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Truncated maximal functions: finiteness, comparison estimates and the good-set bound

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p < \infty$ and let $φ \in S (R^{n})$ with $\int φ \neq 0$ . For $0 < ϵ \leq 1 / 2$ , $L > 0$ , $T > 0$ and $f \in S^{'} (R^{n})$ define the truncated maximal functions $M_{φ, 0}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \frac{∣ (f * φ_{t}) (x) ∣ t^{L}}{(t + ϵ + ϵ ∣ x ∣)^{L}},$ $M_{φ, 1}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \sup_{∣ x - y ∣ < t} \frac{∣ (f * φ_{t}) (y) ∣ t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}},$ $M_{φ, T}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \sup_{y \in R^{n}} \frac{∣ (f * φ_{t}) (x - y) ∣}{(1 + ∣ y ∣ / t)^{T}} \frac{t^{L}}{(t + ϵ + ϵ ∣ x - y ∣)^{L}},$ $M_{N}^{ϵ, L} f (x) = \sup_{ψ \in F_{N}} M_{ψ, 1}^{ϵ, L} f (x),$ where $M_{φ}^{0} f$ , $M_{φ}^{*, 1} f$ and $F_{N}$ are those of Radial and nontangential maximal functions of a tempered distribution and Grand maximal test class of order N and the grand maximal function. The displayed radial and aperture-one truncation formulas also apply to every Schwartz test $ψ$ , including tests of zero integral; the nonzero-integral hypothesis is needed only for estimates involving the fixed comparison kernel $φ$ . Then:

For a nonnegative Borel function $g$ , write $\tilde{M} g (x)$ for the supremum of its centered ball averages, with the nonnegative Lebesgue integral allowed to equal $+ \infty$ . If $g \in L_{l o c}^{1}$ , then $\tilde{M} g = M g$ for the centered maximal operator of The centered and uncentered Hardy-Littlewood maximal functions.

For every $f \in S^{'}$ and every $0 < p < \infty$ there is $L_{0} = L_{0} (f, n, φ, p) < \infty$ such that for every $L \geq L_{0}$ and every $0 < ϵ \leq 1 / 2$ the function $M_{φ, 1}^{ϵ, L} f$ belongs to $L^{p} (R^{n})$ and satisfies $M_{φ, 1}^{ϵ, L} f (x) \leq C ϵ^{- K} (1 + ∣ x ∣)^{- M}$ with some finite $C, K, M > 0$ depending on $f, L, p$ (so the quantity $∥ M_{φ, 1}^{ϵ, L} f ∥_{p}$ is finite).
For every $T > 0$ and $L > 0$ there is $N_{1}$ such that for every $N \geq N_{1}$ , every $0 < ϵ \leq 1 / 2$ , every $f \in S^{'}$ and every $x$ , $M_{N}^{ϵ, L} f (x) \leq C_{1} M_{φ, T}^{ϵ, L} f (x)$ with $C_{1} = C_{1} (n, φ, T, L)$ independent of $ϵ$ and $f$ .
For every $T > 0$ , setting $q = n / T$ and assuming $0 < q < p$ , one has for every $0 < ϵ \leq 1 / 2$ , $L > 0$ and $f \in S^{'}$ $M_{φ, T}^{ϵ, L} f (x)^{q} \leq \tilde{M} ((M_{φ, 1}^{ϵ, L} f)^{q}) (x) (x \in R^{n}), ∥ M_{φ, T}^{ϵ, L} f ∥_{L^{p}} \leq C_{2} ∥ M_{φ, 1}^{ϵ, L} f ∥_{L^{p}},$ where $\tilde{M}$ is the extended centered average defined above and $C_{2} = C_{2} (n, p, q)$ . The norm inequality is interpreted in the extended sense if its right-hand side is infinite.
For every $p_{0} > 0$ , $T > 0$ , $L > 0$ and $λ > 0$ there is $N_{2}$ such that for every $N \geq N_{2}$ there is $C_{3} = C_{3} (n, φ, p_{0}, T, L, λ, N) < \infty$ with the property that for every $0 < ϵ \leq 1 / 2$ , every $f \in S^{'}$ and every $x$ satisfying $M_{N}^{ϵ, L} f (x) < λ M_{φ, 1}^{ϵ, L} f (x)$ , $M_{φ, 1}^{ϵ, L} f (x) \leq C_{3} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)^{1 / p_{0}} .$
(Untruncated good-set estimate.) For every $p_{0} > 0$ and $λ > 0$ there is $N_{3}$ such that for every $N \geq N_{3}$ there is $C_{4} = C_{4} (n, φ, p_{0}, λ, N) < \infty$ with the property that for every $f \in S^{'}$ and every $x$ with $M_{N} f (x) \leq λ M_{φ}^{*, 1} f (x) < \infty$ , $M_{φ}^{*, 1} f (x) \leq C_{4} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)^{1 / p_{0}} .$ The finiteness $M_{φ}^{*, 1} f (x) < \infty$ is part of the hypothesis: no claim is made at points where $M_{φ}^{*, 1} f (x) = + \infty$ . For every $f$ with $M_{φ}^{*, 1} f$ finite a.e. the estimate therefore holds a.e. on the set $F = {M_{N} f \leq λ M_{φ}^{*, 1} f}$ .

