Abstract

We study an elliptic operator $L:=\mathrm{div}(A\nabla \cdot)$ on the upper half space. It is known that solvability of the Regularity problem in $\dot{W}^{1,p}$ implies solvability of the adjoint Dirichlet problem in $L^{p'}$. Previously, Shen ([1], 2007) established a partial reverse result. In our work, we show that if we assume an $L^1$-Carleson condition on only $|\partial_t A|$ the full reverse direction holds. As a result, we obtain equivalence between solvability of the Dirichlet problem $(D)^*_{p'}$ and the Regularity problem $(R)_p$ under this condition.

As a further consequence, we can extend the class of operators for which the $L^p$ Regularity problem is solvable by operators satisfying the mixed $L^1-L^\infty$ condition introduced in [2]. Additionally in the case of the upper half plane, this class includes operators satisfying this $L^1$-Carleson condition on $|\partial_t A|$.

1 Introduction↩︎

In this work, let $\Omega:=\mathbb{R}^{n+1}_+:=\mathbb{R}^n\times (0,\infty), n\geq 1$ be the upper half space and set $L:=\mathop{\mathrm{div}}(A\nabla\cdot)$ as a uniformly elliptic operator with bounded measurable coefficients. More specifically, $A(x,t)$ is a real not necessarily symmetric $n+1$ by $n+1$ matrix and there exists $\lambda_0>0$ such that \[\begin{align} \lambda_0 |\xi|^2\leq \xi^T A(x,t) \xi \leq \lambda_0^{-1}|\xi|^2 \qquad \textrm{ for all }\xi\in \mathbb{R}^{n+1},\label{eq:DefinitionOfUniformElliptic} \end{align}\tag{1}\] and a.e. $(x,t)=(x_1,...,x_n,t)\in \mathbb{R}^{n+1}_+$. We are interested in the solvability of the Regularity boundary value problem given by \[\begin{cases} Lu=\mathop{\mathrm{div}}(A\nabla u)=0 &\textrm{in }\Omega, \\ \nabla_Tu=\nabla_Tf &\textrm{on }\partial\Omega,\end{cases}\] with boundary data $f$ in the homogeneous Sobolev space $\dot{W}^{1,p}(\partial\Omega)$ (see Definition 1). In contrast to the Dirichlet boundary value problem, where the boundary data belongs to $L^p$ (see Definition 1), the Regularity problem focuses on the behavior of the tangential derivative ($\nabla_T$) of the solution $u$ on the boundary.

The question we would like to address is quite well established and dates back to Dahlberg (cf. [3] and [4]). He showed the existence of solutions to the Dirichlet boundary value problem with boundary data in $L^2$ for the Laplacian on Lipschitz graph domains. Notably, he made the following observation: Finding harmonic functions with $L^2$ boundary data on a Lipschitz graph domain $\Omega$ for \[\begin{cases} \Delta u=0 &\textrm{in }\Omega\\ u=f &\textrm{on } \partial\Omega\end{cases}\qquad \textrm{is equivalent to solving } \qquad \begin{cases} Lu=0 &\textrm{in }\mathbb{R}^n\times(0,\infty)\\ u= f &\textrm{on } \mathbb{R}^n\end{cases}\] for an elliptic operator $L$ depending on the Lipschitz function of $\Omega$. This suggests that solving a simpler elliptic PDE on a more complicated Lipschitz graph domain is equivalent to solving a more complicated elliptic PDE on the simpler domain of the upper half space. Depending on which flattening one uses in the argument, the operator $L$ has some additional properties. Among the most established ones are the Carleson condition (or also Dahlberg-Kenig-Pipher (DKP) condition) and the $t-$independence condition. The literature regarding the DKP condition is extensive, addressing not only the Dirichlet boundary value problem but also a range of different elliptic or parabolic boundary value problems (cf. the survey article [5] and references therein). Similarly, the $t-$independent condition yields solvability of various other boundary value problems (cf. [6], [7], [8], [8], [9], [10], [11], [12]). Broadly speaking, the DKP condition controls the Lipschitz constant of the matrix $A$ close to the boundary and requires it to grow less than the function $1/t$ close to the boundary. The $t-$independence condition on the other hand asserts $A(x,t)=A(x)$, i.e. that $A$ is independent in transversal direction with potentially almost arbitrarily bad behavior in any direction parallel to the boundary.

Both of these structurally different conditions - the DKP and $t$-independent condition - are sufficient for solvability of the Dirichlet and Regularity boundary value problem. This raises the question of whether there are other sufficient conditions or improvements of these two that still allow us to solve the Dirichlet or Regularity boundary value problem. Finding such improvements would reduce the number of operators for which we do not know whether any boundary value problem is solvable. It is also noteworthy that [13] and [14] provide examples of linear uniformly elliptic operators for which the Dirichlet boundary value problem is not solvable. Furthermore, the article [6] demonstrates that if the matrix $A$ is nonsymmetric, we cannot expect to obtain $L^2$ solvability for the Dirichlet problem with $t-$independent coefficients. In these cases, the best we can hope for is solvability with $L^p$ data for potentially large $p$. Given that the solvability range of the regularity problem is dual to that of the Dirichlet problem, we can only expect solvability for the Regularity problem for potentially small $p>1$ in our context.

Although the $t-$independent condition for the Dirichlet problem in the symmetric case was well established by Jerison and Kenig via a “Rellich" identity in [15], the extension from symmetric to nonsymmetric matrices in the Dirichlet and Regularity problem necessitated new tools and took some time. It was only after the Kato conjecture was resovled (cf. [16]) that the Dirichlet boundary value problem (cf. [17]) and Regularity boundary value problem (cf. [8]) could be proved for nonsymmetric matrices under the $t-$independence condition and by use of these Kato tools.

In addition to the two previously mentioned conditions, there is also the Dini-condition from [18], where the authors showed that $t$-independence can be relaxed if we have continuous coefficients. More precisely, they assume a symmetric $A$ with continuous coefficients, a bounded $C^1$-domain $\Omega$, and that the modulus of continuity \[\eta(s)=\sup_{P\in \partial\Omega, 0<r<s}|A_{ij}(P-rV(P))-A_{ij}(P)|\] with outer normal vector field $V$ satisfies the Dini-type condition \[\begin{align} \int_0\frac{\eta(s)^2}{s}ds<\infty.\label{DiniTypeCond} \end{align}\tag{2}\] Under these assumptions, they show $\omega\in B_2(\sigma)\subset A_\infty(\sigma)$, i.e. solvability of the $L^2$ Dirichlet problem. This Dini-type condition has also been shown to be somewhat necessary in the following sense: For a given function $\beta$ with $\int_0 \frac{\beta(s)^2}{s}ds=+\infty$ the article [14] constructs symmetric continuous matrices with modulus $\eta(s)\leq \beta(s)$ which admit elliptic measures $\omega$ that are completely singular with respect to the surface measure. Later, [19] extended condition 2 to include also symmetric matrices with merely bounded and measurable coefficients by examining perturbations from continuous matrices. Lastly, [20] demonstrates that a sufficiently small $L^\infty$ perturbation of a symmetric $t$-independent matrix still allows the resulting operator to solve the Dirichlet problem. This condition is also a generalization of the $t-$independence condition and even applies for complex coefficients, although it is still limited to symmetric matrices. All of these conditions were studied in the context of the Dirichlet but not the regularity problem.

Another condition that was studied in [2] is the mixed $L^1-L^\infty$ condition which states that \[\begin{align} \int_0^\infty \Vert \partial_t A(\cdot, t)\Vert_{L^\infty}dt<\infty. \label{cond:mixedL1LInftyCond} \end{align}\tag{3}\] From [2] we know that if $|\partial_t A|\leq C/t$ and 3 is satisfied, then the Dirichlet problem is solvable for some $1<p<\infty$. It is clear that this condition generalizes the $t-$independence condition, but it differs from both the Dini-condition and the condition in [20]. Even if the Dini-condition (or the condition in [20]) serves as sufficient condition for unbounded domains, for nonsymmetric matrices and for the Regularity problem - which is not established yet - there are examples of matrices in [2] that satisfy either condition but not the other. Interestingly, for $n=1$, the same methods also lead to an improvement. In the upper half plane, we only need to assume an $L^1$ Carleson condition on $|\partial_t A|$ instead of 3 (cf. [2]).

Since these new conditions ensure solvability for the Dirichlet problem, it is natural to ask whether they also enable us to solve the Regularity problem. We will derive this result as a consequence of our main theorem.

To determine the solvability of either boundary value problem, it can be useful to analyze their relationships. In particular, [21] established that on Lipschitz domains solvability of the Regularity problem $(R)_p$ with data in $\dot{W}^{1,p}$ implies solvability of the Dirichlet problem $(D)^*_{p'}$ of the adjoint operator with boundary data in $L^{p'}$ for the dual exponent (see also [22] for the endpoint $p=1$). While their result is only formulated for symmetric operators, it is well established that this proof also applies for nonsymmetric operators. In that sense we can say that the Dirichlet problem is generally easier to solve than the Regularity problem. Interestingly, there is a partial reverse result from [1] which is also stated for symmetric operators only but also holds for nonsymmetric ones. It states that if the Dirichlet problem $(D)^*_{p'}$ is solvable then either the Regularity problem $(R)_p$ is solvable or $(R)_q$ is not solvable for any $1<q<\infty$. Which of the two cases is present though depends on the given PDE. For instance, we know that under the DKP condition or the $t-$independence condition we are in the second case of solvability of both problems in dual ranges of $p$ and we have equivalence between solvability of the Dirichlet problem $(D)^*_{p'}$ and the regularity problem $(R)_{p}$. The main result of this work is that this reverse implication also holds if we assume the $L^1$ Carleson condition on $|\partial_t A|$, as detailed in 7 .

Theorem 1. Assume $L:=\mathrm{div}(A\nabla \cdot)$ is a uniformly elliptic operator with bounded, merely measurable coefficients and let $\Omega=\mathbb{R}^{n+1}_+$. Let $p>1$. If the $L^{p'}$ Dirichlet problem is solvable for the adjoint $L^*$, there exists $C>0$ such that $|\partial_t A|\leq C/t$ and \[\begin{align} \sup_{(y,s)\in B(x,t,t/2)}|\partial_t A|\qquad \textrm{ is a Carleson measure},\label{cond:L1CarlesonCondOnPartialtA} \end{align}\qquad{(1)}\] then the Regularity boundary value problem is solvable for $f\in \dot{W}^{1,p}(\partial\Omega)$.

Since [2] established solvability of the Dirichlet problem for some $1<p<\infty$ under the $L^1-L^\infty$ condition 3 for all dimensions $n\geq 2$ and under the $L^1$ Carleson condition on $|\partial_t A|$ for $n=1$, we can conclude solvability of the Regularity problem as presented in the following two corollaries.

Corollary 1. Assume $L:=\mathrm{div}(A\nabla \cdot)$ is a uniformly elliptic operator with bounded, merely measurable coefficients and let $\Omega=\mathbb{R}^{n+1}_+$. If there exists $C>0$ such that $A$ satisfies $|\partial_t A|\leq C/t$ and 3 , then there exists $p>1$ such that 10 holds and hence the Regularity boundary value problem is solvable for $f\in \dot{W}^{1,p}(\partial\Omega)$.

Corollary 1. Assume $L:=\mathrm{div}(A\nabla \cdot)$ is a uniformly elliptic operator with bounded, merely measurable coefficients and let $\Omega=\mathbb{R}^{2}_+$. If there exists $C>0$ such that $|\partial_t A|\leq C/t$ and \[\sup_{(y,s)\in B(x,t,t/2)}|\partial_t A|\qquad \textrm{ is a Carleson measure}\] then there exists $p>1$ such that 10 holds and hence the Regularity boundary value problem is solvable for $f\in \dot{W}^{1,p}(\partial\Omega)$.

Since the mixed $L^1-L^\infty$ condition 3 implies the $t-$independent condition, Corollary 1 is also an alternative proof of solvability of the Regularity problem for $t$-independent operators. This result has been proved in [8] and our proof only relies on integration by parts techniques and all the tools the resolved Kato conjecture provides including the semi group theory, but does not need single layer potentials like [8].

Overview over the proof↩︎

The main proof is provided in Section 3. This section demonstrates how the proof of Theorem 1 reduces to demonstrating area function bounds for the approximation operator $\mathcal{P}_tf:=e^{-t^2L^t_{||}}f$. Specifically, for $p>1$, which is derived from the solvability of the Dirichlet problem, we need to establish $L^p$ norm bounds of the area functions of $\nabla_{||}\mathcal{P}_tf, \partial_t\mathcal{P}f$, and $t\nabla_{||}\partial_t\mathcal{P}_tf$ in terms of the $L^p$ norm of the gradient of the boundary data $f$.

Before we continue, we would like to point out that, if we choose the matrix $A$ to be $t-$independent and we choose $\mathcal{P}_tf:=e^{-t^2L_{||}}$, we can read the main proof in Section 3 as an alternative version of the proof in [8] because almost all the necessary area function bounds in $L^p$ have already been established for this choice of $\mathcal{P}_t$. The exception is that - to the best of the author’s knowledge - the area function bounds on $\nabla_{||}\mathcal{P}_t$ have not appeared before and were circumvented by other estimates provided by the proof of the Kato conjecture.

In our case, however, establishing these $L^p$ area function bounds is the central component of this work. To do this, we are going to split $\partial_t\mathcal{P}_t$ into the sum of two different operators $W_1f$ and $W_2f$ and examine the area function expressions separately (see Section 2.2). Our goal is to apply the real interpolation method to get $L^p$ bounds for the area function expressions. This reduces matters to proving Hardy-Sobolev space to $L^1$ boundedness of the area functions and boundedness of the corresponding Carleson function in $L^\infty$. The real interpolation method and its application in our case are presented in Section 4. The remaining two sections now deal with proving all the remaining area functions bounds on $\nabla_{||} \mathcal{P}_t, W_1$ and $W_2$ respectively.

We begin in Section 5 by proving a pointwise bound of the Carleson function of $W_2$. This result implies the needed $L^\infty$ bound for the Carleson function, as well as an $L^\beta$ to $L^\beta$ bound for the area function of $W_2$ for $\beta>2$. This $L^\beta$ bound is then utilized to derive the Hardy-Sobolev to $L^1$ bound for the area function of $W_2$. With these bounds in hand, real interpolation yields all $L^p$ to $L^p$ bounds for the area function expressions involving $W_2$.

In Section 6, we turn our attention to the bounds for $\nabla_{||}\mathcal{P}_t$ and $W_1$ whose proofs require all the bounds established in the previous section. We will mostly deal with $\nabla_{||} \mathcal{P}_t$ and $W_1$ simultaneously, since their proofs are very similar and use the same tools. First, we establish $L^2$ to $L^2$ bounds of the area functions of $\nabla_{||}\mathcal{P}_t$ and $W_1$ which rely on the previously established $L^2$ to $L^2$ bound of the area function of $W_2$. Next, we prove the Carleson function bound for $\nabla_{||}\mathcal{P}_t$ and $W_1$, and the Hardy-Sobolev to $L^1$ bound of the area functions of $\nabla_{||}\mathcal{P}_t$ and $W_1$; both of these results depend on the $L^2$ to $L^2$ bound of the area function of $\nabla_{||}\mathcal{P}_t$ or $W_1$ respectively. Finally, we use real interpolation to establish the full range of $L^p$ to $L^p$ boundedness for the area function expressions involving $\nabla_{||}\mathcal{P}_t$ and $W_1$.

We provide this overview of all significant steps to highlight the dependencies between the proofs of these bounds. In particular, understanding the bounds for $W_1$ requires the full comprehension of the bounds on $W_2$ first.

