Abstract
Let $$c_2$$ be the least weak-type $$(1,1)$$ constant for the centered Hardy–Littlewood maximal operator over axis-parallel squares in $$\mathbb R^2$$. A classical covering argument gives $$c_2\le4$$. We describe an AI-assisted construction, prompted by the human insight to adapt the partial-balayage method of Spector and Stockdale [1], that proves $$c_2<4$$.
1 Introduction
For $$f\in L^1(\mathbb R^2)$$ define the centered square maximal operator
The classical covering argument gives $$c_2\le4$$; see Tao [4]. This article aims at explaining the main idea that leads to $$c_2<4$$. The motivating human insight was to adapt the partial-balayage method used by Spector and Stockdale in a different maximal-function problem [1].
Let $$D=\{(u,v):|u|+|v|\le1\}$$ be the unit diamond, of area $$2$$, and define
The linear map $$(u,v)\mapsto(u+v,u-v)$$ carries $$D$$ onto $$[-1,1]^2$$ and has determinant of absolute value $$2$$. The corresponding change of variables preserves the optimal weak-type $$(1,1)$$ constant, so it suffices to bound $$M_D$$. We do this by dominating $$M_D$$ with a suitable convolution maximal operator and proving a weak-type estimate for that operator.
2 Kernel comparison
For $$0<\alpha<2$$, let $$A_\alpha=-|D_u|^\alpha-|D_v|^\alpha$$, with Fourier symbol $$-(|\xi_1|^\alpha+|\xi_2|^\alpha)$$. For an integrable kernel $$K$$, the distribution $$A_\alpha K$$ is defined by duality:
Thus $$A_\alpha K\ge0$$ on $$\mathbb R^2\setminus\{0\}$$ means this pairing is nonnegative for every nonnegative smooth test function supported away from the origin.
Theorem 1. Let $$K\in L^1(\mathbb R^2)$$ be even. Suppose $$K\ge\mathbf1_D$$ a.e. and $$A_\alpha K\ge0$$ on $$\mathbb R^2\setminus\{0\}$$. Then $$c_2\le\frac12\int_{\mathbb R^2}K$$.
Proof. The proof has three steps, and the proof of each step is shown in the lemmas below.
Step 1. Show that $$M_Dg$$ is dominated by $$\frac12\sup_{s>0}K_s*|g|$$, where $$K_s(x)=s^{-2}K(x/s)$$.
Step 2. For a smooth source $$f\ge0$$ and cap $$\kappa$$, we construct a potential $$u$$ and a capped density $$\sigma$$ with $$A_\alpha u=\sigma-f$$. This is the partial balayage approach used in [1].
Step 3. Use the result in Step 2 to establish a weak-type estimate for the operator $$\frac12\sup_{s>0}K_s*(\cdot)$$.
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Lemma 1. If $$K\ge\mathbf1_D$$ a.e., then $$M_Dg\le\frac12\sup_{s>0}K_s*|g|$$.
Proof. Since $$K\ge\mathbf1_D$$, we have $$K_s\ge s^{-2}\mathbf1_{sD}$$. For every $$z\in\mathbb R^2$$ and $$s>0$$, the substitution $$y=z-w$$ and the symmetry of $$D$$ give
Taking the supremum over $$s>0$$ proves the claim.
