Abstract
We formulate two principles for weak-type (1,1) estimates. The first applies to linear operators through a locality identity; the second applies to maximal operators through comparison with a positive kernel. We explain the principles and apply them to obtain the following bounds.
| Operator | Upper bound |
|---|---|
Riesz transform | $$2$$ |
Beurling–Ahlfors transform | $$2$$ |
Second-order Riesz transform, $$n= 2$$ | $$\frac{3\sqrt{6}}{4}$$ |
Second-order Riesz transform, $$n\ge 2$$ | $$\frac1{a_n}+\frac{(n-1)a_n}{n-a_n^2}$$, $$a_n:=\sqrt{\frac{2n}{n+1+\sqrt{(n+1)^2+4(n-2)}}}$$ |
Traceless second-order Riesz transform | $$2\sqrt{1-n^{-1}}$$ |
Leray and gradient projections | $$\approx 1.805$$ |
Centered Hardy–Littlewood maximal operator, $$n=1$$ | $$2$$ |
Centered Hardy–Littlewood maximal operator for balls, $$n=2$$ | $$e$$ |
Centered Hardy–Littlewood maximal operator for axis-parallel squares, $$n=2$$ | $$<3.616$$ |
Centered Hardy–Littlewood maximal operator for balls, $$n\ge 3$$ | $$(n/2)^{n/(n-2)}$$ |
Poisson maximal operator, $$n=1$$ | $$1+\frac2\pi\left(\frac{\sqrt5}{3}-\arctan\frac2{\sqrt5}\right)\approx 1.010$$ |
Poisson maximal operator, $$n=2$$ | $$\approx 1.021$$ |
Poisson maximal operator | $$\frac{\Gamma((n+1)/2)}{\sqrt\pi\,\Gamma(n/2)} \biggl[\frac{2b^{n/2}}{n(1+a)^{(n+1)/2}}+\int_b^\infty\frac{z^{n/2-1}}{(1+z)^{(n+1)/2}}\,dz\biggr]$$ |
Heat maximal operator, $$n=1$$ | $$\approx 1.037087$$ |
Heat maximal operator, $$n=2$$ | $$\approx 1.094052$$ |
Heat maximal operator | $$\frac1{\Gamma(n/2)} \left[ \frac{2(\rho_n a)^{n/2}e^{-a}}{n} +\int_{\rho_n a}^\infty z^{n/2-1}e^{-z}\,dz \right]$$, $$\rho_n= \begin{cases} (n/2)^{2/(n-2)},&n\ne2,\\ e,&n=2. \end{cases}$$ Here $$a$$ is the unique nonzero solution in $$(n/(2\rho_n),n/2)$$ of $$e^{-(\rho_n-1)a}=1-\frac{2a}{n}.$$ |
1 Linear principle
1.1 The decomposition
The following theorem extends [2, Theorem 1.2] to finite-dimensional Hilbert-valued functions and includes the endpoint $$\alpha=2$$.
Theorem 1.1. Let $$E$$ be a finite-dimensional real or complex Hilbert space and $$0<\alpha\le2$$. For $$f\in L^1(\mathbb R^n;E)\cap L^2(\mathbb R^n;E)$$ and $$\kappa>0$$, there exist $$u\in H^\alpha(\mathbb R^n;E)\cap L^1(\mathbb R^n;E)$$ and $$\mu\in L^1(\mathbb R^n;E)\cap L^\infty(\mathbb R^n;E)$$ with
$$f=\mu+(-\Delta)^{\alpha/2}u$$;
$$\|\mu\|_{\infty}\le\kappa$$ and $$\|\mu\|_1\le\|f\|_1$$;
$$\mu=\kappa u/|u|_E$$ almost everywhere on $$\Omega:=\{|u|_E>0\}$$;
$$\kappa|\Omega|\le\|f\|_1$$.
For $$\alpha=2$$, both $$\nabla u$$ and $$D^2u$$ vanish almost everywhere on $$\Omega^c$$, and consequently $$f=\mu$$ there.
Proof. Replace the scalar absolute value in the variational argument of [2, Section 4] by the Hilbert norm, and minimize $$J(v)=\frac12\|(-\Delta)^{\alpha/4}v\|_2^2 +\kappa\|v\|_1-\operatorname{Re}\int\langle f,v\rangle_E$$ over $$H^{\alpha/2}\cap L^1$$. The same compactness argument gives a unique minimizer $$u$$. At $$\alpha=2$$, the energy is $$\frac12\|\nabla v\|_2^2$$.
The Euler condition means that no perturbation $$u+th$$ lowers $$J$$ to first order; $$t$$ is real even when $$E$$ is complex. At $$u\ne0$$, the norm has directional derivative $$\operatorname{Re}\langle u/|u|_E,h\rangle_E$$; at zero its one-sided directional derivative is $$|h|_E$$. Minimality in both directions gives $$(-\Delta)^{\alpha/2}u+\mu=f,\qquad |\mu|_E\le\kappa,\qquad \mu=\kappa u/|u|_E\quad\text{on }\Omega.$$ In the nonnegative scalar case the saturated value is $$\kappa$$. Here it is a vector of length $$\kappa$$ pointing along $$u$$.