Acknowledgements↩︎

The author wishes to express sincere gratitude to Jill Pipher and Martin Dindoš for their valuable and insightful discussions on this subject, which significantly contributed to the refinement of the applied methods

2 Preliminaries↩︎

2.1 Notation and Setting↩︎

We will work on the upper half space $\Omega=\mathbb{R}^{n+1}_+=\mathbb{R}^n\times (0,\infty)$, where we call the last component the $t$-direction, or $t-$component, and use the following definitions:

A boundary cube centered at $x\in\partial\Omega$ with radius $r$ is denoted by $\Delta_r(x)=\Delta(x,r):=B(x,r)\cap\partial\Omega$;
A Carleson region over a boundary ball is written as $T(\Delta_r(x)):=B(x,r)\cap\Omega$;
The surface measure is the $n$-dimensional Hausdorff measure restricted on the boundary of the domain and denoted by $\sigma:=\mathcal{H}^{n}|_{\partial\Omega}$;
The matrix $A(x,t)\in\mathbb{R}^{(n+1)\times(n+1)}$ is bounded and elliptic, i.e. there exists $\lambda>0$ with \[\lambda|\xi|^2\leq \xi^T A(x,t)\xi\leq |\xi|^2\qquad\textrm{ for every }\xi\in \mathbb{R}^{n+1}, \textrm{ and a.e. }(x,t)\in\Omega.\] Here $A$ is potentially nonsymmetric, and we set the different components as \[A(x,t)=\begin{pmatrix} A_{\Vert}(x,t) & b(x,t)\\c(x,t) & d(x,t)\end{pmatrix},\] where $A_{\Vert}(x,t)\in\mathbb{R}^{n\times n}, c(x,t)\in\mathbb{R}^{1\times n}, b(x,t)\in\mathbb{R}^{ n\times 1}, d(x,t)\in\mathbb{R}$. All of these components are functions in $(x,t)$ that area merely measurable. Then we call $L:=\mathrm{div}(A\nabla \cdot)$ and elliptic operator and $L^*:=\mathrm{div}(A^*\nabla \cdot)$ its adjont operator, where $A^*$ is the transpose of $A$. To formulate the $L^1$-Carelson condition ?? we also assume that each component is weakly differentiable in $t$ almost everywhere.
The set $\mathcal{D}_k(\Delta):=\{Q_l;l\}$ is a dyadic decomposition of a boundary ball $\Delta$ which consists of a family of boundary balls $Q_l\subset 3\Delta$ with size comparable to $2^{-k}$, with finite overlap and such that they cover $\Delta$, i.e. $\chi_\Delta\leq \sum_{l}\chi_{Q_l}\leq N\chi_{3\Delta}$ for some fixed $N$ independent of scale $k$;
We define the family of elliptic operators $(L^t_{||})_{t>0}=(\mathrm{div}_{||}(A_{\Vert}(\cdot,t)\nabla_{||} \cdot))_{t>0}$, where the subscript $||$ means that we take the gradient or divergence only with respect to the first $n$ components or the $x-$components only. If clear from context, we might also write $L^t=L_{||}^t$ dropping the subscript. Let us also note here that $L_{||}^t$ is an operator on functions $v:\mathbb{R}^n\to\mathbb{R}$;
A nontangential cone with aperture $\alpha>0$ is given by \[\Gamma_\alpha(x):=\{(y,t)\in\Omega; |x-y|<\alpha t\},\] and $\Gamma_\alpha^\tau(x):=\Gamma_\alpha(x)\cap\{(y,t)\in\Omega; t\leq \tau\}$ denotes the at height $\tau>0$ truncated cone;
The mean-valued nontangential maximal function is defined as \[\tilde{N}_\alpha^{(p)}(F)(x):=\sup_{(y,s)\in\Gamma_\alpha(x)}\Big(\fint_{B(y,s,s/2)}|F(z,\tau)|^pdzd\tau\Big)^{1/p},\] with $p\geq 1$. If we drop the superscript $p$ we mean $p=1$. Furthermore, we set the at height $\tau>0$ truncated version as \[\tilde{N}_\alpha^\tau(F)(x):=\sup_{(y,s)\in\Gamma^\tau_\alpha(x)}\fint_{B(y,s,s/2)}|F(z,\tau)|dzd\tau,\] and the away truncated version for $\tau>0$ as \[\begin{align} \tilde{N}_{\alpha,\tau}(F)(x):=\sup_{(y,s)\in\Gamma_\alpha(x)\cap\{(y,s)\in \Omega; s\geq \tau \}}\fint_{B(y,s,s/2)}|F(z,\tau)|dzd\tau; \label{def:NontangentialMaximalFucntionTruncatedAway} \end{align}\tag{4}\]
The area function of $F\in L^2_{loc}$ is set as \[\begin{align} \mathcal{A}_\alpha(F)(x):=\Big(\int_{\Gamma_\alpha(x)}\frac{|F(x,t)|^2}{t^{n+1}}dxdt\Big)^{1/2},\label{recall:areafunctiondef} \end{align}\tag{5}\] and the square function is set as $S(F)(x):=\mathcal{A}(tF)(x)$;
The Carleson function is defined by \[\begin{align} \mathcal{C}(f)(x):=\Big(\sup_{r>0}\frac{1}{|\Delta(x,r)|}\int_{T(\Delta(x,r))}\frac{|f(y,t)|^2}{t}dydt\Big)^{1/2}\label{recall:carelsonfunctiondef} \end{align}\tag{6}\] for $x\in \partial\Omega=\mathbb{R}^n$;
We say that a measure $\mu:\Omega\to [0,\infty]$ is a Carleson measure if there exists a $C>0$ such that for every boundary ball $\Delta$ \[\mu(T(\Delta))\leq C\sigma(\Delta).\] The smallest such constant $C$ is also called the Carleson norm of $\mu$ and denoted by $\Vert \mu\Vert_{\mathcal{C}}$. Hence the $L^1$-Carleson condition ?? means that \[\begin{align} \sup_{\Delta\subset\partial\Omega\textrm{ boundary ball}}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\sup_{B(x,t,t/2)}|\partial_t A| dxdt\leq C<\infty.\label{cond:ExplanationForCarlesonCondition} \end{align}\tag{7}\]

An important property of the Carleson measure is that for every function $F:\mathbb{R}^{n+1}_+\to\mathbb{R}$, with $\tilde{N}(F)\in L^1(d\sigma)$ we have \[\begin{align} \int_{\mathbb{R}^{n+1}_+}|F(x,t)|d\mu\leq C\Big\Vert \sup_{B(x,t,t/2)}|\partial_t A| dxdt\Big\Vert_\mathcal{C}\int_{\mathbb{R}^{n}}\tilde{N}(F)(x)dx.\label{eq:DualityCarelsonNormNontangential} \end{align}\tag{8}\] This property can be formulated also more generally and in an $L^p$-version and can be found for the nontangential maximal function in Proposition 3 in [23], or Proposition 3.11 or Corollary 3.12 in [24], and for the mean-valued nontangential maximal function in Lemmas A.18 and A.20 in [25].

Let us define the Dirichlet boundary value problem with boundary data in $L^p$.

Definition 1 ($(D)_p^L$). We say the $L^p$ Dirichlet boundary value problem is solvable for $L$ if for all boundary data for $f\in C^\infty_C(\Omega)\cap L^p(\partial\Omega)$ the unique solution $u\in W_{loc}^{1,2}(\Omega)$ of \[\begin{cases} Lu=0&\Omega, \\ u=f&\partial\Omega, \end{cases}\] satisfies \[\begin{align} \Vert \tilde{N}(u)\Vert_{L^p(\partial\Omega)}\lesssim\Vert f\Vert_{\dot{L}_{1}^p(\partial\Omega)},\label{eq:DPinequalityElliptic} \end{align}\tag{9}\] where the implied constants are independent of $u$ and $f$. In this case we also write that $(D)^L_p$ holds for $L$. For the adjoint operator we also write in short $(D)^{L^*}_p=(D^*)_p$.

Furthermore, we call a Borel function $g:\partial\Omega\to\mathbb{R}$ a Hajłasz upper gradient of $f:\partial\Omega\to\mathbb{R}$ if \[|f(X)-f(Y)|\leq |X-Y|(g(X)+g(Y))\qquad\textrm{for a.e. }X,Y\in \partial\Omega.\] We denote the collection of all Hajłasz upper gradients of $f$ as $\mathcal{D}(f)$ and define $\dot{L}_1^p(\partial\Omega)$ by all $f$ with \[\Vert f\Vert_{\dot{L}^p_1(\partial\Omega)}:= \inf_{g\in\mathcal{D}(f)}\Vert g\Vert_{L^p(\partial\Omega)}<\infty.\] This space is also called homogeneous Hajłasz Sobolev space. In the case of a flat boundary we have that $\dot{L}_1^p(\mathbb{R}^n)=\dot{W}^{1,p}(\mathbb{R}^n)$ with comparable norms (see [26]).

We can define the Regularity problem with boundary data in $\dot{L}_1^p(\partial\Omega)$ but in our case this could also be done equivalently with $\dot{W}^{1,p}$.

Definition 1 ($(R)_p^L$). We say the $L^p$ Regularity boundary value problem is solvable for $L$ if for all boundary data for $f\in C^\infty_C(\Omega)\cap \dot{L}^p_{1}(\partial\Omega)$ the unique solution $u\in W_{loc}^{1,2}(\Omega)$ of \[\begin{cases} Lu=0&\Omega, \\ u=f&\partial\Omega, \end{cases}\] satisfies \[\begin{align} \Vert \tilde{N}(\nabla u)\Vert_{L^p(\partial\Omega)}\lesssim\Vert f\Vert_{\dot{L}_{1}^p(\partial\Omega)},\label{eq:RPinequalityElliptic} \end{align}\tag{10}\] where the implied constants are independent of $u$ and $f$. In this case we also write that $(R)^L_p$ holds for $L$.

2.2 The approximation operator $\mathcal{P}_t$↩︎

Recall that $L^t:=L_{||}^t:=\mathrm{div}_{||}(A_{\Vert}(x,t)\nabla_{||}\cdot)$ is an elliptic operator on the right upper block $A_{\Vert}$ for each fixed $t>0$. For this family of operators we can define the approximation operator \[\mathcal{P}_t:=e^{-t^2L_{||}^t}.\] Without the dependence of the operator $L_{||}^t$ on $t$, this $\mathcal{P}_t$ is the ellipticized heat semigroup, and solves the heat equation on the upper half space if $t^2$ is replaced by $t$ . Since this operator does not satisfy any PDE directly, we can decouple the dependencies in $t$ and define \[Wf(x,t,s):=e^{-tL_{||}^s}f(x)\] as the solution to the ("t-independent") heat equation \[\begin{cases} \partial_t Wf(x,t,s)-L_{||}^s Wf(x,t,s)=0, &(x,t)\in \Omega, \\ Wf(x,0,s)=f(x), &x\in\partial\Omega\end{cases}\] for fixed $s>0$. Taking the partial derivative of $Wf$ in $s$ yields \[\begin{cases} \partial_t \partial_sWf(x,t,s)-L_{||}^s \partial_s Wf(x,t,s)=\mathrm{div}_{||}(\partial_s A_{\Vert}(x,s)\nabla_{||}Wf(x,t,s)), &(x,t)\in \Omega, \\ Wf(x,0,s)=0, &x\in\partial\Omega.\end{cases}\] By Duhamel’s principle, we can obtain an explicit formula for $\partial_s Wf(x,t,s)$. Hence we obtain \[\begin{align} \partial_t\mathcal{P}_tf(x)&=[\partial_tWf(x,t^2,s) + \partial_sWf(x,t^2,s)]|_{s=t} \\ &=2tL_{||}^te^{-t^2L_{||}^t}f(x) + \int_0^t2\tau e^{(t^2-\tau^2)L_{||}}\mathrm{div}(\partial_tA_{\Vert}(x,t)\nabla_{||}e^{-\tau^2L_{||}^t}f(x))d\tau \\ &=:W_1f(x,t) + W_2f(x,t). \end{align}\] This argument is similar to the argument presented in [2]. Please note that if the operator $L_{||}^t$ is independent of $t$, then $W_2f\equiv 0$.

When clear from context and when there is no chance of confusing the present derivative with the full one is, we will drop the subscript $||$.

2.3 Properties of the heat semi-group and $\mathcal{P}_t$↩︎

First, we would like to note the $L^2-L^2$ boundedness of the following operators.

Proposition 1 ([27]). Let $L_{||}:=\mathrm{div}_{||}(A_{\Vert}(x)\nabla_{||} \cdot)$ be a $t$-independent operator and let $f\in L^2(\mathbb{R}^n)$. Then for $T_t\in\{e^{-tL_{||}}, t\partial_te^{-tL_{||}}, \sqrt{t}\nabla_{||} e^{-tL_{||}}\}$ we have \[\begin{align} \Vert T_tf\Vert_{L^2}\leq C\Vert f\Vert_{L^2}, \end{align}\] where the implicit constant $C$ only depends on the ellipticity constant of $A_{\Vert}$. Furthermore, we also have for $f\in W^{1,2}(\mathbb{R}^n)$ \[\Vert \nabla_{||} e^{-tL_{||}}f\Vert_{L^2}\leq C\Vert \nabla_{||} f\Vert_{L^2}.\]

We have the following bounds of the kernel $K_t(x,y)$ of the semigroup $e^{tL_{||}}$.

Proposition 1 (Prop 4.3 in [27] or Theorem 6.17 in [28]). For any $l\in\mathbb{N}$ there exists $C=C(n,\lambda,l), \beta=\beta(n,\lambda)>0$ such that \[\begin{align} |\partial_t^l K_t(x,y)|\leq C_l t^{-\frac{n}{2}-l}e^{-\beta\frac{|x-y|^2}{t}} \end{align}\] for all $x,y\in\mathbb{R}^n$.

The kernel bounds give rise to the following local bounds.

Proposition 1 (Prop 11 in [12] and Cor. 5.6 in [2], proof of Lemma 6.4 in [2]). Let $x\in \partial\Omega=\mathbb{R}^n$ and $(y,t)\in\Gamma_\alpha(x)$, then there exists $C=C(n,\lambda,\alpha)>0$ such that

(i) \[\begin{align} W_1f(y,t)=tL_{||}e^{-t^2L_{||}^t}f(y,t)\leq C M[\nabla_{||}f](x); \end{align}\]

(ii) \[\begin{align} e^{-t^2L_{||}^t}(f-(f)_{\Delta_{\alpha t}(x)})(y,t)\leq C M[\nabla_{||}f](x); \end{align}\]

(iii) \[\begin{align} \fint_{\Delta_{\alpha t/2}(x)}|\nabla_{||} e^{-t^2L_{||}^t}f(z,\tau)|^2dz d\tau\leq C M[\nabla_{||}f]^2(x)\label{eq:NablaBounded}; \qquad\textrm{and} \end{align}\qquad{(2)}\]

(iv) \[\begin{align} \fint_{\Delta_{\alpha t/2}(x)}|s^2\nabla_{||} L_{||}e^{-t^2L_{||}^t}f(z,\tau)|^2dz d\tau\leq C M[\nabla_{||}f]^2(x)\label{eq:NablaLBounded}. \end{align}\qquad{(3)}\]

The proof of ?? is not provided here but works completely analogously to the proof of Lemma 6.4 in [2], if we replace the operator $e^{-t^2L_{||}}f$ by $tL_{||}e^{-t^2L_{||}}f$.

A direct consequence of Proposition 1 is the following:

Corollary 1. For $f\in W^{1,p}(\partial \Omega), p>1$, it holds that

(i) \[\Vert \tilde{N}^{(2)}(\nabla_{||}\mathcal{P}_sf)\Vert_{L^p(\partial\Omega)}, \Vert \tilde{N}(\nabla_{||}\mathcal{P}_sf)\Vert_{L^p(\partial\Omega)}\lesssim\Vert \nabla_{||} f\Vert_{L^p(\partial\Omega)},\textrm{ and}\]

(ii) \[\Vert \tilde{N}(W_1\mathcal{P}_sf)\Vert_{L^p(\partial\Omega)}\lesssim\Vert \nabla_{||} f\Vert_{L^p(\partial\Omega)}.\]

Furthermore, we have off-diagonal estimates

Proposition 1 (Off-diagonal estimates, Prop. 3.1 in [29]). For $T\in \{e^{-tL},\sqrt{t}e^{-tL},t\partial_te^{-tL}\}$ there exists $C,\alpha>0$ such that \[\Vert Th\Vert_{L^2(E)}\lesssim e^{-\frac{\alpha d(E,F)^2}{t}}\Vert h\Vert_{L^2(\mathbb{R}^n)}\] for all $h\in L^2(\mathbb{R}^n)$ with $\mathrm{supp}(h)\subset F$.

First, we note that since $W_1f$ and $W_2f$ satisfy some PDE, we have a Cacciopolli inequality and Cacciopolli-type inequality respectively.

Proposition 1 (Lemma 4.2 and Lemma 4.3 in [2]). Let $B(x,t,4r)\subset \Omega$, then \[\begin{align} \int_{B(x,t,r)} |\nabla_{||}W_1f(y,s)|^2 dyds\lesssim \frac{1}{r^2}\int_{B(x,t,2r)} |W_1f(y,s)|^2 dyds,\label{lemma:w95tSqFctBound1} \end{align}\qquad{(4)}\] and \[\begin{align} \fint_{B(x,t,r)} |\nabla_{||}W_2f(y,s)|^2 dyds&\lesssim \frac{1}{r^2}\fint_{B(x,t,2r)} |W_2f(y,s)|^2 dyds\nonumber \\ &\qquad + \Vert\partial_sA_{\Vert}(y,s)\Vert_{L^\infty(B(x,t,2r))}^2M[\nabla_{||}f]^2(x).\label{eq:secondsummand} \end{align}\qquad{(5)}\]

The statement of Lemma 4.3 in [2] is slightly different to above proposition. However, we can easily see that the second summand in ?? results from the trivial $L^\infty$ bound on $\partial_tA_{\Vert}$ and ?? , while we use smoothness of the semi group and the resulting Cacciopolli inequality on time slices for the first term of ?? .

3 Reduction of proof of Theorem 1 to area function bounds↩︎

We present the proof of Theorem 1 here. We are going to use certain area function bounds that we will collect and establish in the following sections.

To begin with, let $h$ be a quasi dualising function given by the following lemma.

Lemma 1 (Lemma 2.8 in [30]). There exists a function $h:\mathbb{R}^{n+1}_+\to\mathbb{R}^{n+1}_+$ with compact support in $\Omega=\mathbb{R}^{n+1}_+$ such that \[\Vert \tilde{N}(\nabla u)\Vert_{L^p(\partial\Omega)}\lesssim \int_\Omega \nabla u \cdot h dxdt,\] and \[\int_\Omega F \cdot h dxdt\lesssim \Vert \tilde{N}(F)\Vert_{L^p(\partial\Omega)}\] for every vector valued function $F:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$.