The point of the truncation is that $M_{φ, 1}^{ϵ, L} f$ is finite and integrable, so the good-set argument of the last item can be run without an a priori finiteness assumption on $M_{φ}^{*, 1} f$ ; as $ϵ ↓ 0$ the truncated functions increase pointwise to the untruncated ones. Countable Choice is assumed through dyadic deconvolution and the measure-theoretic estimates used below.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p < \infty$ , $φ \in S$ with $\int φ \neq 0$ , $0 < ϵ \leq 1 / 2$ , $L, T > 0$ , $f \in S^{'}$ ; the maximal functions of Radial and nontangential maximal functions of a tempered distribution and Grand maximal test class of order N and the grand maximal function.

[F1]

Finite-seminorm bound: there are integers $N_{0}, M \geq 0$ and $C$ with $∣ ⟨ f, h ⟩ ∣ \leq C \max_{∣ α ∣ \leq N_{0}, ∣ β ∣ \leq M} \sup_{z} ∣ z^{α} \partial^{β} h (z) ∣$ for all $h \in S$ . Applying this to $h (z) = φ_{t} (y - z)$ gives $∣ (f * φ_{t}) (y) ∣ \leq C_{φ} t^{- (n + M)} (1 + ∣ y ∣)^{N_{0}}$ when $0 < t \leq 1$ and $∣ (f * φ_{t}) (y) ∣ \leq C_{φ} t^{N_{0} - n} (1 + ∣ y ∣)^{N_{0}}$ when $t \geq 1$ : for small scales the largest derivative seminorm is bounded by $t^{- n - M}$ , while for large scales it is bounded by $t^{- n}$ and the polynomial weight contributes at most $t^{N_{0}}$ (Finite seminorm bound characterizes tempered distributions, Schwartz space and its seminorms).

[F2]

The centered Hardy-Littlewood maximal operator is defined for $L_{l o c}^{1}$ inputs and satisfies $∥ M g ∥_{L^{r}} \leq C_{n, r} ∥ g ∥_{L^{r}}$ for $1 < r < \infty$ ; also $M (h_{1} + h_{2}) \leq M h_{1} + M h_{2}$ (The centered and uncentered Hardy-Littlewood maximal functions, The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ ).

[F3]

The deconvolution lemma: for $φ$ with $\int φ \neq 0$ and every pair of positive integers $L^{'}, N^{'}$ , there are $C_{d e c}, M_{d e c}, s_{0} > 0$ such that for every $ψ \in S$ there are $η^{j} \in S$ with $ψ = \sum_{j \geq 0} η^{j} * φ_{s_{0} 2^{- j}}$ in $S$ and $∥ η^{j} ∥_{S_{N^{'}}} \leq C_{d e c} 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ Any smaller positive value of $s_{0}$ also works, with constants depending on that value (Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel).

