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Riesz kernel size, difference and spherical-cancellation bounds

Statement

Assume Countable Choice, let $n \geq 1$ and $1 \leq j \leq n$ , and let $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ with $c_{n} = Γ ((n + 1) / 2) / π^{(n + 1) / 2}$ be the Riesz kernel of Riesz transforms on Euclidean space. Then:

$∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ for every $x \neq 0$ ;
with $C_{n} : = c_{n} 2^{n + 1} (3 n + 4)$ one has $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ whenever $x \neq 0$ and $∣ h ∣ \leq ∣ x ∣ / 2$ ; and
$\int_{S^{n - 1}} K_{j} (r ω) d σ (ω) = 0$ for every $r > 0$ , where $σ$ is the polar surface measure of Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma on the unit sphere $S^{n - 1} = {ω \in R^{n} : ∣ ω ∣ = 1}$ .

The constant $C_{n}$ is explicit and depends only on $n$ ; at $n = 1$ it reads $C_{1} = 28 c_{1} = 28 / π$ . These are the raw size, first-difference and cancellation estimates that a later singular-integral treatment consumes; no Calderón–Zygmund kernel definition is invoked here.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $1 \leq j \leq n$ , and the Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ with $0 < c_{n} < \infty$ .

[F1]

The Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ has $0 < c_{n} < \infty$ , is smooth, odd and homogeneous of degree $- n$ on $R^{n} ∖ {0}$ , so $K_{j} (r ω) = c_{n} r^{- n} ω_{j}$ whenever $r > 0$ and $∣ ω ∣ = 1$ . Riesz transforms on Euclidean space

[F2]

If $N$ is a norm on a real vector space and $u, w$ are vectors, then $∣ N (u) - N (w) ∣ \leq N (u - w)$ . The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$

[F3]

For $n \geq 1$ and $x \in R^{n}$ the Euclidean norm satisfies $∥ x ∥_{\infty} \leq ∥ x ∥_{2}$ , hence $∣ x_{j} ∣ \leq ∥ x ∥_{2} = ∣ x ∣$ for every coordinate $j$ . The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$

[F4]

Mean value theorem: a real function continuous on a closed interval and differentiable on its interior has a point whose derivative equals the average rate of change. The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$

[F5]

For every natural $m \geq 1$ the function $s \mapsto s^{- m}$ is differentiable on $R ∖ {0}$ with derivative $- m s^{- m - 1}$ . For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term

[F6]

Polar coordinates: for $n \geq 1$ and every Borel $h : R^{n} \to [0, \infty]$ , $\int_{R^{n}} h d λ_{n} = \int_{0}^{\infty} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ , and $σ$ is a finite Borel measure on $S^{n - 1}$ . Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma

[F7]

Linear change of variables for Lebesgue measure, in particular $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ for invertible linear $T$ . A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not

Proof

technique · direct

1.1F1F3givenalgebra

Fix $x \neq 0$ . By [F1] the kernel is $K_{j} (x) = c_{n} x_{j} ∣ x ∣^{- (n + 1)}$ , so $∣ K_{j} (x) ∣ = c_{n} ∣ x_{j} ∣ ∣ x ∣^{- (n + 1)} \leq c_{n} ∣ x ∣^{- n}$ by the coordinate bound $∣ x_{j} ∣ \leq ∣ x ∣$ of [F3] and the positivity $∣ x ∣ > 0$ .

1.2F2F3givenalgebra

Fix $x \neq 0$ and $h$ with $∣ h ∣ \leq ∣ x ∣ / 2$ and put $A : = ∣ x - h ∣$ . By [F2] applied to the Euclidean norm, $∣ A - ∣ x ∣ ∣ \leq ∣ h ∣$ , so $∣ x ∣ / 2 \leq A \leq 3 ∣ x ∣ / 2$ ; by [F3], $∣ h_{j} ∣ \leq ∣ h ∣$ and $∣ (x - h)_{j} ∣ \leq A \leq 3 ∣ x ∣ / 2$ . In particular $A > 0$ and the kernel is defined at both arguments.