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Lemma 2 (Partial-balayage). Let $$f$$ be a nonnegative function in $$C_c^\infty(\mathbb R^2)$$ and $$\kappa$$ a positive constant. There exist nonnegative functions $$u$$ and $$\sigma$$ such that, with $$\Omega=\{u>0\}$$,
Proof. For a real-valued $$v\in L^1(\mathbb R^2)$$, define the Dirichlet energy, possibly equal to $$+\infty$$, by
$$\mathcal E$$ has the following bilinear representation for $$v,w\in L^1(\mathbb R^2)$$ of finite Dirichlet energy:
Here $$e_j$$ is the $$j$$th coordinate vector, and $$c_\alpha>0$$ is chosen so that $$c_\alpha\int_{\mathbb R}(1-\cos(s\tau))|s|^{-1-\alpha}\,ds=|\tau|^\alpha$$. To justify the formula, note first that $$v\in L^1$$ implies $$\widehat v$$ is bounded near the origin; finite energy controls $$\widehat v$$ away from the origin. Hence $$v\in L^2$$, and likewise $$w\in L^2$$. Plancherel and Tonelli, applied first with $$w=v$$, give the formula for $$\mathcal E(v,v)$$; polarization gives the displayed bilinear identity.
We will prove the desired result by finding a minimizer in $$\mathcal C$$ of
Step 1: coercivity and existence of a minimizer
For every admissible $$v$$, the fractional Sobolev inequality, followed by Hölder’s and Young’s inequalities, gives
where $$p=4/(2-\alpha)>2$$ and $$p'=4/(2+\alpha)$$; see [2]. Consequently,
Thus $$\inf J>-\infty$$, and a minimizing sequence $$(u_n)$$ is bounded in energy and in $$L^1$$. Interpolation also bounds its $$L^2$$ norm. Hence it is bounded in $$H^{\alpha/2}(\mathbb R^2)$$. Weak compactness and local Rellich compactness yield a subsequence converging weakly in $$H^{\alpha/2}(\mathbb R^2)$$, strongly in $$L^2_{\mathrm{loc}}$$, and almost everywhere to $$u$$. Since $$u_n\ge0$$, we have $$u\ge0$$ almost everywhere. Fatou’s lemma gives $$\int u\le\liminf_n\int u_n$$ and thus $$u\in L^1$$, and weak lower semicontinuity of the squared $$L^2$$ norm of $$(|\xi_1|^\alpha+|\xi_2|^\alpha)^{1/2}\widehat u_n$$ gives $$\mathcal E(u,u)\le\liminf_n\mathcal E(u_n,u_n)$$. Finally, $$\int fu_n\to\int fu$$ by local $$L^2$$ convergence and compact support of $$f$$. Thus $$J(u)\le\liminf_n J(u_n)=\inf J$$, proving that $$u$$ is a minimizer.
Step 2: derivation of the first variational inequality
Fix $$v\in\mathcal C$$, put $$w=v-u$$, and set $$u_t=u+tw=(1-t)u+tv$$ for $$0\le t\le1$$. Convexity of $$\mathcal C$$ makes $$u_t$$ admissible. Minimality gives $$J(u_t)-J(u)\ge0$$. Therefore, for $$t>0$$,
Letting $$t\downarrow0$$ proves the first variational inequality
Step 3: the Lewy–Stampacchia estimate
Taking $$v=u+\varphi$$ in (1), for $$\varphi\ge0$$, gives
Thus, $$A_\alpha u\le\kappa-f$$ as a distribution. We now aim to obtain a lower bound. Fix $$t>0$$ and put $$w=(t\varphi-u)_+$$. Positive-part truncation preserves integrability and finite energy, so $$v=(u-t\varphi)_+=u-t\varphi+w$$ is admissible. Equation (1) gives
Write $$h=t\varphi-u$$, so $$w=h_+$$. Applying the elementary inequality $$(a-b)(a_+-b_+)\ge(a_+-b_+)^2$$ to the bilinear representation of $$\mathcal E$$ gives
Using this truncation inequality and Cauchy–Schwarz, we obtain
Since $$0\le t\varphi-w\le t\varphi$$ and $$f\ge0$$, we also have
Substituting these bounds into the variational inequality above gives
Dividing by $$t>0$$ yields $$\mathcal E(u,\varphi)\le\int f\varphi+\frac t4\mathcal E(\varphi,\varphi)$$. Letting $$t\downarrow0$$ therefore proves $$\mathcal E(u,\varphi)\le\int f\varphi$$. Together, the two bounds prove
Define $$\sigma=f+A_\alpha u$$, which is a distribution. By (2), it is positive and bounded above by the distribution induced by the constant function $$\kappa$$. A positive distribution is a nonnegative Radon measure; the bound $$\sigma\le\kappa\,dx$$ then makes it absolutely continuous. Its Radon–Nikodym density, also denoted by $$\sigma$$, satisfies $$0\le\sigma\le\kappa$$ almost everywhere. Thus $$A_\alpha u=\sigma-f$$ holds distributionally with a bounded function on the right.