[F4]

For every $a \in R^{n}$ and $r > 0$ , $λ (B (a, r)) = c_{n} r^{n}$ with $c_{n} = ω_{n - 1} / n > 0$ ; this follows from the centred-ball formula and translation invariance. Consequently, $B (x - y, t) \subseteq B (x, ∣ y ∣ + t)$ and $λ (B (x, ∣ y ∣ + t)) / λ (B (x - y, t)) = (1 + ∣ y ∣ / t)^{n}$ (Sphere and ball measures scale in Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Open ball, closed ball and sphere in a metric space).

[F5]

Translate bound for the test seminorm: for $ψ \in S$ and $h \in R^{n}$ one has $P_{N} (ψ (\cdot + h)) \leq (1 + ∣ h ∣)^{N} P_{N} (ψ)$ , because $(1 + ∣ w ∣) \leq (1 + ∣ h ∣) (1 + ∣ w + h ∣)$ pointwise; hence for the translated derivative kernel $Ψ^{z} (w) = (\partial_{j} φ) (w + (z - x) / t)$ and $∣ (z - x) / t ∣ \leq r + 1$ with $r \leq 1$ one has $P_{N} (Ψ^{z}) \leq 3^{N} P_{N} (\partial_{j} φ) \leq 3^{N} P_{N + 1} (φ) < \infty$ . Combining the componentwise bounds for $0 \leq j < n$ gives $∣ \nabla (f * φ_{t}) (z) ∣ \leq \sqrt{n}$ times this bound after applying the grand maximal estimate; write $c_{N, φ} : = \sqrt{n} 3^{N} P_{N + 1} (φ)$ for the resulting gradient constant (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

[F6]

For every $r > 1$ , $L^{r} (R^{n}) \subset L_{l o c}^{1} (R^{n})$ : if $K$ is compact, it is bounded and has finite measure, and Holder gives $\int_{K} ∣ g ∣ \leq λ (K)^{1 - 1 / r} ∥ g ∥_{L^{r}}$ (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F7]

The radial and aperture-one nontangential maximal functions are Borel measurable for every $f \in S^{'}$ (Measurability and lower semicontinuity of the smooth maximal functions). For fixed $ϵ, L$ , the truncated aperture-one function is Borel: each strict superlevel set is the union of the open balls $B (y, t)$ indexed by the admissible witnesses whose weighted convolution value exceeds that level. The truncated grand maximal function is Borel by the same open-ball superlevel argument, with tests also indexed over $F_{N}$ . The tangential function is Borel because it is a supremum, over fixed witnesses, of continuous functions of $x$ ; smoothness of each convolution is Tempered convolution is smooth with polynomial growth.

Proof technique: weighted distance estimates, the dyadic deconvolution comparison and a mean-value argument on the good set.

