Lecture Notes 4 (UCLA, Fall 2006) (standard reference, not scraped)
Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249) (standard reference, not scraped)
Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)