Step 4: equality on the positivity set
Taking $$v=0$$ and $$v=2u$$ in (1) gives opposite inequalities, hence
The identity $$A_\alpha u=\sigma-f$$ also gives $$\mathcal E(u,u)=\int u(f-\sigma)$$. To justify this test, approximate $$u$$ by smooth compactly supported functions in both $$L^1$$ and the energy norm, using cutoffs and convolution; the pairings converge because $$f-\sigma$$ is bounded. Subtracting the identities yields
The integrand is nonnegative, so $$\sigma=\kappa$$ almost everywhere on $$\Omega=\{u>0\}$$.
Step 5: conservation of mass
Choose a smooth radial cutoff $$0\le\chi\le1$$, equal to one on the unit ball, compactly supported, and nonincreasing in the radius. Then $$\chi_R(x)=\chi(x/R)\uparrow1$$, and homogeneity gives $$\|A_\alpha\chi_R\|_\infty=R^{-\alpha}\|A_\alpha\chi\|_\infty$$. Testing $$A_\alpha u=\sigma-f$$ against $$\chi_R$$ gives
Monotone convergence, using $$\sigma\ge0$$, proves both its integrability and $$\int\sigma=\int f$$. Finally, Step 4 and conservation of mass yield the area estimate:
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Lemma 3 (Kernel transfer from the obstacle). Under the hypotheses of Theorem 1, for every nonnegative $$f\in C_c^\infty(\mathbb{R}^2)$$ and $$\lambda>0$$,
Proof. Choose $$\kappa=\lambda/\int_{\mathbb{R}^2}K$$. The denominator is positive since $$K\ge\mathbf{1}_D$$. Lemma 2 supplies $$u,\sigma$$ and $$\Omega=\{u>0\}$$ with
We will prove that $$K_s*f\le\lambda$$ almost everywhere outside $$\Omega$$ for every scale, and then bound the superlevel set by $$|\Omega|$$.
Step 1: the positive minimum principle for $$B$$
Define the operator $$B$$ on smooth compactly supported functions by $$B\varphi=K*(A_\alpha\varphi)$$. Its distributional convolution kernel is $$A_\alpha K$$. We first prove that $$B$$ satisfies the positive-minimum principle. That is, we assume $$\varphi\in C_c^\infty(\mathbb{R}^2)$$, $$\varphi\ge0$$, and $$\varphi(x_0)=0$$, and we want to prove that $$B\varphi(x_0)\ge0$$. Since convolution and the Fourier multiplier commute with translations, we may assume without loss of generality that $$x_0=0$$. Evenness of $$K$$ gives $$B\varphi(0)=\langle A_\alpha K,\varphi\rangle$$. The assumed sign of $$A_\alpha K$$ applies only to tests supported away from zero. To remove this restriction, choose $$\eta\in C_c^\infty(\mathbb{R}^2)$$ with $$0\le\eta\le1$$ and $$\eta=1$$ near zero, and put $$g_\varepsilon(x)=\eta(x/\varepsilon)\varphi(x)$$. Since zero is a minimum of $$\varphi$$, we have $$\nabla\varphi(0)=0$$. Taylor’s theorem gives $$|\varphi(x)|\le C|x|^2$$ and $$|\nabla\varphi(x)|\le C|x|$$ near zero. The product rule therefore yields
$$A_\alpha g_\varepsilon$$ has the following integral representation:
where $$c_\alpha$$ is a positive constant. The numerator is bounded by $$Ct^2$$ for $$|t|\le\varepsilon$$, using the second-derivative bound, and by $$C\varepsilon^2$$ for $$|t|>\varepsilon$$, using the bound on $$g_\varepsilon$$. Hence
Both integrals are finite because $$0<\alpha<2$$. Since $$K\in L^1$$,
The function $$\varphi-g_\varepsilon=(1-\eta(\cdot/\varepsilon))\varphi$$ is nonnegative and vanishes near zero. Therefore $$\langle A_\alpha K,\varphi-g_\varepsilon\rangle\ge0$$, and passage to the limit proves $$B\varphi(0)\ge0$$.