Proof

technique · direct

1.1F1F7givenalgebra

Finiteness bound. Let $N_{0}, M, C_{φ}$ be as in [F1], and choose $L_{0} = L_{0} (f, n, φ, p)$ so large that $L_{0} > n + M$ and $L_{0} > N_{0} + n / p$ . Fix $L \geq L_{0}$ . For a witness $(y, t)$ with $0 < t < 1 / ϵ$ and $∣ x - y ∣ < t$ , write $W = \frac{∣ (f * φ_{t}) (y) ∣ t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}}$ . If $t \leq 1$ , [F1] and $t + ϵ + ϵ ∣ y ∣ \geq ϵ (1 + ∣ y ∣)$ give $W \leq C_{φ} t^{L - (n + M)} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L} \leq C_{φ} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L},$ because $L > n + M$ . If $t \geq 1$ , the large-scale bound in [F1] gives $W \leq C_{φ} t^{L + N_{0} - n} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L} \leq C_{φ} ϵ^{- (L + \max {0, L + N_{0} - n})} (1 + ∣ y ∣)^{N_{0} - L},$ because when $L + N_{0} - n \geq 0$ one has $t^{L + N_{0} - n} \leq ϵ^{- (L + N_{0} - n)}$ , while when $L + N_{0} - n < 0$ the factor is at most $1$ . Thus in both cases $W \leq C^{'} ϵ^{- K} (1 + ∣ y ∣)^{- (L - N_{0})}$ for some finite $K$ . If $∣ x ∣ \geq 2 / ϵ$ , then $∣ y ∣ \geq ∣ x ∣ - t \geq ∣ x ∣ / 2$ , so this is at most $C^{''} ϵ^{- K} (1 + ∣ x ∣)^{- (L - N_{0})}$ . If $∣ x ∣ \leq 2 / ϵ$ , then $1 + ∣ x ∣ \leq 1 + 2 / ϵ \leq 3 / ϵ$ ; absorbing the resulting factor $ϵ^{- (L - N_{0})}$ gives the same spatial-decay form with a possibly larger finite power of $ϵ^{- 1}$ . Taking the supremum over witnesses yields $M_{φ, 1}^{ϵ, L} f (x) \leq C ϵ^{- K^{'}} (1 + ∣ x ∣)^{- (L - N_{0})} .$ Because $L > N_{0} + n / p$ , this bound belongs to $L^{p} (R^{n})$ , uniformly for each fixed $0 < ϵ \leq 1 / 2$ . By [F7] the maximal function is Borel, so its $L^{p}$ norm is defined.

1.2F2F4F6F7algebra

Tangential dominated by aperture one. Fix $x, y$ and $0 < t < 1 / ϵ$ , and put $q = n / T > 0$ and $g = (M_{φ, 1}^{ϵ, L} f)^{q}$ , which is nonnegative Borel by [F7]. For every $z \in B (x - y, t)$ the definition gives $\frac{∣ (f * φ_{t}) (x - y) ∣ t^{L}}{(t + ϵ + ϵ ∣ x - y ∣)^{L}} \leq M_{φ, 1}^{ϵ, L} f (z)$ . Raising to the $q$ -th power, averaging with the integral allowed to be infinite, and enlarging the ball using [F4] gives $∣ (f * φ_{t}) (x - y) ∣^{q} \frac{t^{L q}}{(t + ϵ + ϵ ∣ x - y ∣)^{L q}} \leq \frac{1}{λ (B (x - y, t))} \int_{B (x - y, t)} g \leq (1 + \frac{∣ y ∣}{t})^{n} \tilde{M} g (x) .$ Since $n = T q$ , division by $(1 + ∣ y ∣ / t)^{T q}$ and taking the supremum over $y, t$ proves the pointwise estimate for every $L > 0$ . For the norm estimate assume $q < p$ . If $∥ M_{φ, 1}^{ϵ, L} f ∥_{p} = \infty$ , the extended norm inequality is trivial. Otherwise $g \in L^{p / q}$ , and [F6] gives $g \in L_{l o c}^{1}$ , so $\tilde{M} g = M g$ . Applying [F2] with $r = p / q > 1$ yields $∥ M_{φ, T}^{ϵ, L} f ∥_{p}^{q} \leq ∥ M g ∥_{p / q} \leq C_{n, p, q} ∥ g ∥_{p / q} = C_{n, p, q} ∥ M_{φ, 1}^{ϵ, L} f ∥_{p}^{q} .$ Taking $q$ -th roots and renaming the constant proves assertion 3.