Now let $v\in W^{1,2}_{loc}(\Omega)$ be the solution to the Poisson-Dirichlet problem of the adjoint operator \[\begin{align} \begin{cases} L^*v=\mathrm{div}(h) &\textrm{in }\Omega,\\ v=0 &\textrm{on }\partial\Omega.\end{cases}\label{PDEForv} \end{align}\tag{11}\] Since we assume solvability of the Dirichlet problem for the adjoint $L^*$ and $1<p'<\infty$, we know that $\omega^*\in B_{p}(\sigma)$ holds under our assumptions, where $B_{p}(\sigma)$ is the reverse Hölder class. Hence, we obtain the following bounds for $v$:

Proposition 1 ([30]). Let $v$ be given by 11 . If the elliptic measure of the adjoint $\omega^*\in B_{r}(\sigma)$ for some $1<r<\infty$, then \[\begin{align} \Vert S(v)\Vert_{L^r(\partial\Omega)},\Vert \tilde{N}(v)\Vert_{L^r(\partial\Omega)}, \Vert \tilde{N}^{(2)}(\delta|\nabla v|)\Vert_{L^r(\partial\Omega)}, \Vert \tilde{N}(\delta|\nabla v|)\Vert_{L^r(\partial\Omega)}\lesssim C.\label{eq:BoundsOnV} \end{align}\qquad{(6)}\]

Under our assumptions, this proposition holds for $r=p$ and for this choice of $p$ we have by integration by parts and using that $Lu=0$ \[\begin{align} \Vert \tilde{N}(\nabla u)\Vert_{L^p}&\lesssim \int_\Omega \nabla u \cdot h dxdt= \int_\Omega A\nabla u\cdot \nabla v dxdt - \int_{\partial\Omega}A^*\nabla v \cdot u\nu dx\nonumber \\ &=-\int_{\partial \Omega}u b\cdot\nabla_{||}v + ud\partial_t v dx\tag{12} \\ &=\int_{\partial \Omega}u(x,0) \Big(\int_0^\infty \partial_t\big(b(x,t)\cdot\nabla_{||}v(x,t) + d(x,t)\partial_t v(x,t)\big) dt\Big) dx.\tag{13} \end{align}\] Without loss of generality we can approximate all involved components of the matrix $A$ and $h$ by smooth functions so that the PDE 11 can be used pointwise. The following arguments are independent of this approximation and hence we can take the limit in the very end, but we omit this argument here. Since $\int_0^\infty \partial_th(x,t)dt=0$, we can continue with \[\begin{align} \eqref{eq:mainProof1}&=-\int_{\partial \Omega}u \Big(\int_0^\infty \mathrm{div}_{||}\big(A_{\Vert}^*(x,t)\cdot\nabla_{||}v(x,t)+c(x,t)\partial_tv(x,t)\big)dt \\ &\qquad - \int_0^\infty\mathrm{div}_{||}(h(x,t)) + \partial_t h(x,t) dt\Big) dx \\ &=\int_{\partial \Omega}\nabla_{||}u(x,0) \Big(\int_0^\infty A_{\Vert}(x,t)\cdot\nabla_{||}v(x,t)+c(x,t)\partial_tv(x,t)dt\Big)dx \\ &\qquad - \int_{\Omega}\nabla_{||}u(x,0)\cdot h(x,t)dxdt. \end{align}\] The last term yields by Lemma 1 \[\begin{align} \Big|\int_{\Omega}\nabla_{||}u(x,0)\cdot h(x,t)dxdt\Big|=\Big|\int_{\Omega}\nabla_{||}f(x)\cdot h(x,t)dxdt\Big|\lesssim \Vert M[\nabla_{||} f]\Vert_{L^p}\lesssim \Vert \nabla_{||} f\Vert_{L^p}. \end{align}\] For the first term we can set \[V(x,s):=\int_s^\infty A_{\Vert}(x,t)\cdot\nabla_{||}v(x,t) + c(x,t)\partial_t v(x,t)dt,\] and get \[\begin{align} &\int_{\partial \Omega}\nabla_{||}u(x,0) \Big(\int_0^\infty A_{\Vert}(x,t)\cdot\nabla_{||}v(x,t) + c(x,t)\partial_t v(x,t)dt\Big)dx \\ &\qquad=\int_{\partial\Omega}\nabla_{||}f(x) V(x,0) dx =- \int_{\Omega}\partial_s(\nabla_{||}\mathcal{P}_sf(x) V(x,s))dxds \\ &\qquad= -\int_{\Omega}\partial_s\nabla_{||}\mathcal{P}_sf(x)\cdot V(x,s) + \nabla_{||}\mathcal{P}_sf(x)\cdot \partial_s V(x,s) dxds. \end{align}\] We introduce $\partial_s(s)=1$ and use integration by parts to obtain \[\begin{align} &= \int_{\Omega}\partial_{s}^2\nabla_{||}\mathcal{P}_sf(x)\cdot V(x,s)sdxds + 2\int_{\Omega}\nabla_{||}\partial_s\mathcal{P}_sf(x) \cdot\partial_s V(x,s)sdxds \\ &\qquad + \int_{\Omega}\nabla_{||}\mathcal{P}_sf(x) \cdot\partial_{s}^2 V(x,s) s dxds \\ & =:I+II+III. \end{align}\] First, we observe that $|\partial_s V(x,s)|\lesssim |\nabla v(x,s)|$ and hence \[\begin{align} |II|&\lesssim \Vert \mathcal{A}(s\partial_s\nabla_{||}\mathcal{P}_sf)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}}\lesssim \Vert \mathcal{A}(s\partial_s\nabla_{||}\mathcal{P}_sf)\Vert_{L^p}, \end{align}\] where we used ?? . To complete bounding $|II|$ by $\Vert\nabla_{||}f\Vert_{L^p}$, we need the area function bound $\Vert \mathcal{A}(s\partial_s\nabla_{||}\mathcal{P}_sf)\Vert_{L^p}\lesssim \Vert \nabla_{||}f\Vert_{L^p}$, which will be established in Lemma 2.

Next, we have by integration by parts

\[\begin{align} III&=\int_{\Omega}\nabla_{||}\mathcal{P}_sf(x) \cdot \partial_s(A_{||}(x,s)\nabla_{||}v(x,s) + c(x,s)\partial_sv(x,s))s dxds \\ &\lesssim \int_{\Omega}\nabla_{||}\mathcal{P}_sf(x) \cdot \partial_sA_{||}(x,s)\nabla_{||}v(x,s) s - L^s\mathcal{P}_sf(x) \partial_s v(x,s) s dxds \\ &\qquad - \int_{\Omega} \nabla_{||}\partial_s\mathcal{P}_sf(x)\cdot c(x,s)\partial_sv(x,s)s +\nabla_{||}\mathcal{P}_sf(x) \cdot c(x,s)\partial_sv(x,s) dxds \\ &\lesssim \Vert\partial_sA\Vert_{\mathcal{C}}\Vert\tilde{N}(\nabla_{||}\mathcal{P}_s f)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}}+\Vert\mathcal{A}(sL^s\mathcal{P}_s f)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}} \\ &\qquad+\Vert\mathcal{A}(\nabla_{||}\partial_s\mathcal{P}_s f)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}} + \Vert\mathcal{A}(\nabla_{||}\mathcal{P}_s f)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}}. \end{align}\] For the first term we used a stopping time argument. These are considered to be standard in this area, but for convenience we would like to refer the reader to the integral $II_{31}$ in [31], where the authors perform a stopping time argument on an expression of exactly this form.

To bound $III$ by $\Vert \nabla_{||} f\Vert_{L^p}$, we need the area function bounds \[\Vert\mathcal{A}(\nabla_{||}\partial_s\mathcal{P}_sf)\Vert_{L^p},\Vert\mathcal{A}(\partial_s\mathcal{P}_sf)\Vert_{L^p}, \Vert\mathcal{A}(\nabla_{||}\mathcal{P}_sf)\Vert_{L^p}\lesssim\Vert \nabla_{||} f\Vert_{L^p},\] which will be established in Lemma 2. To establish these, first, we will also have to prove $\Vert\mathcal{A}(sL^s\mathcal{P}_sf)\Vert_{L^p}=\Vert\mathcal{A}(W_1f)\Vert_{L^p}\lesssim\Vert \nabla_{||} f\Vert_{L^p}$, which we will do in Corollary 1. Lastly for $III$, we can note that $\Vert\tilde{N}(\nabla_{||}\mathcal{P}_s f)\Vert_{L^p}\lesssim \Vert \nabla_{||} f\Vert_{L^p}$ holds by Corollary 1.

Finally, for $I$ \[\begin{align} |I|&=\Big|\int_{\Omega}s\nabla_{||}\partial_{s}^2\mathcal{P}_sf(x)\cdot\Big(\int_s^\infty A_{\Vert}(x,t)\nabla_{||}v(x,t) + c(x,t)\partial_tv(x,t)dt\Big) dx\Big| \\ &=\Big|\int_{\Omega}\nabla_{||}\Big(\int_0^ts\partial^2_{s}\mathcal{P}_sf(x)ds\Big)\cdot(A_{\Vert}(x,t)\nabla_{||}v(x,t) + c(x,t)\partial_tv(x,t)) dxdt\Big| \\ &=\Big|\int_{\Omega}\nabla_{||}\big(t\partial_{t}\mathcal{P}_tf(x)-\mathcal{P}_tf(x)+f(x)\big)\cdot\big(A_{\Vert}(x,t)\nabla_{||}v(x,t) + c(x,t)\partial_tv(x,t)\big) dxdt\Big| \\ &\lesssim \Vert\mathcal{A}(t\nabla_{||}\partial_t \mathcal{P}_tf)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}} \\ &\qquad + \Big|\int_{\Omega}\nabla_{||}\big(\mathcal{P}_tf(x)-f(x)\big)\cdot\big(A_{\Vert}(x,t)\nabla_{||}v(x,t) + c(x,t)\partial_tv(x,t)\big) dxdt\Big|. \end{align}\] For the last term we have by using the PDE for $v$ 11 and that $(\mathcal{P}f-f)|_{\partial\Omega}=0$ \[\begin{align} &\Big|\int_{\Omega}\nabla_{||}\big(\mathcal{P}_tf(x)-f(x)\big)\cdot\big(A_{\Vert}(x,t)\nabla_{||}v(x,t) + c(x,t)\partial_tv(x,t)\big) dxdt\Big| \\ &\qquad= \Big|\int_{\Omega}\partial_t\mathcal{P}_tf(x)\cdot\big(b(x,t)\nabla_{||}v(x,t) + d(x,t)\partial_tv(x,t)\big) \\ &\qquad \qquad + (\nabla_{||}\mathcal{P}_tf - \nabla_{||}f)(x)\cdot h_{||} + \partial_t \mathcal{P}_tf(x)\cdot h_{n}(x,t) dxdt\Big| \\ &\qquad \lesssim \Vert\mathcal{A}(\partial_t \mathcal{P}_tf)\Vert_{L^p}\Vert S(v)\Vert_{L^{p'}} + \Vert\tilde{N}(\partial_t \mathcal{P}_tf)\Vert_{L^p}+\Vert\tilde{N}(\nabla_{||} \mathcal{P}_tf)\Vert_{L^p} \\ &\qquad\lesssim \Vert\mathcal{A}(\partial_t \mathcal{P}_tf)\Vert_{L^p} + \Vert \nabla_{||}f\Vert_{L^p}. \end{align}\] Here we used again the established bounds in ?? , Corollary 1 and that $\Vert\tilde{N}(\partial_t \mathcal{P}_tf)\Vert_{L^p}\lesssim \Vert\mathcal{A}(\partial_t\mathcal{P}_tf)\Vert_{L^p}$ to reduce the estimate to the area function bound on $\partial_t\mathcal{P}$. This bound appeared already and we are going to prove it in Lemma 2.

Hence we reduced the proof of Theorem 1 to proving the three $L^p$ area function bounds in the next lemma (and additionally Corollary 1 which is part of the proof of Lemma 2):

Lemma 2. Let $1<p<\infty$. For $T_tf\in \Big\{\nabla_{||}\mathcal{P}_tf, \partial_t\mathcal{P}_tf, t\nabla_{||}\partial_t\mathcal{P}_tf\Big\}$, there exists $C>0$ such that \[\Vert \mathcal{A}(T_tf)\Vert_{L^p(\partial\Omega)}\leq C \Vert \nabla_{||}f\Vert_{L^p(\partial\Omega)}\] for every $f\in \dot{L}^{p}_1(\partial\Omega)$.

To prove this lemma, we are going to need the method of real interpolation, which we introduce in the following section.

4 Tent spaces and real interpolation↩︎

Let $A_0,A_1$ be two normed vector spaces of functions $a:\mathbb{R}^n\to\mathbb{R}$. For each $a\in A_0 + A_1$ we define the $K-$functional of real interpolation by \[K(a,t,A_0,A_1)=\inf_{a=a_0+a_1}\Vert a_0\Vert_{A_0} + t\Vert a_1\Vert_{A_1}.\] For $0<\theta<1,1\leq q\leq \infty$, we denote by $(A_0,A_1)_{\theta,q}$ the real interpolation space between $A_0$ and $A_1$ defined as \[(A_0,A_1)_{\theta,q}=\Big\{a\in A_1+A_0:\Vert a\Vert_{\theta,q}=\Big(\int_0^\infty (t^{-\theta}K(a,t,A_0,A_1))^q\frac{\mathrm{d}t}{t}\Big)^{1/q}<\infty \Big\}.\] According to Theorem 3.1.2 in [32], $K$ can be seen as an exact interpolation functor, which means that if an operator $T$ is bounded from $A_0\to B_0$ and from $A_1\to B_1$ for linear normed vector spaces of functions, then $T:(A_0,A_1)_{\theta,q}\to (B_0,B_1)_{\theta,q}$ is a bounded linear operator with $\Vert T\Vert\leq C\Vert T\Vert^{1-\theta}_{A_0\to B_0}\Vert T\Vert^{\theta}_{A_1\to B_1}$.

Let us introduce the Hardy-Sobolev space like in [22] or [33] and [34].

Definition 1. Let $1<\beta\leq\infty$. We call a function $a:\mathbb{R}^n\to\mathbb{R}$ a homogeneous Hardy-Sobolev $\beta$-atom associated to a boundary ball $\Delta\subset\mathbb{R}^n$ if

(i) $\mathrm{supp}(a)\subset \Delta$;

(ii) $\Vert \nabla a\Vert_{L^\beta(\mathbb{R}^n)}\leq |\Delta|^{-\frac{1}{\beta'}}$; and

(iii) $\Vert a\Vert_{L^1(\mathbb{R}^n)}\leq l(\Delta).$

If $f$ can be written as \[\begin{align} f=\sum_{j=1}^\infty\lambda_ja_j\label{eq:AtomicHardyDecomposition} \end{align}\tag{14}\] for $\beta$-atoms $a_j$ and coefficients $\lambda_j\in\mathbb{R}$ with $\sum_{j=1}^\infty|\lambda_j|<\infty$, we say that $f\in \dot{HS}^{1,\beta}_{atom}$, where $\Vert f\Vert_{\dot{HS}^{1,\beta}_{atom}}:=\inf\sum_{j=1}^\infty|\lambda_j|$ with an infimum that is taken over all choices of decompositions 14 .

Now, we have the following real interpolation result.

Proposition 1 (Thm 0.4 in [34]). For every $\beta\in(1,\infty]$ and $1<p<\infty$ the real interpolation space is \[(\dot{HS}^{1,\beta}_{atom}, \dot{W}^{1,\infty})_{1-1/p,p}=\dot{W}^{1,p}.\]

On the other hand, recall the definition of the area function 5 and of the Carleson function 6 and define the tent spaces over $\mathbb{R}^n$ with parameter $1\leq p<\infty$ as \[\begin{align} T^{p,2}(\Omega):=\{F\in L^2_{loc};\Vert F\Vert_{T^{p,2}}:=\Vert \mathcal{A}(F)\Vert_{L^p(\partial\Omega)}<\infty\} \end{align}\] and \[\begin{align} T^{\infty,2}(\Omega):=\{F\in L^2_{loc};\Vert F\Vert_{T^{\infty,2}}:=\Vert \mathcal{C}(F)\Vert_{L^\infty(\partial\Omega)}<\infty\}. \end{align}\]

By [23] we have

Proposition 1 (Theorem 4’ in [23]). For every $1<p<\infty$ the real interpolation space is \[(T^{1,2},T^{\infty,2})_{1-1/p,p}=T^{p,2}.\]

4.1 Proof of Lemma 2↩︎

Since for $1<p<\infty$ the two spaces $\dot{W}^{1,p}(\mathbb{R}^n)$ and $\dot{L}^{p}_1(\mathbb{R}^n)$ are the same modulo constants with comparable norms, these real interpolation results can be used to proof Lemma 2. First, we break the operator $\partial_t\mathcal{P}_tf$ up into the sum of $W_1f$ and $W_2f$ as introduced in Section 2.2 and discuss the necessary bounds separately.

Specifically, we establish the following two corollaries, which immediately give the proof of Lemma 2.

Corollary 1. For every $1<p<\infty$ and $f\in \dot{L}^{p}_1(\partial\Omega)=\dot{W}^{1,p}(\mathbb{R}^n)$ we have \[\Vert \mathcal{A}(W_2f)\Vert_{L^p(\partial\Omega)}, \Vert \mathcal{A}(t\nabla_{||}W_2f)\Vert_{L^p(\partial\Omega)}\leq C\Vert \nabla_{||} f\Vert_{L^p(\partial\Omega)}.\]

Corollary 1. For every $f\in \dot{L}^{p}_1(\partial\Omega)=\dot{W}^{1,p}(\mathbb{R}^n)$ we have \[\Vert \mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^p(\partial\Omega)},\Vert \mathcal{A}(W_1f)\Vert_{L^p(\partial\Omega)}, \Vert \mathcal{A}(t\nabla_{||}W_1f)\Vert_{L^p(\partial\Omega)}\leq C\Vert \nabla_{||} f\Vert_{L^p(\partial\Omega)}.\]

It remains to prove these corollaries. By the method of real interpolation it suffices to establish the $L^\infty$ to Carleson function bound $\Vert \mathcal{C}(T_tf)\Vert_{L^\infty}\lesssim \Vert \nabla_{||} f\Vert_{L^\infty}$ and the Hardy-Sobolev space to $L^1$ bound $\Vert \mathcal{A}(T_tf)\Vert_{L^1}\lesssim \Vert \nabla_{||} f\Vert_{\dot{HS}^{1,\beta}_{atom}}$ for each of the five operators $T_t\in\{\nabla_{||}\mathcal{P}_t,W_1,W_2,t\nabla_{||} W_1,t\nabla_{||} W_2\}$. Hence, we need to establish the following four lemmas to conclude Corollary 1 and Corollary 1.