1.3F1F6F7givenalgebra

Let $A_{n} : = \int_{S^{n - 1}} ω_{j} d σ (ω)$ and $H (x) : = 1_{{1 < ∣ x ∣ < 2}} x_{j}$ . The function $H$ is Borel and $\int_{R^{n}} ∣ H ∣ d λ_{n} < \infty$ ; the map $x \mapsto - x$ is a linear bijection with $∣ \det ∣ = 1$ , so [F7] gives $\int_{R^{n}} H (- x) d λ_{n} (x) = \int_{R^{n}} H (x) d λ_{n} (x)$ , while $H (- x) = - H (x)$ gives $\int H d λ_{n} = - \int H d λ_{n}$ , that is, $\int_{R^{n}} H d λ_{n} = 0$ . On the other hand [F6] applied to the nonnegative and the negative part of $H$ gives $\int_{R^{n}} H d λ_{n} = \int_{1}^{2} r^{n - 1} (\int_{S^{n - 1}} r ω_{j} d σ (ω)) d r = A_{n} \int_{1}^{2} r^{n} d r$ with $\int_{1}^{2} r^{n} d r > 0$ , so $A_{n} = 0$ . Hence for every $r > 0$ the homogeneity [F1] gives $\int_{S^{n - 1}} K_{j} (r ω) d σ (ω) = c_{n} r^{- n} \int_{S^{n - 1}} ω_{j} d σ (ω) = c_{n} r^{- n} A_{n} = 0$ .

2.1step 1.2F1F4F5givenalgebra

Keep $x \neq 0$ and $∣ h ∣ \leq ∣ x ∣ / 2$ as in 1.2, put $B : = ∣ x ∣$ and $ψ (s) : = s^{- (n + 1)}$ ; by [F5] with $m = n + 1 \geq 2 \geq 1$ one has $ψ^{'} (s) = - (n + 1) s^{- (n + 2)}$ on $(0, \infty)$ . The interval with endpoints $A$ and $B$ lies in $[B / 2, \infty)$ by 1.2. If $A = B$ , then $∣ ψ (A) - ψ (B) ∣ = 0$ and the following bound is immediate. If $A \neq B$ , [F4] on the interval with ordered endpoints $\min (A, B) < \max (A, B)$ gives a point $s$ with $ψ (A) - ψ (B) = ψ^{'} (s) (A - B)$ and therefore $∣ ψ (A) - ψ (B) ∣ \leq (n + 1) (B / 2)^{- (n + 2)} ∣ A - B ∣ \leq (n + 1) 2^{n + 2} ∣ h ∣ B^{- (n + 2)}$ . Insert $\pm (x - h)_{j} ψ (B)$ into the difference and expand: $K_{j} (x - h) - K_{j} (x) = c_{n} [(x - h)_{j} (ψ (A) - ψ (B)) - h_{j} ψ (B)]$ , so by 1.2 and the preceding bound, and by $ψ (B) = B^{- (n + 1)}$ , $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq c_{n} [\frac{3}{2} (n + 1) 2^{n + 2} + 2^{n + 1}] ∣ h ∣ B^{- (n + 1)} = C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ with $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ , since $\frac{3}{2} (n + 1) 2^{n + 2} = 3 (n + 1) 2^{n + 1}$ and $3 (n + 1) + 1 = 3 n + 4$ .

3.1step 1.1step 1.3step 2.1given∎

The three assertions are proved: $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ for $x \neq 0$ is 1.1; the difference bound with the stated constant is 2.1, whose hypothesis $∣ h ∣ \leq ∣ x ∣ / 2$ keeps both arguments nonzero as recorded in 1.2; and the vanishing of every spherical integral $\int_{S^{n - 1}} K_{j} (r ω) d σ (ω)$ , $r > 0$ , is 1.3.

Depends on

Dependency tree · 2 levels

Sources

Loukas Grafakos, Classical Fourier Analysis, third edition