1.3F3F5algebra

Truncated grand maximal dominated by the truncated tangential maximal function. Fix $T > 0$ , $L > 0$ and choose integers $L^{'} > L + T$ and $N^{'} > L + T + n$ , for example $L^{'} = ⌊ L + T ⌋ + 1$ and $N^{'} = ⌊ L + T + n ⌋ + 1$ . Apply [F3] with these parameters, obtaining $C_{d e c}, M_{d e c}, s_{0}$ ; by the scaling clause of [F3] we may assume $s_{0} \leq 1$ , and we take an integer $N \geq \max (M_{d e c}, N^{'})$ so that $s t \leq t < 1 / ϵ$ for every $j$ and $∥ ψ ∥_{S_{M_{d e c}}} \leq 1$ for $ψ \in F_{N}$ . Write $M_{*} = M_{d e c}$ . For $ψ \in F_{N}$ , $t \in (0, 1 / ϵ)$ and $x$ write the deconvolution $ψ = \sum_{j} η^{j} * φ_{s_{0} 2^{- j}}$ and set $s = s_{0} 2^{- j}$ . If $M_{φ, T}^{ϵ, L} f (x) = + \infty$ , assertion 2 is immediate. Otherwise associativity follows from Schwartz parameter pairing and integral interchange applied to $H (w) (z) = η_{t}^{j} (w) φ_{s t} (x - w - z)$ : each seminorm is bounded by an integrable polynomial weight times $∣ η_{t}^{j} (w) ∣$ . Continuity of $f$ passes the Schwartz deconvolution partial sums to the scalar limit. Thus, with $w$ the integration variable, $∣ (f * ψ_{t}) (x) ∣ \leq \sum_{j} \int_{R^{n}} ∣ (f * φ_{s t}) (x - w) ∣ ∣ η_{t}^{j} (w) ∣ d w .$ By definition of $M_{φ, T}^{ϵ, L} f (x)$ , for every $w$ and $s, t > 0$ with $s t < 1 / ϵ$ , $∣ (f * φ_{s t}) (x - w) ∣ \leq M_{φ, T}^{ϵ, L} f (x) (1 + \frac{∣ w ∣}{s t})^{T} \frac{(s t + ϵ + ϵ ∣ x - w ∣)^{L}}{(s t)^{L}} .$ Multiplying by $t^{L} / (t + ϵ + ϵ ∣ x ∣)^{L}$ and using the triangle inequality $∣ x - w ∣ \leq ∣ x ∣ + ∣ w ∣$ , together with $ϵ / (t + ϵ) \leq 1 / t$ (valid since $ϵ \leq 1$ and $t \leq 1 / ϵ$ ), gives $(\frac{s t + ϵ + ϵ ∣ x - w ∣}{s t})^{L} (\frac{t}{t + ϵ + ϵ ∣ x ∣})^{L} \leq (\frac{1}{s} + \frac{∣ w ∣}{s t})^{L} .$ Therefore, substituting $w = t u$ and using $N^{'} > L + T + n$ so $(1 + ∣ u ∣)^{L + T - N^{'}}$ is integrable, $∣ (f * ψ_{t}) (x) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ x ∣)^{L}} \leq M_{φ, T}^{ϵ, L} f (x) \sum_{j} \int_{R^{n}} (1 + \frac{∣ u ∣}{s})^{T} (\frac{1}{s} + \frac{∣ u ∣}{s})^{L} ∣ η^{j} (u) ∣ d u \leq M_{φ, T}^{ϵ, L} f (x) \sum_{j} C_{T, L, N^{'}} s^{- (T + L)} ∥ η^{j} ∥_{S_{N^{'}}} \leq C^{'} M_{φ, T}^{ϵ, L} f (x) \sum_{j} 2^{j (T + L)} 2^{- j n L^{'}} = C_{r a d} M_{φ, T}^{ϵ, L} f (x),$ since $s \leq 1$ , [F3] gives $∥ η^{j} ∥_{S_{N^{'}}} \leq C 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ , and $n L^{'} > T + L$ makes the geometric series converge. The constants are independent of $N$ , $ϵ$ , $ψ$ , $t$ and $x$ ; the constant is independent of $ϵ$ because no weight with $ϵ$ remains. The same calculation for an arbitrary Schwartz test $θ$ retains the factor $∥ θ ∥_{S_{M_{*}}}$ on the right. For a cone witness $∣ z - x ∣ < t$ and $ψ \in F_{N}$ , put $h = (z - x) / t$ and $θ (w) = ψ (w + h)$ , so $(f * θ_{t}) (x) = (f * ψ_{t}) (z)$ . Since $∣ h ∣ < 1$ and $N \geq M_{*}$ , the weighted derivative inequality gives $∥ θ ∥_{S_{M_{*}}} \leq 2^{M_{*}}$ , independently of $N$ . Also $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ z ∣)$ since $∣ x - z ∣ < t$ and $ϵ \leq 1$ . Thus $∣ (f * ψ_{t}) (z) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ z ∣)^{L}} \leq 2^{L + M_{*}} C M_{φ, T}^{ϵ, L} f (x) .$ Taking the supremum over these cone witnesses and tests proves assertion 2, with $C_{1} = C_{1} (n, φ, T, L)$ independent of $N$ and $ϵ$ .