Lemma 3. Let $f\in W^{1,\infty}(\mathbb{R}^n)$. Then

(i) $\Vert C(W_2f)\Vert_{L^\infty}=\sup_{\Delta\subset\partial\Omega}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|W_2f(x,t)|^2}{t}dxdt\lesssim \Vert\nabla_{||}f\Vert_{L^\infty}$, and

(ii) $\Vert C(t\nabla W_2f)\Vert_{L^\infty}=\sup_{\Delta\subset\partial\Omega}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|t\nabla_{||} W_2f(x,t)|^2}{t}dxdt\lesssim \Vert\nabla_{||}f\Vert_{L^\infty}$.

Lemma 4. Let $f\in \dot{HS}_{atom}^{1,\beta}(\mathbb{R}^n)$ for $2<\beta<\infty$. Then we have

(i) $\Vert\mathcal{A}(W_2f)\Vert_{L^1(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{\dot{HS}_{atom}^{1,\beta}(\mathbb{R}^n)},$ and

(ii) $\Vert\mathcal{A}(t\nabla W_2f)\Vert_{L^1(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{\dot{HS}_{atom}^{1,\beta}(\mathbb{R}^n)}$.

Lemma 5. Let $f\in W^{1,\infty}(\mathbb{R}^n)$, then

(i) $\Vert C(\nabla\mathcal{P}_tf)\Vert_{L^\infty}=\sup_{\Delta\subset\partial\Omega}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|\nabla_{||}\mathcal{P}_tf(x,t)|^2}{t}dxdt\lesssim \Vert\nabla_{||}f\Vert_{L^\infty}$,

(ii) $\Vert C(W_1f)\Vert_{L^\infty}=\sup_{\Delta\subset\partial\Omega}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|W_1f(x,t)|^2}{t}dxdt\lesssim \Vert\nabla_{||}f\Vert_{L^\infty}$, and

(iii) $\Vert C(t\nabla W_1f)\Vert_{L^\infty}=\sup_{\Delta\subset\partial\Omega}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|t\nabla_{||} W_1f(x,t)|^2}{t}dxdt\lesssim \Vert\nabla_{||}f\Vert_{L^\infty}$.

Lemma 6. Let $f\in \dot{HS}_{atom}^{1,2}(\mathbb{R}^n)$. Then

(i) $\Vert\mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^1(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{\dot{HS}^{1,2}(\mathbb{R}^n)}$

(ii) $\Vert\mathcal{A}(W_1f)\Vert_{L^1(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{\dot{HS}^{1,2}(\mathbb{R}^n)}$

(iii) $\Vert\mathcal{A}(t\nabla W_1f)\Vert_{L^1(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{\dot{HS}_{atom}^{1,2}(\mathbb{R}^n)}$

These lemmas are proved in the next two sections. First, we will deal with the bounds involving $W_2$, i.e. Lemma 3 and Lemma 4, before we are able to turn to the bounds involving $\nabla\mathcal{P}_t$ and $W_1$, namely Lemma 5 and Lemma 6. We grouped the bounds for $\nabla\mathcal{P}_t$ and $W_1$ together because their proofs are very similar to each other.

5 Area function bounds on $W_2$↩︎

5.1 Carleson function bounds and proof of Lemma 3↩︎

We begin with obtaining pointwise Carleson function bounds for $W_2f$.

Lemma 7. Let $f\in W^{1,1}_{loc}(\mathbb{R}^n)$. Then

(i) $C(W_2f)^2(z):=\sup_{\Delta(z)}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|W_2f(x,t)|^2}{t}dxdt\lesssim M[M[|\nabla_{||}f|^2]](z)$, and

(ii) $C(t\nabla W_2f)^2(z):=\sup_{\Delta(z)}\frac{1}{\sigma(\Delta)}\int_{T(\Delta)}\frac{|t\nabla_{||} W_2f(x,t)|^2}{t}dxdt\lesssim M[M[|\nabla_{||}f|^2]](z)$,

where the suprema are taken over all boundary balls centered at $z$.

Proof. For [item:PointwiseLocalSquareBoundW2] we let $\Delta$ be a boundary ball with center $z\in \partial\Omega$. Then Minkowski’s inequality yields

\[\begin{align} &\int_{T(\Delta)}\frac{|W_2f(x,s)|^2}{s}dxds \\ &\qquad = \int_0^{\infty}\frac{1}{s}\Big\Vert\int_0^{s} \tau e^{-(s^2-\tau^2)L_{||}^s}\mathrm{div}(\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f)d\tau\Big\Vert_{L^2(\Delta)}^2 ds \\ &\qquad\leq \int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} \Vert\tau e^{-(s^2-\tau^2)L_{||}^s}\mathrm{div}(\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f)\Vert_{L^2(\Delta)} d\tau\Big)^2 dx ds. \end{align}\]

For the inner $L^2$ norm, let us use off-diagonal estimates ( Proposition 1) and cut-off functions $\eta_k:=\chi_{2^k\Delta\setminus2^{k-1}\Delta}$ such that $\sum_k\chi_{2^k\Delta\setminus 2^{k-1}\Delta} + \chi_{2\Delta}=1$ to obtain \[\begin{align} &\Vert\tau e^{-(s^2-\tau^2)L_{||}^s}\mathrm{div}(\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f)\Vert_{L^2(\Delta)}\nonumber \\ &\leq \sum_{k\geq 2}\Vert\tau e^{-(s^2-\tau^2)L_{||}^s}\mathrm{div}(\eta_k\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f)\Vert_{L^2(\Delta)}\nonumber \\ &\qquad + \Vert\tau e^{-(s^2-\tau^2)L_{||}^s}\mathrm{div}(\chi_{2\Delta}\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f)\Vert_{L^2(\Delta)}\nonumber \\ &\leq \sum_{k\geq 2}\frac{\tau}{\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f\Vert_{L^2(2^k\Delta\setminus 2^{k-1}\Delta)}\nonumber \\ &\qquad + \frac{\tau}{\sqrt{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f\Vert_{L^2(2\Delta)}=I+II.\nonumber \end{align}\] We continue for the first term with the same ideas, in particular the off-diagonal estimates, and the pointwise bound $|\partial_sA|\leq \frac{1}{s}$, and get \[\begin{align} I&\leq \sum_{k\geq 2}\sum_{m\geq 2}\frac{\tau}{\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}(\eta_{k+m}(f-(f)_{2^k\Delta}))\Vert_{L^2(2^k\Delta\setminus 2^{k-1}\Delta)}\nonumber \\ & \qquad + \sum_{k\geq 2}\frac{\tau}{\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}(\chi_{2^{k+1}\Delta}(f-(f)_{2^k\Delta}))\Vert_{L^2(2^k\Delta\setminus 2^{k-1}\Delta)}\nonumber \\ &\leq \sum_{k\geq 2}\sum_{m\geq 2}\frac{1}{\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2} - c\frac{2^{2(m+k-1)}l(\Delta)^2}{\tau^2}}\Vert\partial_s A_{\Vert}(\cdot,s)\Vert_\infty\Vert(f-(f)_{2^k\Delta})\Vert_{L^2(2^{k+m}\Delta)}\nonumber \\ & \qquad + \sum_{k\geq 2}\frac{1}{\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(\cdot,s)\Vert_{L^\infty}\Vert f-(f)_{2^k\Delta}\Vert_{L^2(2^{k+1}\Delta)}.\nonumber \\ &\leq |\Delta|^{1/2}\Big(\sum_{k\geq 2}\sum_{m\geq 2}\frac{2^{(k+m)(\frac{n}{2}+1)}l(\Delta)}{s\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2} - c\frac{2^{2(m+k-1)}l(\Delta)^2}{\tau^2}}M[|\nabla f|^2]^{1/2}\tag{15} \\ & \qquad + \sum_{k\geq 2}\frac{\tau 2^{(k+1)\frac{n}{2}}l(\Delta)}{s\sqrt{s^2-\tau^2}}e^{-c\frac{2^{2k}l(\Delta)^2}{s^2-\tau^2}}M[|\nabla f|^2]^{1/2}\Big).\tag{16} \end{align}\] Here we used Poincaré in the last line, noting that for the first term we can observe that \[\begin{align} &\Vert f-(f)_{2^k\Delta}\Vert_{L^2(2^{k+m}\Delta)}\label{observationf-MVonLargerDomain} \\ &\qquad\leq \Vert f-(f)_{2^{k+m}\Delta}\Vert_{L^2(2^{k+m}\Delta)} + \sum_{l=1}^m \Vert(f)_{2^{k+l}\Delta}-(f)_{2^{k+l-1}\Delta}\Vert_{L^2(2^{k+m}\Delta)}\nonumber \\ &\qquad\lesssim 2^{(k+m)}l(\Delta)\Vert \nabla_{||}f\Vert_{L^2(2^{k+m}\Delta)} + \sum_{l=1}^m 2^{k+l}l(\Delta)\Vert\nabla_{||}f\Vert_{L^2(2^{k+m}\Delta)}\nonumber \\ &\qquad\lesssim 2^{(k+m)}l(\Delta)\Vert \nabla_{||}f\Vert_{L^2}\lesssim 2^{(k+m)(\frac{n}{2}+1)}l(\Delta)|\Delta|^{1/2}M[|\nabla_{||}f|^2]^{1/2}(z),\nonumber \end{align}\tag{17}\] We consider the sums in 15 and 16 as Riemann sums of integral. For instance for 16 , the sum can be bounded by the integrals \[\int_0^\infty\frac{\tau}{s\sqrt{s^2-\tau^2}}x^{\frac{n}{2}-1}e^{-cx^2\frac{l(\Delta)^2}{s^2-\tau^2}}dx,\] which after the change of variables $y=\frac{l(\Delta)^2}{s^2-\tau^2}x$ gives a convergent integral \[\frac{\tau\sqrt{s^2-\tau^2}}{sl(\Delta)\sqrt{s^2-\tau^2}}\int_0^\infty y^{\frac{n}{2}-1}e^{-y^2}dy\lesssim \frac{C}{l(\Delta)}.\] Similar arguments for the other sum in 15 lead to the conclusion \[I\lesssim M[|\nabla_{||} f|^2]^{1/2}(z)|\Delta|^{1/2}\Big(\frac{1}{l(\Delta)}+ \frac{\tau}{l(\Delta)\sqrt{s^2-\tau^2}}\Big).\]

Hence we obtain \[\begin{align} &\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} I d\tau\Big)^2 dx ds \\ &\lesssim\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} M[|\nabla_{||} f|^2]^{1/2}(z)|\Delta|^{1/2}\Big(\frac{1}{l(\Delta)}+ \frac{\tau}{l(\Delta)\sqrt{s^2-\tau^2}}\Big) d\tau\Big)^2 ds \\ &\lesssim |\Delta| M[|\nabla_{||} f|^2](z). \end{align}\]

For $II$ we have for a scale $k$ such that $2^{-k}\leq \tau \leq 2^{-k+1}$ with Proposition 1 \[\begin{align} II&= \frac{\tau}{\sqrt{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f\Vert_{L^2(2\Delta)} \\ &= \frac{\tau}{\sqrt{s^2-\tau^2}}\Big(\sum_{Q\in \mathcal{D}_k(3\Delta)}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f\Vert_{L^2(Q)}^2\Big)^{1/2} \\ &= \frac{\tau}{\sqrt{s^2-\tau^2}}\Big(\sum_{Q\in \mathcal{D}_k(3\Delta)}\sup_{(x,s)\in Q}|\partial_s A_{\Vert}(x,s)|^2\inf_{(x,s)\in Q} |M[\nabla f](x,s)|^2\Big)^{1/2} \\ &= \frac{\tau}{\sqrt{s^2-\tau^2}}\Big(\int_{3\Delta}\sup_{(y,t)\in B(x,s,s/2)}|\partial_t A_{\Vert}(y,t)|^2 |M[\nabla f]|^2 dx\Big)^{1/2}. \end{align}\]

Hence we obtain \[\begin{align} &\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} II d\tau\Big)^2 dx ds \\ &\leq\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} \frac{\tau}{\sqrt{s^2-\tau^2}}\Vert\partial_s A_{\Vert}(x,s)\nabla e^{-\tau^2L_{||}^s}f\Vert_{L^2(2\Delta)} d\tau\Big)^2 dx ds \\ &\leq\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} \frac{\tau}{\sqrt{s^2-\tau^2}}\Big(\int_{3\Delta}\sup_{(y,t)\in B(x,s,s/2)}|\partial_t A_{\Vert}|^2 M[\nabla f]^2 dx\Big)^{1/2} d\tau\Big)^2 dx ds \\ &\leq\int_0^{l(\Delta)}\int_{3\Delta}\sup_{(y,t)\in B(x,s,s/2)}|\partial_t A_{\Vert}|^2s M[\nabla f]^2 dx ds. \end{align}\] Making use of 8 we continue with \[\begin{align} &\lesssim \Big\Vert\sup_{(y,t)\in B(x,s,s/2)}|\partial_t A_{\Vert}(y,t)|^2s \Big\Vert_{\mathcal{C}} \int_\Delta \tilde{N}(M[|\nabla f|]^2) dx \\ &\lesssim \Big\Vert\sup_{(y,t)\in B(x,s,s/2)}|\partial_t A_{\Vert}(y,t)|^2s \Big\Vert_{\mathcal{C}} |\Delta| M[M[|\nabla f|]^2](z). \end{align}\] We would like to note here that 8 does not contain the mean-valued nontangential maximal function on the right hand side, but since the density of the Carleson measure contains a supremum over balls of radius half the distance to the boundary, the standard stopping time argument that yields 8 allows to take a mean value in the nontangential maximal function. We omit the details here.

In total we get for [item:PointwiseLocalSquareBoundW2] \[\begin{align} C(W_2f)(z)&=\sup_{\Delta=\Delta(z)}\frac{1}{|\Delta|}\int_{T(\Delta)}\frac{|W_2f(x,s)|^2}{s}dxds \\ &\lesssim \frac{1}{|\Delta|}\Big(\int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} I d\tau\Big)^2 dx ds + \int_0^{l(\Delta)}\frac{1}{s}\Big(\int_0^{s} II d\tau\Big)^2 dx ds\Big) \\ &\lesssim M[M[|\nabla f|]^2](z). \end{align}\]

For [item:PointwiseLocalSquareBoundNablaW2] we have by Cacciopolli type inequality Proposition 1 \[\begin{align} &\int_{T(\Delta)}\frac{|t\nabla W_2f(x,t)|^2}{t}dxdt \lesssim \int_{T(2\Delta)}\frac{|W_2f(x,t)|^2}{t}dxdt \\ &\qquad + \int_{2\Delta}\int_0^{2l(\Delta)}\sup_{(y,s)\in B(x,t,t/2)}|\partial_s A_{\Vert}(y,s)|^2t M[|\nabla_{||}f|^2] dxdt \\ &\lesssim \Big\Vert\sup_{(y,s)\in B(x,t,t/2)}|\partial_s A_{\Vert}(y,s)|^2 t \Big\Vert_{\mathcal{C}} \int_\Delta \tilde{N}(M[|\nabla f|]^2) dx \\ &\lesssim |\Delta| M[M[|\nabla_{||} f|^2]](z). \end{align}\] ◻

As a corollary we obtain the Carleson measure bound that we need in the real interpolation argument for Corollary 1.

Proof of Lemma 3. The only observation needed is that if $f\in W^{1,\infty}(\mathbb{R}^n)$, then $M[M[|\nabla_{||} f|^2]]\leq \Vert \nabla_{||} f\Vert_{L^\infty}^2$. Then the statement follows from Lemma 7. ◻

Another corollary is the area function bound with $\beta>2$ which will be needed for the proof of Lemma 4.

Corollary 1. Let $f\in W^{1,\beta}(\partial\Omega)$ for $2<\beta<\infty$. Then we have

(i) $\Vert\mathcal{A}(W_2f)\Vert_{L^\beta(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{L^\beta(\mathbb{R}^n)},$ and

(ii) $\Vert\mathcal{A}(t\nabla W_2f)\Vert_{L^\beta(\mathbb{R}^n)}\lesssim \Vert \nabla_{||}f\Vert_{L^\beta(\mathbb{R}^n)}$.

Proof. By Theorem 6.1 in [24] we know $\Vert \mathcal{A}(g)\Vert_{L^\beta(\mathbb{R}^n)}\lesssim_\beta \Vert \mathcal{C}(g)\Vert_{L^\beta(\mathbb{R}^n)}$, if $2<\beta<\infty$. Combining this with Lemma 7 and $L^p$ boundedness of the Hardy-Littlewood maximal function, we obtain \[\Vert \mathcal{A}(W_2f)\Vert_{L^\beta}\lesssim_\beta \Vert \mathcal{C}(W_2f)\Vert_{L^\beta}\lesssim \big(\int_{\partial\Omega} M[M[|\nabla f|^2]]^{\beta/2}dx\big)^{1/\beta}\lesssim \Vert \nabla_{||} f\Vert_{L^\beta}.\] A completely analogous argument works for $t\nabla W_2$. ◻

5.2 Hardy-Sobolev bound (Proof of Lemma 4)↩︎

Proof of Lemma 4. First, we note that it is enough to show \[\Vert\mathcal{A}(W_2f)\Vert_{L^1(\mathbb{R}^n)}, \Vert\mathcal{A}(t\nabla W_2f)\Vert_{L^1(\mathbb{R}^n)}\leq C\] for all homogeneous Hardy-Sobolev $1/2$-atoms $f$ associated with $\Delta$, whence we assume that $f$ is such an atom going forward.