For the $$L^1$$ estimate, choose $$w=u/|u|_E$$ on $$\Omega$$ and $$w=\mu/|\mu|_E$$ elsewhere, taking zero when both vanish. The positive averaging semigroup $$P_t=e^{-t(-\Delta)^{\alpha/2}}$$ satisfies $$|u|_E-P_t|u|_E \le\operatorname{Re}\langle w,u-P_tu\rangle_E.$$ Divide by $$t$$ and pass to the limit against nonnegative smooth test functions. Since $$(u-P_tu)/t\to f-\mu$$ in $$L^2_{\rm loc}$$, this gives $$(-\Delta)^{\alpha/2}|u|_E\le |f|_E-|\mu|_E \quad\text{in distributions}.$$ Test with $$\chi_R=\chi(\cdot/R)$$, where $$0\le\chi\le1$$ is smooth, compactly supported, and equals one near zero. Since $$\|(-\Delta)^{\alpha/2}\chi_R\|_\infty=O(R^{-\alpha})$$, letting $$R\to\infty$$ gives $$\|\mu\|_1\le\|f\|_1$$. Saturation then gives $$\kappa|\Omega|\le\|f\|_1$$.
Now $$\mu\in L^1\cap L^\infty\subset L^2$$, so Plancherel gives $$u\in H^\alpha$$. At $$\alpha=2$$, applying the Sobolev zero-set property to $$u$$ and then to its first derivatives gives $$\nabla u=D^2u=0$$ on $$\Omega^c$$, hence $$\mu=f$$ there. For scalar $$f\ge0$$, replacing $$u$$ by $$|u|$$ does not increase $$J$$, so $$u\ge0$$. Testing the decomposition with $$\chi_R$$ also gives $$\int\mu=\int f$$, which together with the $$L^1$$ bound implies $$\mu\ge0$$. ◻
For nonnegative scalar data, this is the partial balayage construction: $$f$$ is redistributed into the capped density $$\mu$$, while the odometer $$u$$ records the redistribution through $$(-\Delta)^{\alpha/2}u=f-\mu$$. We now apply Theorem 1.1 to obtain our first principle.
Principle 1 (Linear). Let $$E$$ and $$F$$ be finite-dimensional real or complex Hilbert spaces, and let $$T:L^2(\mathbb R^n;E)\to L^2(\mathbb R^n;F)$$ be a bounded linear operator. Suppose that $$\|T\|_{2\to2}\le M$$ and, for some $$0<\alpha\le2$$, the composition $$T(-\Delta)^{\alpha/2}$$ is local in the following sense: for every $$u\in H^\alpha(\mathbb R^n;E)$$, $$T(-\Delta)^{\alpha/2}u=0$$ almost everywhere on $$\{u=0\}.$$ Then $$\|T\|_{1\to1,\infty}\le2M$$.
In the applications below, $$T(-\Delta)^{\alpha/2}$$ is a differential operator; the required vanishing follows from the Sobolev zero-set property.
Proof. For $$f\in L^1\cap L^2$$, use the decomposition $$f=\mu+(-\Delta)^{\alpha/2}u$$. Outside $$\Omega$$ one has $$Tf=T\mu+T(-\Delta)^{\alpha/2}u=T\mu$$ by locality, so Chebyshev’s inequality gives $$\begin{aligned} \lambda\bigl|\{|Tf|>\lambda\}\bigr| \le\lambda|\Omega|+\lambda^{-1}\|T\mu\|_2^2\le\left(\frac{\lambda}{\kappa} +\frac{M^2\kappa}{\lambda}\right)\|f\|_1. \end{aligned}$$ Here $$\|\mu\|_2^2\le\kappa\|\mu\|_1\le\kappa \|f\|_1$$. For $$M>0$$, choose $$\kappa=\lambda/M$$. The case $$M=0$$ is trivial. The estimate extends to $$L^1$$ by density. ◻
An identity component can improve the bound.
Principle 1.2 (Variant of Principle 1). Suppose $$E=F$$ and $$T=cI+S$$ for some $$c\ne0$$, with $$\|S\|_{2\to2}\le M$$. Assume the same locality condition as in Principle 1. Then $$\|T\|_{1\to1,\infty} \le\inf_{0<a<1/|c|} \left\{\frac1a+\frac{M^2a}{(1-|c|a)^2}\right\}.$$
Proof. Choose $$0<\kappa<\lambda/|c|$$ and write $$f=\mu+(-\Delta)^{\alpha/2}u$$ using the decomposition. Outside $$\Omega$$ one has $$|Tf|=|T\mu|=|c\mu+S\mu| \le |c|\kappa+|S\mu|.$$ Hence, up to a null set, $$\{|Tf|>\lambda\} \subseteq \Omega\cup\{|S\mu|>\lambda-|c|\kappa\}.$$ The restriction on $$\kappa$$ makes the latter threshold positive. Applying Chebyshev’s inequality gives $$\begin{aligned} \lambda|\{|Tf|>\lambda\}| \le \lambda|\Omega| +\frac{\lambda\|S\mu\|_2^2}{(\lambda-|c|\kappa)^2}\le \left(\frac{\lambda}{\kappa} +\frac{\lambda M^2\kappa}{(\lambda-|c|\kappa)^2}\right)\|f\|_1. \end{aligned}$$ Here $$\|S\mu\|_2^2\le M^2\|\mu\|_2^2 \le M^2\kappa\|\mu\|_1 \le M^2\kappa\|f\|_1$$. Set $$\kappa=a\lambda$$, where $$0<a<1/|c|$$. The estimate becomes $$\lambda|\{|Tf|>\lambda\}| \le\left(\frac1a+\frac{M^2a}{(1-|c|a)^2}\right)\|f\|_1.$$ ◻
2 Applications of the linear principle
2.1 Riesz transform
Denote the Riesz transform by $$R=\nabla(-\Delta)^{-1/2}$$. Its Fourier multiplier is $$i\xi/|\xi|$$, so Plancherel gives $$\|R\|_{2\to2}=1$$. $$R(-\Delta)^{1/2}u=\nabla u$$. Principle 1 therefore gives $$\|R\|_{1\to1,\infty}\le2.$$ This is exactly Theorem 1.1 in [2].