1.4F5F7F4algebra

Good-set bound. Fix $p_{0} > 0$ , $λ > 0$ , $N$ large enough for the estimate below, $0 < ϵ \leq 1 / 2$ , and $x$ with $M_{N}^{ϵ, L} f (x) < λ M_{φ, 1}^{ϵ, L} f (x)$ . This condition forces $M_{φ, 1}^{ϵ, L} f (x) < \infty$ : put $A = P_{N} (φ) > 0$ and $φ_{0} = φ / A \in F_{N}$ . For every aperture-one witness $(t, y)$ at $x$ , $v = (x - y) / t$ has $∣ v ∣ < 1$ , and the translated kernel $Ψ (w) = φ_{0} (w - v)$ satisfies $P_{N} (Ψ) \leq 2^{N}$ by [F5]. Thus $Ψ / 2^{N} \in F_{N}$ , $(f * Ψ_{t}) (x) = (f * (φ_{0})_{t}) (y)$ , and the denominator comparison $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ y ∣)$ gives $M_{N}^{ϵ, L} f (x) \geq 2^{- N - L} A^{- 1} M_{φ, 1}^{ϵ, L} f (x) .$ In particular an infinite right-hand side would force an infinite left-hand side, contrary to the strict good-set inequality. If the aperture-one quantity is positive, its finiteness lets us choose $t \in (0, 1 / ϵ)$ and $y$ with $∣ x - y ∣ < t$ such that $M_{φ, 1}^{ϵ, L} f (x) \leq 2 ∣ h (y) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}}, h = f * φ_{t} .$ If the aperture-one quantity is zero, the strict good-set inequality is impossible because $M_{N}^{ϵ, L} f (x) \geq 0$ . With $c = 2^{L} c_{N, φ}$ , [F5] gives $t \sup_{∣ z - y ∣ < r t} ∣ \nabla h (z) ∣ \leq c M_{N}^{ϵ, L} f (x) \frac{(t + ϵ + ϵ ∣ y ∣)^{L}}{t^{L}} (0 < r \leq 1) .$ Indeed, $t \partial_{j} h (z) = (f * Ψ_{t}^{z}) (x)$ for the translated derivative test function $Ψ^{z}$ , whose translation length is at most $2$ ; its $P_{N}$ seminorm is at most $c_{N, φ}$ . Also $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ y ∣)$ because $∣ x - y ∣ < t$ and $ϵ \leq 1$ . The mean value theorem and the good-set hypothesis now give $∣ h (x^{'}) - h (y) ∣ \leq c r λ M_{φ, 1}^{ϵ, L} f (x) \frac{(t + ϵ + ϵ ∣ y ∣)^{L}}{t^{L}} (x^{'} \in B (y, r t)) .$ Choose $0 < r \leq 1$ so $c r λ \leq 1 / 4$ . The saturation estimate yields $∣ h (x^{'}) ∣ \geq \frac{1}{4} M_{φ, 1}^{ϵ, L} f (x) (t + ϵ + ϵ ∣ y ∣)^{L} / t^{L} \geq \frac{1}{4} M_{φ, 1}^{ϵ, L} f (x)$ on $B (y, r t)$ , and $∣ h (x^{'}) ∣ \leq M_{φ}^{0} f (x^{'})$ . Thus, using the ball inclusion and [F4], $M_{φ, 1}^{ϵ, L} f (x)^{p_{0}} \leq 4^{p_{0}} (\frac{1 + r}{r})^{n} \frac{1}{λ (B (x, (1 + r) t))} \int_{B (x, (1 + r) t)} (M_{φ}^{0} f)^{p_{0}} .$ By [F7], $(M_{φ}^{0} f)^{p_{0}}$ is Borel, so the extended average defined before assertion 1 applies even if it is not locally integrable. The displayed average is at most $\tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)$ , so assertion 4 follows with $C_{3} = 4 ((1 + r) / r)^{n / p_{0}}$ .