We begin with showing [item:HardySquareBound]. We split the integral into a local and a far away part \[\begin{align} \Vert\mathcal{A}(W_2f)\Vert_{L^1(\mathbb{R}^n)}=\Vert\mathcal{A}(W_2f)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)} + \Vert\mathcal{A}(W_2f)\Vert_{L^1(5\Delta)}. \end{align}\] For the local part we have by Hölder’s inequality, Corollary 1 and the properties of the Hardy-Sobolev space (cf. Definition 1) that \[\begin{align} \Vert\mathcal{A}(W_2f)\Vert_{L^1(5\Delta)}\lesssim \Vert\mathcal{A}(W_2f)\Vert_{L^\beta(5\Delta)}|\Delta|^{1/\beta'}\lesssim \Vert\nabla_{||} f\Vert_{L^\beta(\mathbb{R}^n)}|\Delta|^{1/\beta'}\lesssim 1. \end{align}\]

First, we note that the kernel bounds in Proposition 1 imply \[\begin{align} &\sup_{(x,t)\in\Delta_{t/2}(y)} |tL^te^{-t^2L^t}f(x)|^2dx\nonumber \\ &\qquad\lesssim \begin{cases} \frac{1}{t^{2n+2}}e^{-c\frac{\mathrm{dist}(\Delta_{t/2}(y),\Delta)^2}{t^2}}\Vert f\Vert_{L^1(\Delta)}^2 &\textrm{if }\mathrm{dist}(\Delta_{t/2}(y),\Delta)>0, \\ \frac{1}{t^{2n+2}}\Vert f\Vert_{L^1(\Delta)}^2& \mathrm{else}.\end{cases}\label{eq:CasesForL941BoundofTermInProof} \end{align}\tag{18}\]

For the away part we are going to distinguish several cases. We are going to split the integral over $\mathbb{R}^n$ into integrals over annuli of the form $2^{j+1}\Delta\setminus 2^{j}\Delta$ for $j\geq 2$ and bound all cones in each annulus uniformly. However, each cone with tip in $2^{j+1}\Delta\setminus 2^{j}\Delta$ is itself split into a a close and a far away part $\Gamma=\Gamma^{2^{j-1}l(\Delta)}\cup \Gamma\setminus\{t\leq 2^{j-1}l(\Delta)\}$.

To start with, we fix $y\in\partial\Omega$ with $y\in 2^{j+1}\Delta\setminus 2^j\Delta$. Let us also fix $t$ and we look at the term \[\int_{\Delta_{t/2}(y)}\Big|\int_0^t2\tau e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)d\tau\Big|^2dx.\]

First assume that $0\leq t\leq 2^{j-1}l(\Delta)$. Then $\Delta_{t/2}(y)\subset \Delta_{2^{j-2}l(\Delta)}(y):=\tilde{\Delta}$, and we have for a dualising function $g\in L^2(\tilde{\Delta})$ \[\begin{align} &\Vert\tau e^{-(t^2-\tau^2)L_{||}^t}\mathrm{div}(\partial_t A_{\Vert}(x,t)\nabla e^{-\tau^2L_{||}^t}f)\Vert_{L^2(\tilde{\Delta})}\nonumber \\ &=\tau\int_{\mathbb{R}}\partial_t A_{\Vert}(x,t)\nabla e^{-\tau^2L_{||}^t}f\cdot \nabla e^{-(t^2-\tau^2)L_{||}^t} g dx\nonumber \\ &\lesssim \tau\Vert\partial_tA_{\Vert}\Vert_\infty\big(\Vert \nabla e^{-(t^2-\tau^2)L_{||}^t} g\Vert_{L^2(2\tilde{\Delta})} \Vert \nabla e^{-\tau^2L_{||}^t}f\Vert_{L^2(2\tilde{\Delta})}\nonumber \\ &\qquad+ \Vert \nabla e^{-(t^2-\tau^2)L_{||}^t} g\Vert_{L^2(2\Delta)} \Vert \nabla e^{-\tau^2L_{||}^t}f\Vert_{L^2(2\Delta)}\nonumber \\ &\qquad+ \Vert \nabla e^{-(t^2-\tau^2)L_{||}^t} g\Vert_{L^2(\mathbb{R}^n\setminus 2\tilde{\Delta})} \Vert \nabla e^{-\tau^2L_{||}^t}f\Vert_{L^2(\mathbb{R}^n\setminus 2\Delta)}\big).\label{eq:ProofUsefulEquation1} \end{align}\tag{19}\] Let us note that by using a smooth cut-off function $\eta$ with $\mathrm{supp}(\eta)\subset\frac{3}{2}\tilde{\Delta}$ and $\eta\equiv 1$ on $\tilde{\Delta}$ we have \[\int_{\tilde{\Delta}}|\nabla e^{-\tau^2L_{||}^t}f|^2dx\lesssim \int_{2\tilde{\Delta}} W_1f \cdot \mathcal{P}_tf\eta^2 dx + \int_{2\tilde{\Delta}} \nabla e^{-\tau^2L_{||}^t}f\cdot\nabla \eta \mathcal{P}_tf\eta dx\] which implies $\int_{\tilde{\Delta}}|\nabla e^{-\tau^2L_{||}^t}f|^2dx\lesssim \int_{\tilde{\Delta}}|W_1f|^2dx + \frac{1}{t^2}\int_{\tilde{\Delta}}|\mathcal{P}_tf|^2dx$. This observation holds not only for the set $\tilde{\Delta}$, but also for integrals over other sets and enlargements thereof. For each of those terms we can use a similar observation to 18 and combine this with Proposition 1 to bound $\eqref{eq:ProofUsefulEquation1}$ by \[\begin{align} &\Vert\partial_tA_{\Vert}\Vert_\infty\Big[\Big(\int_{2\tilde{\Delta}}\big(\frac{\tau}{\sqrt{t^2-\tau^2}\tau^{n+1}}e^{-c\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}}\Vert f\Vert_{L^1(\Delta)}\big)^2dx\Big)^{1/2}\Vert g\Vert_{L^2} \\ &\qquad + \Big(\int_{2\Delta}\big(\frac{1}{\sqrt{t^2-\tau^2}^{n+1}}e^{-c\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}}\Vert g\Vert_{L^1(\tilde{\Delta})}\big)^2dx\Big)^{1/2}\Vert f\Vert_{L^2} \\ &\qquad + \tau\Big(\int_{\mathbb{R}^n\setminus 2\tilde{\Delta}}\big(\frac{1}{\sqrt{t^2-\tau^2}^{n+1}}e^{-c\frac{\mathrm{dist}(x,\tilde{\Delta})^2}{t^2-\tau^2}}\Vert g\Vert_{L^1(\tilde{\Delta})}\big)^2dx\Big)^{1/2} \\ &\qquad\qquad \cdot\Big(\int_{\mathbb{R}^n\setminus 2\Delta}\big(\frac{1}{\tau^{n+1}}e^{-c\frac{\mathrm{dist}(x,\Delta)^2}{\tau}}\Vert f\Vert_{L^1(\Delta)}\big)^2dx\Big)^{1/2} \Big]. \end{align}\] Further, we can use Hölder’s inequality and Poincaré’s inequality and continue with \[\begin{align} &\Vert\partial_tA_{\Vert}\Vert_\infty\Big[\frac{\tau 2^{jn/2}l(\Delta)^n}{\sqrt{t^2-\tau^2}\tau^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}}\Vert f\Vert_{L^2(\Delta)}\Vert g\Vert_{L^2} \\ &\qquad + \frac{2^{jn/2}l(\Delta)^n}{\sqrt{t^2-\tau^2}^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}}\Vert g\Vert_{L^2(\tilde{\Delta})}\Vert f\Vert_{L^2} \\ &\qquad + \frac{2^{jn/2}l(\Delta)^{n/2}}{\sqrt{t^2-\tau^2}^{n+1}}\Vert g\Vert_{L^2(\tilde{\Delta})}\Big(\int_{2^{j-1}l(\Delta)}^\infty r^{n-1}e^{-\frac{r^2}{t^2-\tau^2}}dr\Big)^{1/2} \\ &\qquad\qquad\cdot\frac{l(\Delta)^{n/2}}{\tau^{n}}\Vert f\Vert_{L^2(\Delta)}\Big(\int_{l(\Delta)}^\infty r^{n-1}e^{-\frac{r^2}{\tau^2}}dr\Big)^{1/2} \Big] \\ &\lesssim \Vert\partial_tA_{\Vert}\Vert_\infty\Vert \nabla_{||}f\Vert_{L^2(\Delta)}\Vert g\Vert_{L^2}\Big[\frac{\tau 2^{jn/2}l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}\tau^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}} + \frac{2^{jn/2}l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}} \\ &\qquad + \frac{2^{jn/2}l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}^{n/2+1}\tau^{n/2}}\Big(\int_{2^{j-1}l(\Delta)/\sqrt{t^2-\tau^2}}^\infty r^{n-1}e^{-r^2}dr\Big)^{1/2}\Big(\int_{l(\Delta)/\tau}^\infty r^{n-1}e^{-r^2}dr\Big)^{1/2}\Big]. \end{align}\] The integral expressions are a little bit delicate. If $n$ is even, than $\int_{a}^\infty r^{n-1}e^{-r^2}dr=P(a)e^{-a^2}$ where $P$ is a polynomial of degree $n-2$. If $n$ is odd, we can bound the integral above by the same expression with $n+1$ instead, since we only integrate over numbers greater than $1$. Hence, we can bound the first integral by $\frac{2^{(j-1)(n-2)}l(\Delta)^{n-2}}{\sqrt{t^2-\tau^2}^{n-2}}e^{-\frac{2^{2(j-1)}l(\Delta)}{t^2-\tau^2}}$ or $\frac{2^{(j-1)(n-1)}l(\Delta)^{n-1}}{\sqrt{t^2-\tau^2}^{n-1}}e^{-\frac{2^{2(j-1)}l(\Delta)}{t^2-\tau^2}}$. Since $\frac{2^{j-1}l(\Delta)}{\sqrt{t^2-\tau^2}}\geq 1$, we can bound both expressions by $\frac{2^{n(j-1)}l(\Delta)^{n}}{\sqrt{t^2-\tau^2}^{n}}e^{-\frac{2^{2(j-1)}l(\Delta)}{t^2-\tau^2}}$. Similarly, we obtain for the second integral the bound $\frac{l(\Delta)^n}{\tau^n}e^{-\frac{l(\Delta)^2}{\tau^2}}$. Plugging in these bounds yields \[\begin{align} &\lesssim \Vert\partial_tA_{\Vert}\Vert_\infty\Vert\nabla_{||}f\Vert_{L^2}\Big[\frac{\tau 2^{jn/2}l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}\tau^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}} + \frac{2^{jn/2}l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}} \\ &\qquad + \frac{2^{jn}l(\Delta)^{2n+1}}{\sqrt{t^2-\tau^2}^{n+1}\tau^{n}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}}e^{-\frac{l(\Delta)^2}{\tau^2}} \Big] \\ &\lesssim \frac{1}{t}\Big[\frac{l(\Delta)^{n/2+1}}{\sqrt{t^2-\tau^2}\tau^{n}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}} +\frac{ l(\Delta)^{n/2+1}}{\sqrt{t^2-\tau^2}^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}} \\ &\qquad + \frac{2^{jn}l(\Delta)^{3n/2+1}}{\sqrt{t^2-\tau^2}^{n+1}\tau^{n}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}}e^{-\frac{l(\Delta)^2}{\tau^2}}\Big]. \end{align}\] In the last step we used $|\partial_tA|\leq \frac{C}{t}$ and that $f$ is a Hardy-Sobolev atom which yields $\Vert \nabla_{||} f\Vert_{L^2(\mathbb{R}^n)}\leq |\Delta|^{1/2-1/\beta} \Vert \nabla_{||} f\Vert_{L^\beta(\mathbb{R}^n)}\leq |\Delta|^{-1/2}$ by Hölder inequality.

Now, we can estimate the integral over a truncated cone for a point $y\in\partial\Omega$ with $y\in 2^{j+1}\Delta\setminus2^j\Delta$. We have \[\begin{align} &\Big(\int_{\Gamma^{2^jl(\Delta)}(y)}\frac{|W_2(f)|^2}{t^{n+1}}dxdt\Big)^{1/2}\nonumber \\ &=\Big(\int_0^{2^{j-1}l(\Delta)} \frac{1}{t^{n+1}}\int_{\Delta_{t/2}(y)}\Big|\int_0^t2\tau e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)d\tau\Big|^2dx dt\Big)^{1/2}\nonumber \\ &\lesssim \Big(\int_0^{2^{j-1}l(\Delta)} \frac{1}{t^{n+3}}\Big(\int_0^t\frac{2^{jn/2}l(\Delta)^{n/2+1}}{\sqrt{t^2-\tau^2}\tau^{n}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{\tau^2}}\nonumber \\ &\qquad +\frac{2^{jn/2}l(\Delta)^{n/2+1}}{\sqrt{t^2-\tau^2}^{n+1}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}} + \frac{2^{jn}l(\Delta)^{3n/2+1}}{\sqrt{t^2-\tau^2}^{n+1}\tau^{n}}e^{-\frac{2^{2(j-1)} l(\Delta)^2}{t^2-\tau^2}}e^{-\frac{l(\Delta)^2}{\tau^2}}d\tau\Big)^2 dt\Big)^{1/2}.\label{eq:InArgumentEquation1} \end{align}\tag{20}\] Here we use the observation that a function of type $(0,t]\to\mathbb{R}, \rho\mapsto \frac{1}{\rho^n}e^{-\frac{c}{\rho^2}}$ is maximized in $\rho=t$ if $t\leq c$ and in $\rho=c$ if $t\geq c$. Applying this observation to each of the expressions in 20 gives \[\begin{align} \eqref{eq:InArgumentEquation1}&\lesssim \Big(\int_0^{2^{j-1}l(\Delta)} \Big(\frac{2^{jn}l(\Delta)^{n+2}}{t^{3n+3}} + \frac{2^{2jn}l(\Delta)^{3n+2}}{t^{5n+3}}\Big)\nonumber \\ &\qquad\qquad \cdot e^{-2\frac{2^{2(j-1)} l(\Delta)^2}{t^2}}\Big(\int_0^t\frac{1}{\sqrt{t^2-\tau^2}}d\tau\Big)^2 dt\Big)^{1/2}\nonumber \\ &\lesssim \Big(\int_0^{2^{j-1}l(\Delta)}\Big(\frac{1}{2^{j(2n+3)}l(\Delta)^{2n+1}} + \frac{1}{2^{j(3n+3)}l(\Delta)^{2n+1}}\Big) dt\Big)^{1/2}\nonumber \\ &\lesssim \Big(\frac{1}{2^{j(n+1)}}+\frac{1}{2^{j(n+n/2+1)}}\Big)\frac{1}{l(\Delta)^{n}} \lesssim \frac{1}{2^{j(n+1)}}\frac{1}{l(\Delta)^{n}}.\label{eq:Proofshortconeheight} \end{align}\tag{21}\]

On the other hand we have for the away part of the cone by Minkowski inequality

\[\begin{align} &\Big(\int_{\Gamma(y)\setminus\{t\leq 2^{j-1}l(\Delta)\}}\frac{|W_2(f)|^2}{t^{n+1}}dxdt\Big)^{1/2}\nonumber \\ &=\Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{n+1}}\int_{\Delta_{t/2}(y)}\Big|\int_0^t2\tau e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)d\tau\Big|^2dx dt\Big)^{1/2}\nonumber \\ &\lesssim\Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{n+1}}\Big(\int_0^{t/2}\tau \Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))}d\tau\Big)^2 dt\Big)^{1/2}\nonumber \\ &\qquad+ \Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{n+1}}\Big(\int_{t/2}^t\tau \Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))}d\tau\Big)^2 dt\Big)^{1/2}.\label{eq:NonlocalConeEstimateW2} \end{align}\tag{22}\] We split the inner integral into two integrals, one over small $\tau$ and one over large $\tau$.