2.2 Beurling–Ahlfors transform
The Beurling–Ahlfors transform is $$Bf(z)=-\frac1\pi\operatorname{p.v.}\int_\mathbb C \frac{f(w)}{(z-w)^2}\,dA(w), \qquad \widehat{Bf}(\xi)=\frac{(\xi_1-i\xi_2)^2}{|\xi|^2}\widehat f(\xi).$$ Its multiplier has modulus one and thus $$\|B\|_{2\to2}=1$$. With $$D=\partial_1-i\partial_2$$, $$B(-\Delta u)=-D^2u.$$ Principle 1 gives $$\|B\|_{1\to1,\infty}\le2$$.
Classical Calderón–Zygmund theory gives weak-type $$(1,1)$$ boundedness; Astala–Iwaniec–Martin record the explicit upper bound $$30$$ [3, Theorem 4.5.2]. Guerra [4, Corollary 1.2] proved the sharp constant $$2$$ for real-valued inputs using quasiconvexity and Morse theory.
2.3 Second-order Riesz transforms
For scalar $$f$$, define the matrix-valued second-order Riesz transform $$\mathcal H_nf=(R_jR_kf)_{j,k} =D^2(-\Delta)^{-1}f.$$ The multiplier of $$\mathcal H_n$$ is $$-\xi\otimes\xi/|\xi|^2$$. With the Frobenius norm, $$\|\mathcal H_n\|_{2\to2}=1$$. Define its traceless part by $$\mathcal H_{0,n}f=\mathcal H_nf+\frac fn I.$$ Orthogonality to the identity matrix gives $$|\mathcal H_nf|_F^2=|\mathcal H_{0,n}f|_F^2+\frac{|f|^2}{n}, \qquad \|\mathcal H_{0,n}f\|_2^2=\left(1-\frac1n\right)\|f\|_2^2.$$ Both compositions are local: $$\mathcal H_n(-\Delta u)=D^2u, \qquad \mathcal H_{0,n}(-\Delta u)=D^2u-\frac{\Delta u}{n}I.$$ Principle 1 therefore gives $$\|\mathcal H_n\|_{1\to1,\infty}\le2$$ and $$\|\mathcal H_{0,n}\|_{1\to1,\infty}\le2\sqrt{1-1/n}$$. The trace identity improves the first bound.
Corollary 2.1. For $$n\ge2$$,
$$\|\mathcal H_n\|_{1\to1,\infty} \le\min_{0<a<\sqrt n}\left(\frac1a+\frac{(n-1)a}{n-a^2}\right) =\frac1{a_n}+\frac{(n-1)a_n}{n-a_n^2}, \tag{1}$$
where $$a_n^2=\frac{2n}{n+1+\sqrt{(n+1)^2+4(n-2)}}.$$ In dimension two this upper bound is $$3\sqrt6/4=1.837117307087\ldots$$. For every $$n\ge2$$ it is less than $$2$$, and as $$n\to\infty$$ it equals $$2-n^{-2}+2n^{-3}+O(n^{-4})$$.
Proof. The proof is similar to that of Principle 1.2. Write $$f=\mu-\Delta u$$ with cap $$\kappa$$. Outside $$\Omega=\{|u|>0\}$$, locality and trace orthogonality imply $$|\mathcal H_nf|_F^2=|\mathcal H_n\mu|_F^2 =|\mathcal H_{0,n}\mu|_F^2+\frac{|\mu|^2}{n} \le |\mathcal H_{0,n}\mu|_F^2+\frac{\kappa^2}{n}.$$ For $$0<\kappa<\sqrt n\,\lambda$$, Chebyshev’s inequality yields $$\begin{aligned} \lambda|\{|\mathcal H_nf|_F>\lambda\}| \le \lambda|\Omega| +\frac{\lambda\|\mathcal H_{0,n}\mu\|_2^2}{\lambda^2-\kappa^2/n}\le\left(\frac\lambda\kappa+ \frac{\lambda(n-1)\kappa}{n\lambda^2-\kappa^2}\right)\|f\|_1. \end{aligned}$$ Set $$a=\kappa/\lambda$$. Differentiating the resulting expression gives $$(n-2)a^4+n(n+1)a^2-n^2=0,$$ whose unique positive solution gives the stated minimum. ◻
For real-valued $$f$$ in dimension two, $$Bf=-(\mathcal H_2f)_{11}+(\mathcal H_2f)_{22} +2i(\mathcal H_2f)_{12},\qquad |\mathcal H_2f|_F^2=\frac{|f|^2+|Bf|^2}{2},\qquad |\mathcal H_{0,2}f|_F=\frac{|Bf|}{\sqrt2}.$$ Guerra’s quasiconvexity theorem and sharp Beurling–Ahlfors estimate [4, Theorem 1.1 and Corollary 1.2] imply, for real-valued inputs, $$\|\mathcal H_2\|_{1\to1,\infty} =\|\mathcal H_{0,2}\|_{1\to1,\infty}=\sqrt2.$$ We give the deduction in Appendix A. Thus the method we use gives a sharp bound to $$\mathcal H_{0,2}$$ but not to $$\mathcal H_{2}$$.