1.5givenF7algebra

Removing truncation. For fixed $L, T, N$ and every fixed admissible witness, the weight $t^{L} / (t + ϵ + ϵ ∣ z ∣)^{L}$ increases to $1$ as $ϵ ↓ 0$ , and the permitted scale range increases to all $t > 0$ . Hence the radial, aperture-one and tangential truncated functions increase to their corresponding untruncated suprema. The same argument, also taking the supremum over $ψ \in F_{N}$ , gives $M_{N}^{ϵ, L} f (x) ↑ M_{N} f (x)$ . The strict cone $∣ z - x ∣ < t$ has the same supremum as the closed cone $∣ z - x ∣ \leq t$ , because each convolution is continuous and every boundary point is a limit of interior points at fixed $t$ ; therefore this limit is precisely the supplied nontangential $M_{N}$ .

2.1step 1.4F5F4algebra

Untruncated good-set estimate. Let $x$ satisfy $M_{N} f (x) \leq λ M_{φ}^{*, 1} f (x) < \infty$ , with $N \geq 1$ . If $M_{φ}^{*, 1} f (x) = 0$ , the asserted bound is immediate; otherwise, by definition of the supremum choose $t > 0, y$ with $∣ x - y ∣ < t$ and $∣ h (y) ∣ \geq \frac{1}{2} M_{φ}^{*, 1} f (x)$ , where $h = f * φ_{t}$ . As in step 1.4 but without truncation weights, $t \sup_{∣ z - y ∣ < r t} ∣ \nabla h (z) ∣ \leq c_{N, φ} M_{N} f (x)$ by [F5]. Choose $r \leq 1$ so $c_{N, φ} r λ \leq 1 / 4$ . The mean value theorem gives $∣ h (z) ∣ \geq \frac{1}{4} M_{φ}^{*, 1} f (x)$ on $B (y, r t)$ , and $∣ h (z) ∣ \leq M_{φ}^{0} f (z)$ . Using [F4], the ball inclusion and the extended average $\tilde{M}$ defined in step 1.4 yields $M_{φ}^{*, 1} f (x)^{p_{0}} \leq 4^{p_{0}} (\frac{1 + r}{r})^{n} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x) .$ This is assertion 5 with $N_{3} = 1$ and $C_{4} = 4 ((1 + r) / r)^{n / p_{0}}$ .

3.1step 1.1step 1.3step 1.2step 1.4step 2.1step 1.5∎

Conclusion. Step 1.1 gives the finiteness and pointwise decay of the aperture-one truncated function; step 1.3 gives the pointwise comparison of the truncated grand maximal function with the truncated tangential one, with constants independent of $ϵ$ ; step 1.2 gives the tangential-to-aperture-one comparison via the Hardy-Littlewood maximal operator; step 1.4 gives the truncated good-set bound and step 2.1 the untruncated one. Step 1.5 proves the stated monotone limits.

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Maximal-function characterisations of real Hardy spaces Theorem

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Sources

David Cruz-Uribe SFO, Li-An Daniel Wang, Variable Hardy Spaces, arXiv:1211.6505 (2012) (standard reference, not scraped)
Marcin Bownik, Anisotropic Hardy Spaces and Wavelets, Memoirs of the American Mathematical Society 164 (2003), no. 781 (standard reference, not scraped)