Step 2: application of the Courrège representation theorem
We now show that $$B$$ also satisfies the positive-maximum principle. This means that if $$\varphi\in C_c^\infty(\mathbb{R}^2)$$ attains a nonnegative global maximum at $$x$$, then $$B\varphi(x)\le0$$. Let this maximum be $$m$$. Choose a smooth cutoff $$0\le\chi_R\le1$$ that is equal to one on a ball containing both $$\operatorname{supp}\varphi$$ and $$x$$. Then
The positive-minimum principle proved in Step 1, applied at the point $$x$$, gives $$B\varphi(x)\le mB\chi_R(x)$$. Writing $$\chi_R(y)=\chi(y/R)$$, homogeneity of $$A_\alpha$$ gives
Letting $$R\to\infty$$ proves $$B\varphi(x)\le0$$.
Since $$B$$ is translation invariant and satisfies the positive-maximum principle, the Courrège representation theorem applies [3]:
where $$Q$$ is positive semidefinite, $$c\ge0$$, and $$\nu$$ is a nonnegative measure satisfying
We now remove the terms in the representation one at a time.
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Zeroth-order term. Use the cutoffs $$\chi_R(x)=\chi(x/R)$$. They equal one near the origin, so $$\chi_R(0)=1$$, $$\nabla\chi_R(0)=0$$, and $$D^2\chi_R(0)=0$$. Both derivative terms vanish at $$x=0$$, without using any symmetry of $$\nu$$. The representation there reads
$$ B\chi_R(0)=-c+\int_{\mathbb{R}^2\setminus\{0\}}(\chi_R(z)-1)\,\nu(dz). $$The integral vanishes on $$|z|\le R$$ and its absolute value is at most $$\nu(\{|z|>R\})$$, which tends to zero because $$\nu$$ has finite mass outside the unit ball. The estimate $$\|B\chi_R\|_\infty\to0$$ proved above now gives $$c=0$$.
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First-order term and symmetry of the measure. Let $$(R\varphi)(x)=\varphi(-x)$$. Evenness of $$K$$ and of $$A_\alpha$$ gives $$RBR=B$$. Reflecting the displayed Courrège formula replaces $$b$$ by $$-b$$ and $$\nu$$ by its reflected measure $$\nu^-$$, defined by $$\nu^-(E)=\nu(-E)$$ for every Borel set $$E$$. Since $$B=RBR$$, we may average these two formulas. Their first-order terms $$b\cdot\nabla\varphi$$ and $$-b\cdot\nabla\varphi$$ cancel. The measure becomes $$(\nu+\nu^-)/2$$; we continue to denote it by $$\nu$$. Thus $$\nu$$ is symmetric: $$\nu(E)=\nu(-E)$$ for every Borel set $$E$$.