2.4 Projections
For $$n\ge2$$, the Helmholtz–Leray projection and its complement have multipliers $$\widehat{\mathbb Pf}(\xi)=\left(I-\frac{\xi\otimes\xi}{|\xi|^2}\right) \widehat f(\xi),\qquad \mathbb Q=I-\mathbb P.$$ The fields carry the pointwise Euclidean norm (Hermitian norm for complex fields). These operators extract their divergence-free and gradient parts. Classical singular-integral theory gives weak $$(1,1)$$ boundedness in each dimension [5]. To obtain an explicit uniform bound, use the local identities: $$\mathbb P(-\Delta u)=-\Delta u+\nabla\operatorname{div}u, \qquad \mathbb Q(-\Delta u)=-\nabla\operatorname{div}u.$$ For either projection $$T$$, write $$T=\tfrac12I+S$$. The Fourier multipliers project onto the direction of $$\xi$$ and its orthogonal complement, so Plancherel’s theorem gives $$Tf\perp(I-T)f$$ in $$L^2$$. Hence $$\|Sf\|_2^2 =\frac14\|Tf-(I-T)f\|_2^2 =\frac14\bigl(\|Tf\|_2^2+\|(I-T)f\|_2^2\bigr) =\frac14\|f\|_2^2.$$ Thus orthogonality gives $$\|S\|_{2\to2}=1/2$$. Principle 1.2 gives the dimension-free bound $$\|T\|_{1\to1,\infty}\le\min_{0<a<2}\left(\frac1a+\frac{a}{(2-a)^2}\right) =1.805359306638\ldots.$$ The minimizing $$a_*$$ is the unique root in $$(0,1)$$ of $$a_*^3-2a_*^2+6a_*-4=0, \qquad \|T\|_{1\to1,\infty}\le\frac1{a_*}+\frac{a_*}{(2-a_*)^2}.$$ The gain comes from using pointwise control of the identity component and the smaller $$L^2$$ norm of the remainder.
3 Maximal principle
3.1 The decomposition
Let $$\mathcal A$$ be the generator of a conservative, symmetric Markov convolution semigroup on $$\mathbb R^n$$. This means $$P_tg(x)=\int_{\mathbb R^n}g(x-y)\,d\nu_t(y),$$ where $$\nu_t$$ are probability measures invariant under $$y\mapsto-y$$, satisfying $$\nu_s*\nu_t=\nu_{s+t}$$ and $$\nu_t\rightharpoonup\delta_0$$ as $$t\downarrow0$$. Thus $$P_t$$ preserves positivity, constants, and mass, and is self-adjoint on $$L^2$$. The generator is $$\mathcal Ag=\lim_{t\downarrow0}\frac{P_tg-g}{t} \quad\text{in }L^2,\qquad g\in\operatorname{Dom}(\mathcal A),$$ with Fourier symbol $$-\psi$$, where $$\psi\ge0$$. We assume that, for every bounded, compactly supported $$f\ge0$$ and $$\kappa>0$$, there exist $$u\ge0$$ in $$L^1\cap\operatorname{Dom}(\mathcal A)$$ and a density $$\sigma$$ such that $$f=\sigma-\mathcal A u,\qquad 0\le\sigma\le\kappa,\qquad \sigma=\kappa\text{ on }D=\{u>0\},\qquad \int\sigma=\int f.$$ Here $$\operatorname{Dom}(\mathcal A)$$ is the $$L^2$$ operator domain. In particular, $$|D|\le\|f\|_1/\kappa$$. This is the partial balayage decomposition needed below. It holds for $$\Delta$$ by Theorem 1.1, and $$\mathcal A_\alpha=-\sum_j|D_j|^\alpha$$, $$0<\alpha<2$$, by the same variational argument: their energy is comparable to the fractional Sobolev energy, their semigroups preserve positivity and mass, and $$\|\mathcal A_\alpha\chi(\cdot/R)\|_\infty=O(R^{-\alpha})$$. The details are recorded in [8].
3.2 The estimate
Principle 2 (Maximal). Suppose $$\mathcal A$$ is homogeneous of order $$\alpha>0$$ and admits the decomposition above. Let $$K\ge0$$ be an even $$L^1$$ kernel satisfying $$\mathcal AK\ge0 \quad\text{in distributions on }\mathbb R^n\setminus\{0\}.$$ For $$K_r(x)=r^{-n}K(x/r)$$, define $$M_Kf(x)=\sup_{r>0}(K_r*|f|)(x).$$ Then $$\|M_K\|_{1\to1,\infty}\le\int_{\mathbb R^n}K.$$ Any measurable operator dominated in absolute value by $$M_K$$ satisfies the same estimate.
Proof. We give an outline; details of the kernel argument for the coordinate-stable case can be found in [6, Lemma 3]. Write $$M=\int K$$ and assume $$M>0$$, since $$K=0$$ otherwise. For $$\lambda>0$$ and bounded, compactly supported $$f\ge0$$, take the decomposition at height $$\kappa=\lambda/M$$. It suffices to show that every $$K_r*f$$ is at most $$\lambda$$ outside the same set $$D=\{u>0\}$$.