We now look at the inner $L^2$ norm in just the $x$ components. For $t/2\leq \tau\leq t$ we have by Proposition 1 and Hölder’s inequality \[\begin{align} &\tau\Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))} \\ &\qquad\lesssim \frac{\tau}{\sqrt{t^2-\tau^2}}\Vert\partial_t A\Vert_{\infty}\Vert\nabla e^{-\tau^2L^t}f\Vert_{L^2(\mathbb{R}^n)} \lesssim \frac{\tau}{t\sqrt{t^2-\tau^2}\tau^{n+1}}\Vert f\Vert_{L^1(\mathbb{R}^n)} \\ &\qquad\lesssim \frac{1}{t\sqrt{t^2-\tau^2}t^{n}}\Vert f\Vert_{L^1(\mathbb{R}^n)} \lesssim \frac{l(\Delta)^{n/2+1}}{t\sqrt{t^2-\tau^2}t^{n}}\Vert \nabla_{||}f\Vert_{L^2(\mathbb{R}^n)} \lesssim \frac{l(\Delta)}{t\sqrt{t^2-\tau^2}t^{n/2}}. \end{align}\]

For $0\leq \tau\leq t/2$ we proceed with a dualising function $g\in L^2(\Delta_{t/2}(y))$ and the same ideas as in the local cone part to get \[\begin{align} &\tau\Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))} \\ &\lesssim \tau\Vert\partial_t A\Vert_{\infty}\Big(\Vert \nabla e^{-(t^2-\tau^2)L^t}g \Vert_{L^2(2\Delta)}\Vert\nabla e^{-\tau^2L^t}f\Vert_{L^2(2\Delta)} \\ &\qquad\qquad + \Vert \nabla e^{-(t^2-\tau^2)L^t}g \Vert_{L^2(\mathbb{R}^n)}\Vert\nabla e^{-\tau^2L^t}f\Vert_{L^2(\mathbb{R}^n\setminus 2\Delta)}\Big) \\ &\lesssim\frac{\tau}{t} \Big(\frac{l(\Delta)^{n/2}}{\sqrt{t^2-\tau^2}^{n+1}}\Vert g\Vert_{L^1(\Delta_{t/2}(y))}\Vert \nabla_{||} f\Vert_{L^2(\mathbb{R}^n)} \\ &\qquad\qquad+ \frac{1}{\sqrt{t^2-\tau^2}\tau^{n+1}}\Vert g \Vert_{L^2}\Vert f\Vert_{L^1(\Delta)}\big(\int_{l(\Delta)}^\infty r^{n-1}e^{-\frac{r^2}{\tau^2}}dr\big)^{1/2}\Big) \\ &\lesssim \frac{\tau}{t}\Big(\frac{l(\Delta)^{n/2}}{\sqrt{t^2-\tau^2}^{n+1}}\Vert g\Vert_{L^1(\Delta_{t/2}(y))}\Vert \nabla_{||} f\Vert_{L^2(\mathbb{R}^n)} \\ &\qquad\qquad+ \frac{l(\Delta)^{n/2}}{\sqrt{t^2-\tau^2}\tau^{n+1}}\Vert g \Vert_{L^2}\Vert f\Vert_{L^1(\Delta)}e^{-c\frac{l(\Delta)^2}{\tau^2}}\Big) \\ &\lesssim \frac{\tau}{t}\Big(\frac{l(\Delta)^{n/2}t^{n/2}}{\sqrt{t^2-\tau^2}^{n+1}}\Vert g\Vert_{L^2}\Vert \nabla_{||} f\Vert_{L^2(\mathbb{R}^n)} + \frac{l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}\tau^{n+1}}\Vert g \Vert_{L^2}\Vert \nabla_{||}f\Vert_{L^2(\Delta)}e^{-c\frac{l(\Delta)^2}{\tau^2}}\Big) \\ &\lesssim \frac{\tau}{t}\Big(\frac{l(\Delta)^{n/2}t^{n/2}}{\sqrt{t^2-\tau^2}^{n+1}}\Vert \nabla_{||} f\Vert_{L^2(\mathbb{R}^n)} + \frac{l(\Delta)^{n+1}}{\sqrt{t^2-\tau^2}t^{n+1}}\Vert \nabla_{||}f\Vert_{L^2(\Delta)}\Big) \\ &\lesssim \frac{l(\Delta)^{n/2+1}}{\sqrt{t^2-\tau^2} t^{n/2+1}}\Vert\nabla_{||} f\Vert_{L^2(\mathbb{R}^n)}\lesssim \frac{l(\Delta)}{\sqrt{t^2-\tau^2} t^{n/2+1}}. \end{align}\] Hence we obtained the same bound for both small and large $\tau$, and we can continue 22 with \[\begin{align} &\Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{n+1}}\Big(\int_0^{t/2}\tau \Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))}d\tau\Big)^2 dt\Big)^{1/2} \\ &\qquad+ \Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{n+1}}\Big(\int_{t/2}^t\tau \Vert e^{-(t^2-\tau^2)L^t}\mathrm{div}(\partial_tA_{\Vert}\nabla e^{-\tau^2L^t}f)\Vert_{L^2(\Delta_{t/2}(y))}d\tau\Big)^2 dt\Big)^{1/2} \\ &\lesssim\Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{l(\Delta)^{2}}{t^{2n+3}}\Big(\int_0^{t}\frac{1}{\sqrt{t^2-\tau^2}} d\tau\Big)^2 dt\Big)^{1/2} \lesssim\Big(\int_{2^{j-1}l(\Delta)}^\infty \frac{l(\Delta)^{2}}{t^{2n+3}} dt\Big)^{1/2} \\ &\lesssim \frac{1}{2^{j(n+1)}l(\Delta)^{n}}. \end{align}\] Together with 21 this yields \[\begin{align} \Vert \mathcal{A}(W_2(f))\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)}&\lesssim \sum_{j\geq 2}\int_{2^{j+1}\Delta\setminus 2^j\Delta} \Big(\int_{\Gamma^{2^{j}l(\Delta)}(y)}\frac{|W_2(f)|^2}{t^{n+1}}dxdt\Big)^{1/2} \\ &\qquad\qquad + \Big(\int_{\Gamma(y)\setminus\{t\leq 2^{j-1}l(\Delta)\}}\frac{|W_2(f)|^2}{t^{n+1}}dxdt\Big)^{1/2}dy \\ &\lesssim \sum_{j\geq 2} \big(2^{j+1}l(\Delta)\big)^n\frac{1}{2^{j(n+1/2)}l(\Delta)^n}\leq C. \end{align}\]

Lastly, [item:HardySquareBoundNabla] follows from [item:HardySquareBound] and Cacciopolli type inequality Proposition 1 like previously in [item:PointwiseLocalSquareBoundNablaW2] of Lemma 7. ◻

6 Area function bounds on $\nabla_{||}\mathcal{P}_t$ and $W_1$↩︎

As we saw in the proof of Lemma 4, we would like to split off the local part when proving the Hardy-Sobolev to $L^1$ bound for $W_1$. Therefore, we used $L^\beta-L^\beta$ boundedness of the area function of $W_2$ (see Lemma 1). For $\nabla_{||}\mathcal{P}_t$ and $W_1$ we would like to proceed similarly. Therefore, the first subsection establishes $L^2-L^2$ boundedness of the area functions of $\nabla_{||}\mathcal{P}_t$ and $W_1$.

6.1 Area function of $\nabla_{||}\mathcal{P}_t$ in $L^2$↩︎

Proposition 1. For $f\in W^{1,2}(\mathbb{R}^n)$ it holds that \[\begin{align} \Vert \mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^2(\mathbb{R}^n)}\lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}.\label{eq:L2BoundAreaNablaP} \end{align}\qquad{(7)}\]

The main idea of the proof is inspired by Proposition 5.1 in [27], but since our operator has additional dependencies in the transversal $t-$direction, we require completely new ideas.

Proof. Before we commence with the proof, we observe the following using only the kernel estimates ( Proposition 1) of the operators $e^{-tL},\partial_te^{-tL}$: Let $f$ be a function with $\mathrm{supp}(f)\subset E$ and let $E, F\subset\mathbb{R}^n$ be two disjoint sets with $d:=\mathrm{dist}(E,F)>0$. We call $\tilde{F}:=F+B(0,d/2)$ an enlargement of $F$ and we choose a cut-off function $\psi\in C_0(\tilde{F})$ such that $\psi\equiv 1$ on $F$ and $|\nabla \psi|\lesssim \frac{1}{d}$. Then \[\begin{align} \Vert \nabla e^{-s^2L^s}f\Vert_{L^2(F)}^2&\lesssim \int_{\tilde{F}} A_{\Vert}(x,s)\nabla e^{-s^2L^s}f(x)\cdot \nabla e^{-s^2L^s}f(x)\psi^2(x) dx \\ &\lesssim \int_{\tilde{F}} L^s e^{-s^2L^s}f(x) e^{-s^2L^s}f(x)\psi^2(x) \\ &\qquad + 2\psi(x)\nabla e^{-s^2L^s}f(x)\cdot \nabla \psi(x) e^{-s^2L^s}f(x) dx \\ &\lesssim \Vert L^s e^{-s^2L^s}f\Vert_{L^2(\tilde{F})} \Vert e^{-s^2L^s}f\Vert_{L^2(\tilde{F})} \\ &\qquad + \sigma \Vert \psi\nabla e^{-s^2L^s}f \Vert^2_{L^2(\mathbb{R}^n)} + \frac{1}{\sigma}\Vert \frac{1}{d}e^{-s^2L^s}f\Vert_{L^2(\tilde{F})}^2. \end{align}\] For a sufficiently small choice of $\sigma$ we can hide the third term on the left hand side and use the kernel estimates ( Proposition 1) to conclude for $x\in \tilde{F}$ \[\begin{align} e^{-s^2L^s}f(x)=\int_{E}K_{s^2}(x,y)f(y)dy\lesssim \frac{1}{s^n}e^{-c\frac{d^2}{s^2}}\int_{E}f(y)dy=\frac{1}{s^n}e^{-c\frac{d^2}{s^2}}\Vert f\Vert_{L^1}, \end{align}\] and similarly \[\Vert L^s e^{-s^2L^s}(f)\Vert_{L^2(\tilde{F})}\lesssim \frac{\sqrt{\sigma(\tilde{F})}}{s^{n+2}}e^{-c\frac{d^2}{s^2}}\Vert f\Vert_{L^1},\] whence in total \[\begin{align} \Vert \nabla e^{-s^2L^s}(f)\Vert_{L^2(F)}^2\lesssim \big(\frac{\sigma(\tilde{F})}{s^{2n+2}}+\frac{\sigma(\tilde{F})}{s^{2n}d^2}\big)e^{-c\frac{d^2}{s^2}}\Vert f\Vert_{L^1}^2\label{observation:KernelIntoL1}. \end{align}\tag{23}\]
Now, we us turn to the proof of ?? . Let us fix a small $a>0$ and consider \[\int_a^\infty \int_{\mathbb{R}^n}\frac{|\nabla_{||}\mathcal{P}_sf|^2}{s} ds=:\int_a^\infty I ds.\] For this choice of $a$ there exists a scale $k$ with $2^{-k}\approx a$ and the collection $\mathcal{D}_k$ consists of boundary balls $Q$ with $l(Q)\approx 2^{-k}\approx a$ that cover $\partial\Omega$ in such a way that the collection of $2Q$ have finite overlap, i.e. $1\leq |\sum_{Q\in \mathcal{D}_k} \chi_{Q}|\leq N$ for some $N\in \mathbb{N}$. Note that this $N$ is independent of $a$. Then for $s\geq a$ we have \[\begin{align} s^2I&=\sum_{Q\in\mathcal{D}_k} s\Vert\nabla e^{-s^2 L^s}f\Vert_{L^2(Q)}^2=\sum_{Q\in\mathcal{D}_k}s\Vert\nabla e^{-s^2 L^s}(f-(f)_{2Q})\Vert_{L^2(Q)}^2 \\ &\lesssim\sum_{Q\in\mathcal{D}_k}s\Vert\nabla e^{-s^2 L^s}(\chi_{2Q}(f-(f)_{2Q}))\Vert_{L^2(Q)}^2 \\ &\qquad + \sum_{Q\in\mathcal{D}_k}s\Vert\nabla e^{-s^2 L^s}(\chi_{\mathbb{R}^n\setminus 2Q}(f-(f)_{2Q}))\Vert_{L^2(Q)}^2:=J+K. \end{align}\]

By observation 23 we obtain for the second term \[\begin{align} K\lesssim &\sum_{Q\in\mathcal{D}_k}\Big(\sum_{l\geq 1} \sqrt{s}\Vert\nabla e^{-s^2 L^s}(\chi_{2^{l+1}Q\setminus 2^lQ}(f-(f)_{2Q}))\Vert_{L^2(Q)}\Big)^2 \\ &\lesssim \sum_{Q\in\mathcal{D}_k}\Big(\sum_{l\geq 1}\Big(\frac{a^{n/2}}{s^{n+1/2}}+\frac{a^{n/2-1}}{s^{n-1/2}}\Big)e^{-c\frac{2^{2l}a^2}{s^2}}\Vert f-(f)_{2Q}\Vert_{L^1(2^{l+1}Q\setminus 2^lQ)}\Big)^2 \end{align}\] Analogously to previously in 17 but now for $L^1$ norms, we have by Poincaré inequality \[\Vert f-(f)_{2Q}\Vert_{L^1(2^{l+1}Q\setminus 2^lQ)}\lesssim (l+1)2^{(l+1)(n+1)}a^{n+1}\inf_{x\in Q}M[|\nabla f|](x),\] and hence \[\begin{align} K&\lesssim \sum_{Q\in\mathcal{D}_k}\Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^{2}}}\sqrt{\int_{Q}M[\nabla f]^2(x)dx}\Big)^2 \\ &\lesssim \sum_{Q\in\mathcal{D}_k}\Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^{2}}}\Big) \\ &\cdot\Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^{2}}}\int_{Q}M[\nabla f]^2(x)dx\Big) \\ &\lesssim \Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^{2}}}\Big) \\ &\cdot\Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^{2}}}\sum_{Q\in\mathcal{D}_k}\int_{Q}M[\nabla f]^2(x)dx\Big) \\ &\lesssim \Big(\sum_{l\geq 1}\frac{(l+1)2^{(n+1)(l+1)}a^{n}}{s^{n-1/2}}e^{-c\frac{2^{2l}a^2}{s^2}}\Big)^2\Vert M[\nabla f]\Vert_{L^2(\mathbb{R}^n)}^2. \end{align}\] We can bound $(l+1)\leq 2^l$ and, as previously in the proof of Lemma 4, we can consider the sum over $l$ as Riemann sum of the integral \[\frac{a^{n}}{s^{n-1/2}}\int_1^\infty y^{n+2}e^{-y^2\frac{a^2}{s^2}}dy=\frac{s^{5/2}}{a^2}\int_{a/s}^\infty z^{n+2}e^{-z^2}dz.\] If $n$ is odd, then $\int_{a/s}^\infty z^{n+2}e^{-z^2}dz=P(a/s)e^{-\frac{a^2}{s^2}}$, where $P$ is a polynomial of degree $n-1$. If $n$ is even, we can reduce to the odd case by $\int_{a/s}^\infty z^{n+2}e^{-z^2}dz\leq \frac{s}{a}\int_{a/s}^\infty z^{n+3}e^{-z^2}dz\leq \frac{s}{a}P(a/s)e^{-\frac{a^2}{s^2}}$. Hence for some polynomial $P$ we have \[\frac{s^{5/2}}{a^2}\int_{a/s}^\infty z^{n+2}e^{-z^2}dz\lesssim \frac{s^{7/2}}{a^3}P(a/s)e^{-\frac{a^2}{s^2}}.\]

Thus, we obtain that \[\begin{align} \int_a^\infty \frac{1}{s^2}K ds&\lesssim \int_a^\infty \frac{1}{s^2}\Big(\frac{s^{7/2}}{a^3}P(a/s)e^{-\frac{a^2}{s^2}}\Big)^2\Vert M[\nabla_{||} f]\Vert_{L^2(\mathbb{R}^n)}^2 \\ &\lesssim \Vert\nabla f\Vert_{L^2}^2 \int_a^\infty \frac{s^5}{a^6}P(a/s)e^{-\frac{a^2}{s^2}} ds. \end{align}\] Since $\frac{a}{s}$ stays in the interval $(0,1)$, the polynomial is bounded and we can bound the integral expression independently of the choice of $a$, whence the same bound remains valid when $a$ tends to $0$. Hence this term is bounded above by $\Vert\nabla f\Vert_{L^2}^2$.

For the first term $J$, we abbreviate notation by setting $f^{Q}:=\chi_{2Q}(f-(f)_{2Q})$ and continue with \[\begin{align} J&\lesssim \sum_{Q\in\mathcal{D}_k}s\Vert\nabla e^{-s^2 L^s}f^{Q}\Vert_{L^2(Q)}^2=\sum_{Q\in\mathcal{D}_k}\int_{\mathbb{R}^n}A_{\Vert}\nabla e^{-s^2 L^{s}}f^{Q} \cdot \nabla e^{-s^2 L^{s}}f^{Q} s dx \\ &=\sum_{Q\in\mathcal{D}_k}\int_{\mathbb{R}^n}L^{s}e^{-s^2 L^{s}}f^{Q} \cdot e^{-s^2 L^{s}}f^{Q} s dx \\ &=\sum_{Q\in\mathcal{D}_k}\Big[\int_{\mathbb{R}^n}\partial_s\big(e^{-s^2 L^{s}}f^{Q}\big) \cdot e^{-s^2 L^{s}}f^{Q} dx \\ &\qquad -\int_{\mathbb{R}^n}\Big(\int_0^s2\tau e^{-(s^2-\tau^2) L^{s}}\mathrm{div}(\partial_sA_{\Vert}\nabla e^{-\tau^2L^s}f^{Q})d\tau\Big) e^{-s^2 L^s}f^{Q} dx \\ &=\sum_{Q\in\mathcal{D}_k} J^{s,Q}_1+J^{s,Q}_2. \end{align}\] Here we used that $\partial_s\big(e^{-s^2 L^{s}}f_{Q}\big)$ can be computed like $\partial_s\big(e^{-s^2 L^{s}}f\big)$ in Section 2.2.