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Second-order term. The multiplier of $$-B$$ is $$(|\xi_1|^\alpha+|\xi_2|^\alpha)\widehat K(\xi)$$, whose absolute value is at most $$(|\xi_1|^\alpha+|\xi_2|^\alpha)\left(\int_{\mathbb{R}^2}K\right)$$. The representation, with $$c=0$$ and symmetric $$\nu$$, gives
$$ \begin{aligned} \xi^TQ\xi+\int_{\mathbb{R}^2\setminus\{0\}}(1-\cos(\xi\cdot z))\,\nu(dz) &=(|\xi_1|^\alpha+|\xi_2|^\alpha)\widehat K(\xi). \end{aligned} $$Both terms on the left are nonnegative. If $$Q\ne0$$, choose a unit vector $$e$$ with $$e^TQe>0$$ and set $$\xi=re$$. We would obtain
$$ r^2e^TQe\le\left(\int_{\mathbb{R}^2}K\right)r^\alpha(|e_1|^\alpha+|e_2|^\alpha), $$which is impossible as $$r\to\infty$$ because $$\alpha<2$$. Hence $$Q=0$$.
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Cancellation of the gradient term inside the integral. Fix $$x$$. Since $$\nu(E)=\nu(-E)$$, the substitution $$z\mapsto-z$$ in the integral gives
$$ \begin{aligned} B\varphi(x)&=\int\bigl[\varphi(x+z)-\varphi(x)-\mathbf{1}_{\{|z|<1\}}z\cdot\nabla\varphi(x)\bigr]\,\nu(dz)\\ &=\int\bigl[\varphi(x-z)-\varphi(x)+\mathbf{1}_{\{|z|<1\}}z\cdot\nabla\varphi(x)\bigr]\,\nu(dz). \end{aligned} $$Averaging these two equal expressions cancels the gradient terms inside the integrand and yields
$$ B\varphi(x)=\frac12\int_{\mathbb{R}^2\setminus\{0\}} \bigl(\varphi(x+z)+\varphi(x-z)-2\varphi(x)\bigr)\,\nu(dz). $$
For $$v,w\in C_c^\infty(\mathbb{R}^2)$$, define $$\mathcal E_B(v,w):=-\int_{\mathbb{R}^2}vBw$$. Insert the preceding formula for $$Bw$$ and use the change of variables formula. This gives
This formula also shows $$\mathcal E_B(v,w)=\mathcal E_B(w,v)$$.
As in Lemma 2, apply $$(a-b)(a_+-b_+)\ge(a_+-b_+)^2$$ inside this integral to obtain
It remains to check that the potential $$u$$ belongs to this energy domain. Plancherel’s identity and the Fourier multiplier above give
The Courrège representation makes this integral nonnegative. Moreover, $$\widehat K(\xi)\le\int_{\mathbb{R}^2}K$$. Therefore
Lemma 2 gives $$\mathcal E(u,u)<\infty$$, so $$u$$ has finite $$\mathcal E_B$$ energy. This fact is needed in Step 3: it ensures that the truncation $$(\varepsilon\varphi-u)_+$$ also has finite $$\mathcal E_B$$ energy and that every pairing in the truncation argument is well defined.
Step 3: bound the convolution outside $$\Omega$$
Convolving the obstacle equation with $$K$$ gives
in distributions. The right-hand side is bounded because $$K\in L^1$$ and $$\sigma,f\in L^\infty$$, so $$Bu$$ is represented by a bounded function. We show that this function is nonnegative almost everywhere on $$\Omega^c=\{u=0\}$$.
Fix a nonnegative smooth compactly supported $$\varphi$$ and put $$w=(\varepsilon\varphi-u)_+$$, where $$\varepsilon>0$$. The truncation inequality gives
Dividing this truncation inequality by $$\varepsilon$$ now gives
Moreover,
Dominated convergence therefore gives $$\int_{\Omega^c}(Bu)\varphi\ge0$$. Since this holds for every nonnegative test function, $$Bu\ge0$$ almost everywhere on $$\Omega^c$$. On that set,
Step 4: take the supremum over scales
Every $$K_s$$ is even and nonnegative, has integral $$\int_{\mathbb{R}^2}K$$, and satisfies $$A_\alpha K_s\ge0$$ away from zero by homogeneity. Steps 1–3 therefore apply to $$K_s$$ with the same $$u,\sigma,\Omega$$, giving $$K_s*f\le\lambda$$ almost everywhere on $$\Omega^c$$ for each fixed $$s>0$$.