Put $$\mathcal Bv=K*\mathcal Av$$. The condition $$\mathcal AK\ge0$$ away from the origin, together with a cutoff argument at the origin, gives the positive minimum principle for $$\mathcal B$$: if $$\varphi\in C_c^\infty$$ is nonnegative and $$\varphi(x)=0$$, then $$\mathcal B\varphi(x)\ge0$$. Courrège’s theorem [7] then identifies $$\mathcal B$$ as a conservative, symmetric Markov generator. Its symbol is $$-\psi\widehat K$$, with $$0\le\psi\widehat K\le M\psi$$, so $$u$$ lies in its operator domain.
The Markov property gives $$\mathcal Bu\ge0$$ almost everywhere on $$D^c$$. Indeed, for $$0\le\varphi\in C_c^\infty$$, set $$h_\varepsilon=(\varepsilon\varphi-u)_+$$. The Dirichlet form satisfies $$-\langle\mathcal Bu,h_\varepsilon\rangle \le-\varepsilon\langle\mathcal B\varphi,h_\varepsilon\rangle.$$ Divide by $$\varepsilon$$ and let $$\varepsilon\downarrow0$$, using $$h_\varepsilon/\varepsilon\to\varphi\mathbf1_{D^c}$$ and $$\|h_\varepsilon\|_2\le\varepsilon\|\varphi\|_2$$. Consequently, $$K*f=K*\sigma-\mathcal Bu\le\kappa\int K=\lambda \quad\text{almost everywhere on }D^c.$$ Homogeneity gives $$\mathcal AK_r=r^{-n-\alpha}(\mathcal AK)(\cdot/r)\ge0$$ away from zero, so the same argument applies to every dilation. Continuity of $$r\mapsto K_r$$ in $$L^1$$ extends the conclusion from rational scales to all scales outside one null set. Hence $$\lambda|\{M_Kf>\lambda\}|\le\lambda|D|\le M\|f\|_1.$$ Monotone approximation gives the result for general $$f$$. ◻
Corollary 3.1. For $$\mathcal A$$ as in Principle 2 and an even nonnegative $$\phi\in L^1$$, let $$M_\phi f=\sup_{r>0}\phi_r*|f|, \qquad \phi_r(x)=r^{-n}\phi(x/r).$$ Then $$\begin{aligned} \|M_\phi\|_{1\to1,\infty}\le C_{\mathcal A}(\phi):=\inf\biggl\{ &\int K:\ K\in L^1\text{ is even},\quad K\ge\phi,\quad\mathcal AK\ge0\text{ on }\mathbb R^n\setminus\{0\}\biggr\}. \end{aligned}$$
Proof. Since $$M_\phi\le M_K$$ for every admissible $$K$$, Principle 2 applies; take the infimum of the resulting bounds. ◻
4 Applications of the maximal principle
4.1 Intervals
The normalized tent kernel is the simplest exact example of the maximal principle: $$\tau(x)=(1-|x|)_+,\qquad \int_\mathbb R\tau=1,\qquad \tau''=\delta_{-1}+\delta_1-2\delta_0.$$ Thus its dilation maximal operator $$T_*f(x)=\sup_{r>0}(\tau_r*|f|)(x),\qquad \tau_r(x)=r^{-1}\tau(x/r),$$ has weak-type constant at most one. For the centered Hardy–Littlewood maximal operator on intervals, $$M_cf(x)=\sup_{r>0}\frac1{2r}\int_{x-r}^{x+r}|f(y)|\,dy,$$ the majorant $$K(x)=(1-|x|/2)_+\ge\tfrac12\mathbf1_{[-1,1]}(x), \qquad \int K=2,$$ gives $$\|M_c\|_{1\to1,\infty}\le2$$. This majorant attains $$C_\Delta(\frac12\mathbf1_{[-1,1]})$$ by Corollary 4.1. Melas [9] proved the sharper, optimal value $$\|M_c\|_{1\to1,\infty}=\frac{11+\sqrt{61}}{12}=1.5675208063\ldots$$ using discretization.
4.2 Euclidean balls
The centered Hardy–Littlewood maximal operator over Euclidean balls is $$M_{B,n}f(x)=\sup_{r>0}\frac1{\omega_n r^n}\int_{B(x,r)}|f(y)|\,dy, \qquad \omega_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}.$$ Here the averages are over solid Euclidean balls.
The classical covering argument gives an exponential bound $$2^n$$ [10, Exercise 42]. Stein–Strömberg improved the dimensional order to $$O(n)$$ [11].
Corollary 4.1 (Radial majorants). Put $$\rho_n=\begin{cases}(n/2)^{2/(n-2)},&n\ne2,\\ e,&n=2.\end{cases}$$ Suppose $$\phi(x)=q(|x|^2)$$, with $$q$$ left-continuous at $$a>0$$. If an admissible radial majorant for $$\mathcal A=\Delta$$ is harmonic for $$0<|x|<\sqrt b$$, touches $$\phi$$ at $$|x|=\sqrt a$$, and equals $$\phi$$ for $$|x|\ge\sqrt b$$, where $$b=\rho_n a$$, then it attains $$C_\Delta(\phi)$$.