For $J_2^{s,Q}$ we have by integration by parts and 8 \[\begin{align} &\int_a^\infty J_2^{s,Q}ds \\ &=\int_a^\infty \frac{1}{s^2}\int_{\mathbb{R}^n}\Big(\int_0^s2\tau e^{-(s^2-\tau^2) L^{s}}\mathrm{div}(\partial_sA_{\Vert}\nabla e^{-\tau^2L^s}f^{Q})d\tau\Big) e^{-s^2 L^s}f^{Q} dx ds\nonumber \\ &=\int_{\mathbb{R}^n}\int_a^\infty \frac{1}{s^2}\int_0^s2\tau \partial_sA_{\Vert}\nabla e^{-\tau^2L^s}f^{Q} \cdot\nabla e^{-(s^2-\tau^2) L^{s}}e^{-s^2 L^s}f^{Q} d\tau ds dx\nonumber \\ &\lesssim \Vert\sup_{B(x,t,t/2)} |\partial_sA_{\Vert}|\Vert_{\mathcal{C}} \int_{\mathbb{R}^n}\tilde{N}_{1,a}\Big( \frac{1}{s^2}\int_0^s2\tau |\nabla e^{-\tau^2L^s}f^{Q}||\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}| d\tau\Big) dx\label{eq:nontangnetialexpresssionbound1} \end{align}\tag{24}\] where $\tilde{N}_{1,a}$ the the away truncated version of the nontangential maximal function (cf. 4 ). Let us discuss the appearing nontangential maximal function. For $t\geq a$ and $x\in \mathbb{R}^n$, Hölder yields \[\begin{align} &\fint_{B(x,t,t/2)}\Big(\frac{1}{s^2}\int_0^s2\tau |\nabla e^{-\tau^2L^s}f^{Q}| |\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}| d\tau\Big) dyds\nonumber \\ &\lesssim \fint_{t/2}^{3t/2}\frac{1}{s^2}\int_0^s2\tau \big(\fint_{\Delta(x,t/2)}|\nabla e^{-\tau^2L^s}f^{Q}|^2dy\big)^{1/2} \\ &\cdot\big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}|^2dy\big)^{1/2} d\tau ds.\label{eq:nontangentialneed1} \end{align}\tag{25}\] We can now distinguish two cases: First, we assume that $x\in 8Q$, then by Proposition 1 \[\begin{align} \big(\fint_{\Delta(x,t/2)}|\nabla e^{-\tau^2L^s}f^{Q}|^2dy\big)^{1/2}\leq\frac{1}{t^{n/2}}\big(\int_{\mathbb{R}^n}|\nabla e^{-\tau^2L^s}f^{Q}|^2dy\big)^{1/2}\lesssim \frac{1}{t^{n/2}}\Vert \nabla f\Vert_{L^2(2Q)} \end{align}\] and \[\begin{align} \big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}|^2dy\big)^{1/2} \lesssim \frac{1}{t^{n/2}}\Vert \nabla f\Vert_{L^2(2Q)}. \end{align}\] Hence for 25 we obtain in this case \[\begin{align} &\fint_{B(x,t,t/2)}\Big(\frac{1}{s^2}\int_0^s2\tau |\nabla e^{-\tau^2L^s}f^{Q}| |\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}| d\tau\Big) dyds \\ &\lesssim \frac{1}{t^{n}}\Vert \nabla f\Vert_{L^2(2Q)}^2\fint_{t/2}^{3t/2}\int_0^s\frac{2\tau}{s^2}d\tau ds\lesssim \frac{1}{t^{n}}\Vert \nabla f\Vert_{L^2(2Q)}^2. \end{align}\]

Second, we assume that $x\in \mathbb{R}^n\setminus 8 Q$. Let us assume that \[2^{j+1}a\approx 2^{j+1}l(Q)\geq\mathrm{dist}(x, Q)\geq 2^jl(Q)\approx 2^ja.\] We distinguish two more sub cases depending on whether $t$ is greater or smaller than $2^{j-1}a$. If we assume that $a\leq t\leq 2^{j-1}a$, then $\mathrm{dist}(\Delta(x,t/2),Q)\approx 2^{j-1}a$ and we obtain by off-diagonal estimates ( Proposition 1) and Poincaré’s inequality \[\begin{align} \big(\fint_{\Delta(x,t/2)}|\nabla e^{-\tau^2L^s}f^{Q}|^2dy\big)^{1/2}&\leq\frac{e^{-\frac{2^{2(j-1)}a^2}{\tau^2}}}{\tau t^{n/2}}\Vert f-(f)_{2Q}\Vert_{L^2(2Q)} \\ &\lesssim \frac{ae^{-\frac{2^{2(j-1)}a^2}{\tau^2}}}{\tau t^{n/2}}\Vert \nabla f\Vert_{L^2(2Q)}, \end{align}\] and \[\begin{align} &\big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}|^2dy\big)^{1/2} \\ &\lesssim \big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}(\chi_{2^{j-2}Q} e^{-s^2 L^s}f^{Q})|^2dy\big)^{1/2} \\ &\qquad + \big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}(\chi_{\mathbb{R}^n\setminus 2^{j-2}Q} e^{-s^2 L^s}f^{Q})|^2dy\big)^{1/2} \\ &\lesssim \big(\frac{e^{-\frac{2^{2(j-2)}a^2}{s^2-\tau^2}}}{\sqrt{s^2-\tau^2} t^{n/2}} + \frac{e^{-\frac{2^{2(j-2)}a^2}{s^2}}}{s t^{n/2}}\big)\Vert f-(f)_{2Q}\Vert_{L^2(2Q)} \\ &\lesssim a\big(\frac{e^{-\frac{2^{2(j-2)}a^2}{s^2-\tau^2}}}{\sqrt{s^2-\tau^2} t^{n/2}} + \frac{e^{-\frac{2^{2(j-2)}a^2}{s^2}}}{s t^{n/2}}\big)\Vert \nabla f\Vert_{L^2(2Q)}. \end{align}\] Hence for 25 we obtain in this case the bound \[\begin{align} \fint_{t/2}^{3t/2} \frac{1}{s^2}\int_0^s2\tau a^2\frac{e^{-\frac{2^{2(j-1)}a^2}{\tau^2}}}{\tau t^{n/2}}\big(\frac{e^{-\frac{2^{2(j-2)}a^2}{s^2-\tau^2}}}{\sqrt{s^2-\tau^2} t^{n/2}} + \frac{e^{-\frac{2^{2(j-2)}a^2}{s^2}}}{s t^{n/2}}\big)\Vert \nabla f\Vert_{L^2(2Q)}^2d\tau ds. \end{align}\] Since $0\leq \tau\leq s\approx t$ and the function $\rho\mapsto e^{-\frac{2^{2(j-1)}a^2}{\rho^2}}$ is monotonically increasing, we can bound each of the exponential functions by some constant multiple of $e^{-\frac{2^{2(j-2)}a^2}{t^2}}$. Next, we observe that the function $t\mapsto \frac{1}{t^n}e^{-\frac{2^{2(j-1)}a^2}{t^2}}$ is maximized for $t=2^{j-1}a$. Hence we can continue with \[\begin{align} &\lesssim \fint_{t/2}^{3t/2}\int_0^s\big(a^2\frac{1}{\sqrt{s^2-\tau^2} t^{n+2}} + \frac{1}{ st^{n+2}}\big)e^{-\frac{2^{2(j-2)}a^2}{t^2}}\Vert \nabla f\Vert_{L^2(2Q)}^2d\tau ds \\ &\lesssim \fint_{t/2}^{3t/2}\int_0^s\big(a^2\frac{1}{\sqrt{s^2-\tau^2} 2^{(j-2)(n+3)}a^{n+2}} + \frac{1}{ s2^{(j-2)(n+2)}a^{n+2}}\big)\Vert \nabla f\Vert_{L^2(2Q)}^2d\tau ds \\ &\lesssim 2^{-(j-2)(n+2)}a^{-n}\Vert \nabla f\Vert_{L^2(2Q)}^2. \end{align}\]

For the second sub case, where we assume $t\geq 2^{j-1}a$, we proceed similarly to the case of $x$ close to $Q$. Here, we have for 25 by Hölder and Proposition 1 \[\begin{align} &\fint_{t/2}^{3t/2}\frac{1}{s^2}\int_0^s2\tau \big(\fint_{\Delta(x,t/2)}|\nabla e^{-\tau^2L^s}f^{Q}|^2dy\big)^{1/2} \\ &\cdot\big(\fint_{\Delta(x,t/2)}|\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}|^2dy\big)^{1/2} d\tau ds \\ &\lesssim \fint_{t/2}^{3t/2}\frac{1}{s^2}\int_0^s\frac{2}{\sqrt{s^2-\tau^2}t^n}\Vert f-(f)_{2Q}\Vert_{L^2(2Q)}^2 d\tau ds \\ &\lesssim \frac{a^2}{t^{n+2}}\Vert \nabla f\Vert_{L^2(2Q)}^2 d\tau ds\lesssim 2^{-(j-2)(n+2)}a^{-n}\Vert \nabla f\Vert_{L^2(2Q)}^2. \end{align}\]

In total, we obtain now for the integral over the nontangential maximal function in 24 \[\begin{align} &\int_{8Q}\tilde{N}_{1,a}\Big( \frac{1}{s^2}\int_0^s2\tau |\nabla e^{-\tau^2L^s}f^{Q}||\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}| d\tau\Big) dx \\ &\qquad + \sum_{j\geq 3} \int_{2^{j+1}Q\setminus 2^{j}Q}\tilde{N}_a\Big( \frac{1}{s^2}\int_0^s2\tau |\nabla e^{-\tau^2L^s}f^{Q}||\nabla e^{-(s^2-\tau^2) (L^{s})^*}e^{-s^2 L^s}f^{Q}| d\tau\Big) dx \\ &\lesssim 8|Q| a^{-n}\Vert\nabla f\Vert_{L^2(2Q)}^2 + \sum_{j\geq 3}|2^{j+1}Q| 2^{-(j-2)(n+2)}a^{-n} \Vert\nabla f\Vert_{L^2(2Q)}^2\lesssim \Vert\nabla f\Vert_{L^2(2Q)}^2. \end{align}\]

Hence, we obtain in total for the sum over all $Q$ \[\begin{align} \int_a^\infty \sum_{Q\in\mathcal{D}_k} \frac{1}{s^2}J_2^{s,Q}&= \sum_{Q\in\mathcal{D}_k}\Vert\nabla f\Vert_{L^2(2Q)}^2\lesssim \Vert\nabla f\Vert_{L^2(\mathbb{R}^n)}^2. \end{align}\]

Lastly, for $J_1^{s,Q}$ we have \[\begin{align} \int_a^\infty \frac{1}{s^2}\sum_{Q\in\mathcal{D}_k} J_1^{s,Q}=-\sum_{Q\in\mathcal{D}_k}\int_a^\infty \frac{1}{s^2}\partial_s\big(\Vert e^{-s^2L^s}f_{s,Q}\Vert_{L^2(\mathbb{R}^n)}^2\big)ds. \end{align}\] Since the integrand is nonnegative, we can again use Proposition 1 and Poincaré’s inequality to bound this above by \[\begin{align} \int_a^\infty \frac{1}{s^2}\sum_{Q\in\mathcal{D}_k} J_1^{s,Q}&\lesssim -\frac{1}{a^2}\sum_{Q\in\mathcal{D}_k}\int_a^\infty \partial_s\big(\Vert e^{-s^2L^s}f^{Q}\Vert_{L^2(\mathbb{R}^n)}^2\big)ds \\ &\lesssim \frac{1}{a^2}\sum_{Q\in\mathcal{D}_k}\Vert e^{-a^2L^a}f^{Q}\Vert_{L^2(\mathbb{R}^n)}^2 \\ &\lesssim \frac{1}{a^2}\sum_{Q\in\mathcal{D}_k}\Vert f-(f)_{2Q}\Vert_{L^2(2Q)}^2\lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}^2. \end{align}\] Since these upper bounds are all independent of $a$, if we take the limit when $a$ tends to $0$, we obtain that $\int_0^\infty I ds\lesssim \Vert \nabla_{||}f\Vert_{L^2}^2$. ◻

6.2 Area function of $W_1f$ in $L^2$↩︎

The following proposition is the $L^2$ to $L^2$ area function bound on $W_1$. Lemma 7.7 in [2] proves this bound under a stronger condition on the coefficients on $A$, but the adaptation to the $L^1$ Carleson condition on $\partial_t A$ needs significantly more delicate handling of all terms and new ideas.

Proposition 1. For $f\in W^{1,2}(\mathbb{R}^n)$ it holds that \[\begin{align} \Vert \mathcal{A}(W_1f)\Vert_{L^2(\mathbb{R}^n)}\lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}\label{eq:L2BoundAreaW1} \end{align}\qquad{(8)}\] and \[\begin{align} \Vert \mathcal{A}(t\nabla W_1f)\Vert_{L^2(\mathbb{R}^n)}\lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}.\label{eq:L2BoundAreaNablaW1} \end{align}\qquad{(9)}\]

Proof. Now, let us begin with proving ?? . By ellipticity of $A$, integration by parts, and Hölder’s inequality we have \[\begin{align} &\int_\Omega\frac{|sL^se^{-s^2L^s}f|^2}{s}dxds =-\int_0^\infty\int_{\mathbb{R}^n}A_{\Vert}(x,s)\nabla e^{-s^2 L^s}f(x) \cdot \nabla L^te^{-s^2 L^s}f(x) s dx ds \\ & \lesssim\int_0^\infty C_\sigma\frac{1}{s}\Vert\nabla e^{-s^2 L^s}f\Vert_{L^2(\mathbb{R}^n)}^2 + \sigma s^3\Vert\nabla L^se^{-s^2 L^s}f\Vert_{L^2(\mathbb{R}^n)}^2 dx ds \\ &\lesssim\int_0^\infty C_\sigma\frac{1}{s}\Vert\nabla e^{-s^2 L^s}f\Vert_{L^2(\mathbb{R}^n)}^2 + \sigma s\Vert\nabla e^{-s^2 L^s}(sL^sf)\Vert_{L^2(\mathbb{R}^n)}^2 dx ds \\ & =:\int_0^\infty (C_\sigma I+\sigma II) ds. \end{align}\] Here $\sigma$ is a small constant which later will allow us to hide the integral $II_3$ appearing in the estimate of $II$ on the left hand side. We can see that the term $\int_0^\infty I ds=\Vert \mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^2}$ and this has already been dealt with in Proposition 1. Hence it is enough to bound $\int_0^\infty IIds$.

Similarly to the term $J$ in the proof of Proposition 1, we have \[\begin{align} &II=s\Vert\nabla e^{-s^2 L^s}(sL^s f)\Vert_{L^2(\mathbb{R}^n)}^2 \\ &\qquad=\int_{\mathbb{R}^n}L^{s}e^{-s^2 L^{s}}(sL^sf) \cdot e^{-s^2 L^{s}}(sL^sf) t dx \\ &\qquad=\int_{\mathbb{R}^n}\partial_s(sL^se^{-s^2 L^{t}}f) \cdot sL^se^{-t^2 L^t}f dx \\ &\qquad=\partial_s\int_{\mathbb{R}^n}(sL^se^{-s^2 L^{s}}f)^2 dx -\int_{\mathbb{R}^n}\partial_s (sL^se^{-t^2L^s}f)|_{t=s} sL^se^{-s^2 L^s}f dx \\ &\qquad=\partial_s\int_{\mathbb{R}^n}(sL^se^{-s^2 L^{s}}f)^2 dx -\int_{\mathbb{R}^n}s\mathrm{div}(\partial_sA_{\Vert}\nabla e^{-s^2L^s}f)\cdot sL^se^{-s^2 L^s}f dx \\ &\qquad\qquad-\int_{\mathbb{R}^n}L^s e^{-s^2L^s}f\cdot sL^se^{-s^2 L^s}f dx \\ &\qquad\qquad- \int_{\mathbb{R}^n}sL^s\Big(\int_0^s2\tau e^{-(s^2-\tau^2) L^{s}}\mathrm{div}(\partial_sA_{\Vert}\nabla e^{-\tau^2L^s}f)d\tau\Big) sL^se^{-s^2 L^s}f dx \\ &\qquad=:II_1+II_2+II_3+II_4. \end{align}\] First, we note that $II_3$ can be hidden on the left hand side since the whole integral term $II$ is multiplied by a small constant $\sigma$ which we can choose sufficiently small. Furthermore we obtain by integration by parts for $II_4$ \[\begin{align} II_4&=\int_{\mathbb{R}^n}sL^sW_2(x,s) \cdot sL^se^{-s^2 L^s}f(x) dx \\ &=-\int_{\mathbb{R}^n}sA_{\Vert}(x,s)\nabla_{||}W_2(x,s) \cdot \nabla_{||}sL^s e^{-s^2 L^s}f(x) dx \\ &\lesssim s\Vert\nabla W_2(\cdot,s)\Vert_{L^2}^2 + \sigma s\Vert\nabla L^se^{-s^2 L^s}f\Vert_{L^2}^2. \end{align}\]

We can hide the second term on the left hand side. The first term integrated in $s$, i.e. $\int_0^\infty s\Vert\nabla W_2(\cdot,s)\Vert_{L^2}^2 ds$, is the $L^2$ area function bound Corollary 1 and hence bounded by $\Vert \nabla f\Vert_{L^2}^2$.

Next, we have by integration by parts and 8 \[\begin{align} \int_0^\infty II_2 ds&=\int_0^\infty\int_{\mathbb{R}^n}s\partial_sA_{\Vert}(x,s)\nabla e^{-s^2L^s}f(x)\cdot s\nabla_{||}L^se^{-s^2 L^s}f(x) dx ds \\ &\lesssim \Vert \sup_{B(x,t,t/2)}|\partial_s A_{\Vert}|\Vert_{\mathcal{C}}\int_{\mathbb{R}^n}\tilde{N}(|\nabla e^{-s^2L^s}f(x)||s^2\nabla_{||}L^se^{-s^2 L^s}f|)(x) dx. \end{align}\] For the nontangential maximal function we have a pointwise bound in $x\in\partial\Omega$ by taking the supremum over expressions of the form \[\begin{align} &\fint_{B(x,t,t/2)}|\nabla e^{-s^2L^s}f(x)||s^2\nabla_{||}L^se^{-s^2 L^s}f| dxds \\ &\lesssim \Big(\fint_{B(x,t,t/2)}|\nabla e^{-s^2L^s}f(x)|^2dxds\Big)^{1/2}\Big(\fint_{B(x,t,t/2)}|s^2\nabla_{||}L^se^{-s^2 L^s}f|^2 dxds\Big)^{1/2} \\ &\lesssim M[\nabla_{||} f]^2(x). \end{align}\] Here, again, we used Proposition 1.