For each positive rational scale $$q$$, the preceding argument gives $$(K_q*f)(x)\le\lambda$$ for almost every $$x\in\Omega^c$$. For each positive rational $$q$$, we denote its corresponding exceptional null set by $$N_q$$. Let $$N:=\bigcup_{q\in\mathbb{Q},q>0}N_q$$. On $$\Omega^c\setminus N$$, the inequality holds for every rational $$q>0$$. For each $$x$$, $$(K_s*f)(x)$$ is continuous in $$s>0$$, which can be proved by dominated convergence.
Fix $$x\in\Omega^c\cap N^c$$. Approximating any real $$s>0$$ by positive rational scales and using continuity gives
Thus the superlevel set of the supremum is contained in $$\Omega\cup N$$. Consequently,
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3 Optimization formulation and candidate kernels
Theorem 2. $$c_2<4$$.
By Theorem 1, to prove Theorem 2 it is enough to consider
and to prove that $$C_\alpha<4$$ for some $$\alpha\in(0,2)$$. There are two natural approaches. The first is to choose a specific value of $$\alpha$$, construct a corresponding kernel $$K$$, and compute $$\frac12\int_{\mathbb R^2}K$$. This is the approach taken by the AI agents, which produced the following two concrete examples:
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Let $$\alpha=1$$, $$r=|u|+|v|$$, $$R=3797/2000$$, and $$\varepsilon=159411/200000$$. Define
$$K(u,v)=\frac{(R/r-1)_+}{R-1}-\varepsilon(1-r)_+^2,\qquad r>0.$$The first candidate has been verified in Lean:
$$\frac12\int_{\mathbb R^2}K =\frac{R^2}{R-1}-\frac{\varepsilon}{6}<3.878624.$$Consequently, $$c_2<3.878624$$. The Lean code is available on GitHub, and the certificate is registered on Palomar.
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Let $$\alpha=6/5$$, $$R=7/4$$, and $$r=|u|+|v|$$. Define the cubic cardinal spline
$$\beta(x)=\frac16\sum_{q=0}^4(-1)^q\binom4q(x+2-q)_+^3.$$The kernel is
$$\begin{gathered} K(u,v)=a(r^{-6/5}-R^{-6/5})_+ +\sum_{i,j}C_{ij}\,\beta(16u-i)\beta(16v-j),\\ a=\frac{11248245541992991}{6250000000000000}. \end{gathered}$$The AI-assisted computation gives
$$\begin{aligned} \frac12\int_{\mathbb R^2}K &=\frac{33744736625978973}{12500000000000000}\left(\frac74\right)^{4/5} -\frac{3114374068343977777}{5120000000000000000}\\ &<3.615749. \end{aligned}$$This result has not been verified in Lean because the computation is too large: checking it would require excessive CPU time in the Lean kernel.
The second, more ambitious approach is to determine the exact theoretical infimum $$\inf_{0<\alpha<2}C_\alpha$$. The AI-assisted search has not yet made substantial progress on this harder problem.
References
D. Spector and C. B. Stockdale, The weak-type $$(1,1)$$ bound for the Hardy–Littlewood maximal function is $$O(\sqrt n\log n)$$, arXiv:2609.05377 (2026).
E. Di Nezza, G. Palatucci, and E. Valdinoci, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math. 136 (2012), 521–573. arXiv:1104.4345.
P. Courrège, Générateur infinitésimal d’un semi-groupe de convolution sur $$\mathbb R^n$$, et formule de Lévy-Khinchine, Bull. Sci. Math. (2) 88 (1964), 3–30.
T. Tao, 245A, Notes 5: Differentiation theorems, 2010.