Proof. Rotational averaging preserves the mass and admissibility of any competitor. Write the average as $$k(|x|^2)$$. Subharmonicity means that $$k$$ is convex in the coordinate $$s(z)=\begin{cases} \dfrac{z^{1-n/2}}{1-n/2},&n\ne2,\\[4pt] \log z,&n=2. \end{cases}$$ If $$Z$$ has density $$\frac{n}{2b^{n/2}}z^{n/2-1}$$ on $$(0,b)$$, then $$\mathbb E s(Z)=s(a)$$. Jensen’s inequality gives $$\int_0^b k(z)z^{n/2-1}\,dz \ge\frac{2b^{n/2}}n k(a) \ge\frac{2b^{n/2}}n q(a).$$ Outside $$b$$ use $$k\ge q$$. The stated majorant attains both inequalities, so no admissible kernel has smaller mass. ◻
The following radial majorants give explicit constants: $$\|M_{B,2}\|_{1\to1,\infty}\le e, \qquad \|M_{B,n}\|_{1\to1,\infty}\le\left(\frac n2\right)^{n/(n-2)}\quad(n\ge3).$$ For the planar unit ball use $$K_2(x)=\frac2\pi\log\frac{\sqrt e}{|x|}\, \mathbf1_{\{|x|<\sqrt e\}}.$$ It majorizes $$\pi^{-1}\mathbf1_{B(0,1)}$$, its distributional Laplacian is a positive boundary measure minus an atom at zero, and its mass is $$e$$. For $$n\ge3$$ put $$R_n=\left(\frac n2\right)^{1/(n-2)},\qquad K_n(x)=\frac1{\omega_n} \frac{(|x|^{2-n}-R_n^{2-n})_+}{1-R_n^{2-n}}.$$ The same conditions hold and direct polar integration gives $$\int K_n=R_n^n=\left(\frac n2\right)^{n/(n-2)}.$$ These kernels attain $$C_\Delta(\omega_n^{-1}\mathbf1_{\overline{B(0,1)}})$$ by Corollary 4.1; changing the ball boundary does not affect the averages. The bound is asymptotic to $$n/2$$, whereas Spector–Stockdale prove $$\|M_{B,n}\|_{1\to1,\infty}=O(\sqrt n\log(1+n))$$ by exploiting the heat residual [1, Theorem 1.1].
4.3 Axis-parallel squares
The centered Hardy–Littlewood maximal operator over axis-parallel squares in the plane is $$M_\square f(x)=\sup_{r>0}\frac1{4r^2} \int_{x+[-r,r]^2}|f(y)|\,dy.$$ The invertible map $$(u,v)\mapsto(u+v,u-v)$$ takes the unit diamond $$D_\diamond=\{(u,v):|u|+|v|\le1\}$$ to the unit square. Change of variables preserves a weak-type constant, and the diamond has area two. Use the homogeneous symmetric Markov generator $$\mathcal A_\alpha=-|D_u|^\alpha-|D_v|^\alpha, \qquad \widehat{\mathcal A_\alpha g}(\xi) =-(|\xi_1|^\alpha+|\xi_2|^\alpha)\widehat g(\xi).$$ Corollary 3.1 gives $$\begin{aligned} \|M_\square\|_{1\to1,\infty}\le C_\alpha:=\inf\biggl\{ &\frac12\int_{\mathbb R^2}K:\ K\in L^1(\mathbb R^2)\text{ is even},\\ &K\ge\mathbf1_{D_\diamond},\quad \mathcal A_\alpha K\ge0\text{ on }\mathbb R^2\setminus\{0\}\biggr\}. \end{aligned}$$ Take $$\alpha=6/5$$. A fractional radial profile on the diamond, corrected by a finite combination of cubic splines, gives an admissible kernel with $$\boxed{\|M_\square\|_{1\to1,\infty}\le C_{6/5}<3.615749.}$$ Exact rational certificates verify domination and generator positivity [8]. This is a feasible comparison bound; attainment of the infimum $$C_{6/5}$$ is not established. For comparison, the covering bound is $$4$$. A preceding certified bound $$3.879$$ and a lower bound $$1.68550999\ldots$$ are recorded in optimizationproblems PR 192. The GitHub development and Palomar v3 contain formalizations of that earlier square bound and the ball estimates. The refined bound $$3.615749$$ is supported by the separate exact certificate in the accompanying supplement.
4.4 Poisson maximal
Define $$P_*f(x)=\sup_{t>0}(p_t*|f|)(x),\qquad p_t(x)=\frac{\Gamma((n+1)/2)}{\pi^{(n+1)/2}} \frac{t}{(t^2+|x|^2)^{(n+1)/2}}.$$ Dimension-free weak $$(1,1)$$ boundedness is a classical application of the Hopf–Dunford–Schwartz ergodic theorem [12, pp. 48–49]; see also [13, Theorem 5.5]. An explicit evaluation of this argument gives, in every dimension, $$\|P_*\|_{1\to1,\infty}\le B_P:= \operatorname{erf}\sqrt{\frac32} +\sqrt{\frac{3}{2\pi}}e^{-3/2} =1.070915813140\ldots.$$ Indeed, the subordination density $$m_t(s)=\frac{t}{2\sqrt\pi}s^{-3/2}e^{-t^2/(4s)}$$ peaks at $$s_0=t^2/6$$. Its nonincreasing majorant, constant before $$s_0$$ and equal to $$m_t$$ thereafter, has mass $$B_P$$. Integration by parts dominates $$P_*f$$ by $$B_P$$ times the maximal heat ergodic average, whose weak constant is one.