Together, we obtain for any $a>0$ \[\begin{align} \int_a^\infty IIds&\lesssim\int_a^\infty II_1 ds +\Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}^2 +s\Vert\sup|\partial_s A_{\Vert}|\Vert_{\mathcal{C}}\Vert M[\nabla f]\Vert_{L^2(\mathbb{R}^n)}^2 \end{align}\] where the second term comes from $II_4$. Using the Fundamental Theorem of Calculus, the first term is bounded by \[\begin{align} \int_a^\infty II_1 ds&=\int_a^\infty\partial_s\Big(\int_{\mathbb{R}^n}(sL^se^{-s^2 L^{s}}f)^2 dx\Big) = \Vert aL^ae^{-a^2 L^{a}}f\Vert_{L^2(\mathbb{R}^n)}^2 \lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}^2. \end{align}\] Here we also used Proposition 1. Since all bounds are independent of $a$, in total we get by boundedness of the Hardy-Littlewood maximal function and the $L^1$-Carelson condtion ?? , that $\int_0^\infty IIds \lesssim \Vert \nabla f\Vert_{L^2(\mathbb{R}^n)}$. ◻

6.3 Carleson function bound (Proof of Lemma 5)↩︎

Proof of Lemma 5. The proof for [item:LocalSquareBoundNablaP] and [item:LocalSquareBoundW1] work completely analogously, since we only use off-diagonal estimates ( Proposition 1) and the $L^2-L^2$ boundedness of the area functions ( Proposition 1 and Proposition 1). Thus, it suffices to only present the proof of [item:LocalSquareBoundW1] in the following.

We introduce a smooth cut-off function $\eta\in C_0^\infty(3\Delta)$ with $\eta\equiv 1$ on $2\Delta$ and $|\nabla \eta|\lesssim \frac{1}{l(\Delta)}$ and split the data $f$ into a local and far-away part. We obtain \[\begin{align} &\int_{T(\Delta)}\frac{|W_1f(x,t)|^2}{t}dxdt=\int_{T(\Delta)}\frac{|W_1(f-(f)_{3\Delta})(x,t)|^2}{t}dxdt \\ &\qquad \leq \int_{T(\Delta)}\frac{|W_1((f-(f)_{3\Delta})\eta)(x,t)|^2}{t}dxdt \\ &\qquad \qquad + \int_{T(\Delta)}\frac{|W_1((f-(f)_{3\Delta})(1-\eta))(x,t)|^2}{t}dxdt. \end{align}\] By $L^2-L^2$ boundedness of the area functions ( Proposition 1 or Proposition 1) and Poincaré’s inequality, the first integral is bounded by \[\Vert\nabla( (f-(f)_{3\Delta})\eta)\Vert_{L^2(\mathbb{R}^n)}\lesssim \frac{1}{l(\Delta)}\Vert (f-(f)_{3\Delta}) \Vert_{L^2(3\Delta)} + \Vert \nabla_{||} f\Vert_{L^2(3\Delta)}\lesssim \Vert \nabla_{||} f\Vert_{L^2(3\Delta)}.\] Thus we only need to deal with the second integral, the away part.

We have for a fixed $t\in (0,l(\Delta))$ with off-diagonal estimates ( Proposition 1) for $W_1g(x,t)=t\partial_\tau e^{-\tau L^s}g(x)|_{s=t, \tau=t^2}$ \[\begin{align} &\Vert W_1 ((f-(f)_{3\Delta})(1-\eta))\Vert_{L^2(\Delta)} \\ &\lesssim \Vert W_1 (\chi_{4\Delta}(f-(f)_{3\Delta})(1-\eta))\Vert_{L^2(\Delta)}+\sum_{k\geq 2}\Vert W_1 (\chi_{2^{k+1}\Delta\setminus 2^{k}\Delta}(f-(f)_{3\Delta}))\Vert_{L^2(\Delta)} \\ &\lesssim \frac{1}{t}e^{-c\frac{l(\Delta)^2}{t^2}}\Vert f-(f)_{3\Delta} \Vert_{L^2(4\Delta\setminus 2\Delta)} + \sum_{k\geq 2}\frac{1}{t}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Vert f-(f)_{3\Delta}\Vert_{L^2(2^{k+1}\Delta\setminus 2^k\Delta)}. \end{align}\] As in 17 or in the proof of Proposition 1, Poincaré’s inequality yields \[\begin{align} \Vert f-(f)_{3\Delta}\Vert_{L^2(2^{k+1}\Delta\setminus 2^k\Delta)}&\lesssim (k+1)2^{k+1}l(\Delta)\Vert \nabla_{||}f\Vert_{L^2(\mathbb{R}^n)} \\ &\lesssim 2^{(k+1)(\frac{n}{2}+2)}l(\Delta)^{\frac{n}{2}+1}\Vert \nabla_{||} f\Vert_{L^\infty(\mathbb{R}^n)}. \end{align}\]

Hence, we obtain \[\begin{align} &\frac{1}{t}e^{-c\frac{l(\Delta)^2}{t^2}}\Vert f-(f)_{3\Delta} \Vert_{L^2(4\Delta\setminus 2\Delta)} + \sum_{k\geq 2}\frac{1}{t}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Vert f-(f)_{3\Delta}\Vert_{L^2(2^{k+1}\Delta\setminus 2^k\Delta)} \\ &\lesssim \frac{l(\Delta)^{\frac{n}{2}+1}}{t}\Big(e^{-c\frac{l(\Delta)^2}{t^2}} + \sum_{k\geq 2} (k+1)2^{(k+1)(\frac{n}{2}+1)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Big)\Vert \nabla_{||} f\Vert_{\infty} \\ &\lesssim \frac{l(\Delta)^{\frac{n}{2}+1}}{t}\Big(\sum_{k\geq 1} (k+1)2^{(k+1)(\frac{n}{2}+1)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Big)\Vert \nabla_{||} f\Vert_{\infty} \\ &\lesssim \frac{l(\Delta)^{\frac{n}{2}+1}}{t}\Big(\sum_{k\geq 1} 2^{(k+1)(\frac{n}{2}+2)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Big)\Vert \nabla_{||} f\Vert_{\infty}. \end{align}\]

With this in hand, we return now to \[\begin{align} &\int_{T(\Delta)}\frac{|W_1((f-(f)_{3\Delta})(1-\eta))(x,t)|^2}{t}dxdt \\ &\qquad\lesssim \int_0^{l(\Delta)}\frac{l(\Delta)^{n+2}}{t^3}\Big(\sum_{k\geq 1} 2^{(k+1)(\frac{n}{2}+2)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Big)^2\Vert \nabla_{||} f\Vert_{\infty}^2dt \\ &\qquad \lesssim l(\Delta)^{n+2}\Vert \nabla_{||} f\Vert_{\infty}^2\int_0^{l(\Delta)}\Big(\sum_{k\geq 1} \frac{1}{t^{3/2}}2^{(k+1)(\frac{n}{2}+2)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\Big)^2dt. \end{align}\]

Similar to previous arguments in Lemma 4 and Lemma 1, we can study the function $t\mapsto \frac{1}{t^{3/2}}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}$ for $t\in (0,l(\Delta))$ and observe that it attains its maximum if $t=l(\Delta)$, whence \[\sum_{k\geq 1} \frac{1}{t^{3/2}}2^{(k+1)(\frac{n}{2}+2)}e^{-c\frac{2^{2k}l(\Delta)^2}{t^2}}\lesssim \frac{1}{l(\Delta)^{3/2}}\sum_{k\geq 1} 2^{(k+1)(\frac{n}{2}+2)}e^{-c2^{2k}}.\] We consider the sum as Riemann sums of the corresponding integral \[\int_0^\infty x^{\frac{n}{2}+1}e^{-cx^2}dx,\] which converges. Hence we obtain in total

\[\begin{align} &\int_{T(\Delta)}\frac{|W_1((f-(f)_{3\Delta})(1-\eta))(x,t)|^2}{t}dxdt \\ &\qquad \lesssim l(\Delta)^{n+2}\Vert \nabla_{||} f\Vert_{\infty}^2\int_0^{l(\Delta)}\frac{1}{l(\Delta)^3}dt=|\Delta| \Vert \nabla_{||} f\Vert_{\infty}^2. \end{align}\] ◻

6.4 Hardy-Sobolev bounds (Proof of Lemma 6)↩︎

Proof of Lemma 6. First, we note that it is enough to show \[\Vert\mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^1(\mathbb{R}^n)},\Vert\mathcal{A}(W_1f)\Vert_{L^1(\mathbb{R}^n)}, \Vert\mathcal{A}(t\nabla W_1f)\Vert_{L^1(\mathbb{R}^n)}\leq C\] for all homogeneous Hardy-Sobolev $1/2$-atoms $f$ associated with $\Delta$, whence we assume that $f$ is such an atom going forward.

We begin with showing [item:HardySquareBoundW1]. We split the integral into a local and a far away part \[\begin{align} \Vert\mathcal{A}(W_1f)\Vert_{L^1(\mathbb{R}^n)}=\Vert\mathcal{A}(W_1f)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)} + \Vert\mathcal{A}(W_1f)\Vert_{L^1(5\Delta)}. \end{align}\] For the local part we have by Hölder’s inequality, Proposition 1, and the properties of the Hardy-Sobolev space that \[\begin{align} \Vert\mathcal{A}(W_1f)\Vert_{L^1(5\Delta)}\lesssim \Vert\mathcal{A}(W_1f)\Vert_{L^2(5\Delta)}|\Delta|^{1/2}\lesssim \Vert\nabla_{||} f\Vert_{L^2(\mathbb{R}^n)}|\Delta|^{1/2}\lesssim 1. \end{align}\] For the away part, we make use of observation 18 . We divide $\mathbb{R}^n$ into annuli $2^{j+1}\Delta\setminus 2^j\Delta$ for $j\geq 2$. If we have $y\in 2^{j+1}\Delta\setminus 2^j\Delta$ then \[\begin{align} &\int_{\Gamma(y)}\frac{|W_1(f)|^2}{t^{n+1}}dxdt=\int_0^\infty \frac{1}{t^{n+1}}\int_{\Delta_{t/2}(y)}|tL^te^{-t^2L^t}f(x,t)|^2dx dt \\ &\qquad=\int_0^{2^{j-1}l(\Delta)} \frac{e^{-2\frac{|2^{j}l(\Delta)-\frac{t}{2}|^2}{t^2}}}{t^{2n+3}}\Vert f\Vert_{L^1(\mathbb{R}^n)}^2dx dt + \int_{2^{j-1}l(\Delta)}^\infty \frac{1}{t^{2n+3}}\Vert f\Vert_{L^1(\mathbb{R}^n)}^2 dt \\ &\qquad\lesssim\int_0^{2^{j-1}l(\Delta)} \frac{l(\Delta)^{2}e^{-\frac{2^{2(j-1)}l(\Delta)^2}{t^2}}}{t^{2n+3}}dx dt \\ &\qquad\qquad + \int_{2^{j-1}l(\Delta)}^\infty \frac{l(\Delta)^{2}}{t^{2n+3}} dt \\ &\qquad \lesssim \frac{2^{-j(2n+2)}}{l(\Delta)^{2n}}. \end{align}\] Hence \[\begin{align} \Vert\mathcal{A}(W_1f)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)}&\lesssim \sum_{j\geq 2}\int_{2^{j+1}\Delta\setminus 2^j\Delta}\Big(\int_{\Gamma(y)}\frac{|W_1(f)|^2}{t^{n+1}}dxdt\Big)^{1/2}dy \\ &\lesssim \sum_{j\geq 2}\frac{2^{jn}l(\Delta)^n}{2^{j(n+1)}l(\Delta)^n}\leq C, \end{align}\] and [item:HardySquareBoundW1] follows. Next, [item:HardySquareBoundNablaW1] follows from [item:HardySquareBoundW1] and Proposition 1.

Lastly, for [item:HardySquareBoundNablaP] we can follow the idea of part [item:HardySquareBoundW1] to split \[\begin{align} \Vert\mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^1(\mathbb{R}^n)}=\Vert\mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)} + \Vert\mathcal{A}(\nabla_{||}\mathcal{P}_tf)\Vert_{L^1(5\Delta)} \end{align}\] into a local and far away part. As before, the local part is bounded with Hölder and the $L^2-L^2$ area function bound Proposition 1.

For the away part, again, we want to make use of observation 18 . First, we can note that 18 also holds true for the operator $\frac{\mathcal{P}_t}{t}$, since it satisfies the same kernel bounds as $W_1=tL^t\mathcal{P}_t$ (see Proposition 1). Hence on each cone $\Gamma_\alpha(x)$, by introducing a cut-off function $\eta\in C_0^\infty(\Gamma_{2\alpha}(x))$ with $0\leq \eta \leq 1$, $\eta\equiv 1$ on $\Gamma_{\alpha}(x)$ and $|\nabla \eta|\lesssim \frac{1}{t}$, we obtain \[\begin{align} \int_{\Gamma_\alpha(x)}\frac{|\nabla_{||}\mathcal{P}_tf|^2}{t^{n+1}}dxdt&\lesssim \int_{\Gamma_{2\alpha}(x)}\frac{A\nabla_{||}\mathcal{P}_tf\cdot \nabla_{||} \mathcal{P}_t f \eta^2}{t^{n+1}}dxdt \\ &\lesssim \int_{\Gamma_{2\alpha}(x)}\frac{W_1f\, \frac{\mathcal{P}_t f}{t} \eta^2}{t^{n+1}} + \frac{|\nabla_{||}\mathcal{P}_tf|\eta\, \frac{|\mathcal{P}_t f|}{t}}{t^{n+1}} dxdt. \end{align}\] Using Hölder’s inequality yields \[\begin{align} \mathcal{A}_\alpha(\nabla\mathcal{P}_tf)^2(x)\lesssim \mathcal{A}_{2\alpha}(W_1 f)(x)\mathcal{A}_{2\alpha}\Big(\frac{\mathcal{P}_t}{t}\Big)(x) + \mathcal{A}_{2\alpha}(\eta\nabla \mathcal{P}_t f)(x)\mathcal{A}_{2\alpha}\Big(\frac{\mathcal{P}_t}{t}\Big)(x) \\ \lesssim \mathcal{A}_{2\alpha}(W_1 f)^2(x) + \mathcal{A}_{2\alpha}\Big(\frac{\mathcal{P}_t}{t}\Big)^2(x) + \sigma\mathcal{A}_{2\alpha}(\eta\nabla \mathcal{P}_t f)^2(x) + \frac{1}{\sigma}\mathcal{A}_{2\alpha}\Big(\frac{\mathcal{P}_t}{t}\Big)^2(x) \end{align}\] For a sufficiently small choice of $\sigma>0$ we can hide the third term on the left hand side, and obtain for the far away part \[\Vert\mathcal{A}(\nabla\mathcal{P}_tf)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)}\lesssim \Vert\mathcal{A}(W_1 f)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)} + \Vert\mathcal{A}\Big(\frac{\mathcal{P}_tf}{t}\Big)\Vert_{L^1(\mathbb{R}^n\setminus 5\Delta)}\] Now, we can proceed with both terms as for the away part in [item:HardySquareBoundW1] which only required observation 18 . The increased apertures do not cause any problems, since we can just show $\Vert\mathcal{A}_{\alpha/2}(\nabla \mathcal{P}_tf)\Vert_{L^1}\lesssim \Vert\nabla _{||}f\Vert_{\dot{HS}_{atom}^{1,2}(\mathbb{R}^n)}$ for half the aperture and then use Proposition 4.5 in [24]. This result provides comparability of the $L^p$-norms of area functions with different apertures, and hence allows us to conclude $\Vert\mathcal{A}_{\alpha}(\nabla \mathcal{P}_tf)\Vert_{L^1}\lesssim \Vert\nabla _{||}f\Vert_{\dot{HS}_{atom}^{1,2}(\mathbb{R}^n)}$ for the original aperture. ◻

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Equivalence between solvability of the Dirichlet and Regularity problem under an \(L^1\) Carleson condition on \(\partial_t A\)

Abstract

1 Introduction↩︎

Overview over the proof↩︎

Acknowledgements↩︎

2 Preliminaries↩︎

2.1 Notation and Setting↩︎

2.2 The approximation operator \(\mathcal{P}_t\)↩︎

2.3 Properties of the heat semi-group and \(\mathcal{P}_t\)↩︎

3 Reduction of proof of Theorem 1 to area function bounds↩︎

4 Tent spaces and real interpolation↩︎

4.1 Proof of Lemma 2↩︎

5 Area function bounds on \(W_2\)↩︎

5.1 Carleson function bounds and proof of Lemma 3↩︎

5.2 Hardy-Sobolev bound (Proof of Lemma 4)↩︎

6 Area function bounds on \(\nabla_{||}\mathcal{P}_t\) and \(W_1\)↩︎

6.1 Area function of \(\nabla_{||}\mathcal{P}_t\) in \(L^2\)↩︎

6.2 Area function of \(W_1f\) in \(L^2\)↩︎

6.3 Carleson function bound (Proof of Lemma 5)↩︎

6.4 Hardy-Sobolev bounds (Proof of Lemma 6)↩︎

References↩︎

Subjects

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