The Laplacian majorant gives the following bound.
Corollary 4.2. With $$\rho_n$$ as in Corollary 4.1, let $$a$$ be the unique nonzero solution in $$(n/(3\rho_n),n/3)$$ of $$\frac{1-a/n}{(1+a)^{(n+3)/2}} =\frac{1}{(1+\rho_n a)^{(n+1)/2}},\qquad b=\rho_n a.$$ Then
$$\begin{aligned} \|P_*\|_{1\to1,\infty}\le \frac{\Gamma((n+1)/2)}{\sqrt\pi\,\Gamma(n/2)} \biggl[&\frac{2b^{n/2}}{n(1+a)^{(n+1)/2}}+\int_b^\infty\frac{z^{n/2-1}}{(1+z)^{(n+1)/2}}\,dz\biggr]. \end{aligned} \tag{2}$$
For $$n=1$$, this is $$\|P_*\|_{1\to1,\infty}\le 1+\frac2\pi\left(\frac{\sqrt5}{3}-\arctan\frac2{\sqrt5}\right) =1.009949307880\ldots.$$ The right-hand side of (2) tends to $$B_P=1.070915813140\ldots$$ as $$n\to\infty$$.
Proof. Write $$\beta=n/2$$, $$q(z)=(1+z)^{-(n+1)/2}$$, and $$z=|x|^2$$. For $$z<b$$, replace $$q$$ by its radial harmonic tangent at $$a$$: $$k(z)=\begin{cases} q(a)+\dfrac{a q'(a)}{1-\beta} \bigl((z/a)^{1-\beta}-1\bigr),&n\ne2,\\[6pt] q(a)+a q'(a)\log(z/a),&n=2. \end{cases}$$ For $$z\ge b$$, keep $$k(z)=q(z)$$. The equation for $$a$$ makes the branches continuous. The Poisson profile has negative Laplacian for $$z<n/3$$ and positive Laplacian for $$z>n/3$$. Thus its graph is concave and then convex in the harmonic radial coordinate: the tangent majorizes the inner profile, and the derivative jumps upward at the outer joining point. Consequently $$K(x)=\frac{\Gamma((n+1)/2)}{\pi^{(n+1)/2}}k(|x|^2) \ge p_1(x),\qquad \Delta K=\nu-c\delta_0, \quad \nu\ge0,\ c>0.$$ Corollary 3.1 bounds the maximal operator by $$\int K$$. Integrating the inner branch, using $$\rho_n^{1-\beta}=1/\beta$$ when $$n\ne2$$, gives $$\int_0^b k(z)z^{\beta-1}\,dz=\frac{b^\beta q(a)}\beta;$$ the logarithmic case gives the same identity. The majorant attains $$C_\Delta(p_1)$$ by Corollary 4.1. Polar integration now yields (2). In dimension one $$a=1/5$$ and $$b=4/5$$, which gives the stated elementary expression. Finally, $$a/n\to1/3$$ and $$b/n\to1/3$$. Stirling’s formula gives the limit $$\sqrt{3/(2\pi)}e^{-3/2}$$ for the inner contribution. In the outer integral, the substitution $$y=(n/2)/z$$ gives the limiting contribution $$\pi^{-1/2}\int_0^{3/2}y^{-1/2}e^{-y}\,dy =\operatorname{erf}\sqrt{3/2}$$. ◻
4.5 Heat maximal
Define $$H_*f(x)=\sup_{t>0}(h_t*|f|)(x),\qquad h_t(x)=(4\pi t)^{-n/2}e^{-|x|^2/(4t)}.$$
For each $$n\ge1$$, use $$\rho_n$$ from Corollary 4.1 and let $$a=a(n)$$ be the unique nonzero solution in $$(n/(2\rho_n),n/2)$$ of $$e^{-(\rho_n-1)a}=1-\frac{2a}{n}.$$ Corollary 3.1 gives $$\|H_*\|_{1\to1,\infty}\le \frac1{\Gamma(n/2)} \left[ \frac{2(\rho_n a)^{n/2}e^{-a}}{n} +\int_{\rho_n a}^\infty z^{n/2-1}e^{-z}\,dz \right].$$
To construct the majorant, put $$z=|x|^2$$ and replace $$q(z)=e^{-z}$$ for $$0<z<\rho_n a$$ by its harmonic tangent at $$a$$: $$k(z)=e^{-a} \begin{cases} 1+\dfrac{2a}{n-2}\bigl((z/a)^{1-n/2}-1\bigr),&n\ne2,\\[6pt] 1-a\log(z/a),&n=2. \end{cases}$$ For $$z\ge\rho_n a$$, keep $$k(z)=e^{-z}$$, and set $$K(x)=\pi^{-n/2}k(|x|^2)$$. The root equation makes the two branches meet. The profile is concave and then convex in the harmonic radial coordinate, because the sign of its Laplacian is the sign of $$z-n/2$$. The tangent lies above the inner profile and meets its convex outer branch with a nonnegative flux jump. Therefore $$K\ge h_{1/4}$$ and $$\Delta K$$ is a positive finite measure minus an atom at zero. Polar integration gives the displayed bound. The majorant attains $$C_\Delta(h_{1/4})$$ by Corollary 4.1.
In dimension one, $$a(1)$$ satisfies $$e^{-3a(1)}=1-2a(1)$$, and the bound is $$\|H_*\|_{1\to1,\infty}\le 4\sqrt{\frac{a(1)}\pi}\,e^{-a(1)} +\operatorname{erfc}(2\sqrt{a(1)}) =1.037087059430\ldots,$$ where $$\operatorname{erfc}(r)=\frac2{\sqrt\pi}\int_r^\infty e^{-s^2}\,ds$$. In dimension two, $$a(2)$$ satisfies $$e^{-(e-1)a(2)}=1-a(2)$$, and $$\|H_*\|_{1\to1,\infty}\le [1+(e-1)a(2)]e^{-a(2)}=1.094052157640\ldots.$$ For $$n=3$$, the bound is $$1.150845692298\ldots$$.
As $$n\to\infty$$, $$a(n)=\frac n2-1+o(1),\qquad \rho_n a(n)=\frac n2+\log\frac n2-1+o(1).$$ Stirling’s formula shows that the resulting upper bound is asymptotic to $$\sqrt n/(2\sqrt\pi)$$. This growth is worse than the $$O(\log(1+n))$$ heat bound proved by Spector–Stockdale [1, Theorem 1.2].
A Sharp planar bound for the second-order Riesz transform
We deduce the sharp planar matrix estimate from Guerra’s quasiconvexity theorem [4, Theorem 1.1].
Corollary A.1. For every real-valued $$f\in L^1(\mathbb R^2)$$ and $$t>0$$, $$t\bigl|\{|f|+|Bf|>t\}\bigr|\le2\|f\|_1.$$ Consequently, $$\|\mathcal H_2\|_{1\to1,\infty} =\|\mathcal H_{0,2}\|_{1\to1,\infty}=\sqrt2$$.
Proof. Identify $$\mathbb R^2$$ with $$\mathbb C$$ and write a real matrix as $$Az=a_+z+a_-\overline z$$. Then $$\|A\|_{\rm op}=|a_+|+|a_-|$$ and $$\det A=|a_+|^2-|a_-|^2$$. Guerra’s integrand is $$L(A)= \begin{cases} \det A,& |a_+|+|a_-|\le1,\\ 2|a_+|-1,& |a_+|+|a_-|>1. \end{cases}$$ Its definition gives the following strengthening of [4, (6.1)]: $$L(A/t)\le\frac{2|a_+|}{t} -\mathbf1_{\{|a_+|+|a_-|>t\}}.$$ Indeed, equality holds when $$|a_+|+|a_-|>t$$; otherwise the indicator vanishes and $$\det(A/t)\le |a_+|^2/t^2\le |a_+|/t$$.
First take real $$f\in C_c^\infty(\mathbb R^2)$$ and set $$u(x)=\frac1\pi\int_{\mathbb R^2}\log|x-y|\,f(y)\,dy, \qquad \Delta u=2f.$$ For $$A=D^2u$$, the complex coefficients satisfy $$a_+=\frac{u_{11}+u_{22}}2=f,\qquad a_-=\frac{u_{11}-u_{22}}2+i u_{12},\qquad |a_-|=|Bf|.$$ We claim that Guerra’s theorem implies $$\int_{\mathbb R^2}L(D^2u/t)\,dx\ge0.$$ To justify its use for this noncompact potential, choose a smooth cutoff $$\eta_R$$ equal to one on $$B(0,R)$$ and zero outside $$B(0,2R)$$, and apply the theorem at the zero matrix to $$\eta_Ru/t$$. At infinity, $$u=O(\log|x|),\qquad \nabla u=O(|x|^{-1}),\qquad D^2u=O(|x|^{-2}).$$ On the cutoff annulus, therefore, $$D^2(\eta_Ru)=O(R^{-2}\log R)$$. For fixed $$t$$ and sufficiently large $$R$$, this lies in the determinant branch of $$L$$, whose integral over that annulus is $$O(R^{-2}\log^2R)=o(1)$$. Also $$L(D^2u/t)=O(|x|^{-4})$$ at infinity, so it is integrable. Letting $$R\to\infty$$ proves the claim.
Integrating the pointwise inequality now gives $$0\le\int L(D^2u/t) \le\frac2t\|f\|_1-\bigl|\{|f|+|Bf|>t\}\bigr|.$$ Approximation in $$L^1$$, using the weak-type continuity of $$B$$, extends this estimate to every real $$f\in L^1$$. By the identities in Section 2.3, $$|\mathcal H_2f|_F =\frac{\sqrt{|f|^2+|Bf|^2}}{\sqrt2} \le\frac{|f|+|Bf|}{\sqrt2}.$$ Taking $$t=\sqrt2\lambda$$ yields $$\lambda\bigl|\{|\mathcal H_2f|_F>\lambda\}\bigr| \le\sqrt2\|f\|_1.$$ Conversely, $$|\mathcal H_2f|_F\ge |Bf|/\sqrt2$$, so the sharp constant $$2$$ for $$B$$ on real inputs [4, Corollary 1.2] forces $$\|\mathcal H_2\|_{1\to1,\infty}\ge\sqrt2$$. Finally, $$|\mathcal H_{0,2}f|_F=|Bf|/\sqrt2$$ gives the same sharp constant for the traceless transform. ◻
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