## 1 Introduction

In this paper we prove a weak type (1, 1) theorem for the maximal operator associated to a general Ornstein–Uhlenbeck semigroup. We extend the proof given by the third author in 1983 in a symmetric context. Our setting is the following.

In $$\mathbb {R}^n$$ we will consider the semigroup generated by the elliptic operator

\begin{aligned} \mathcal L= \frac{1}{2} \sum _{i,j=1}^n q_{ij} \frac{\partial ^2}{\partial x_i \partial x_j} + \sum _{i,j=1}^n b_{ij} x_i\frac{\partial }{ \partial x_j}, \end{aligned}

or, equivalently,

\begin{aligned} \mathcal L= \frac{1}{2}\, \mathrm {tr} \big ( Q\nabla ^2 \big )+\langle Bx, \nabla \rangle ,\end{aligned}

where $$\nabla$$ is the gradient and $$\nabla ^2$$ the Hessian. Here $$Q = (q_{ij})$$ is a real, symmetric and positive definite $$n\times n$$ matrix, indicating the covariance of $$\mathcal L$$. The real $$n\times n$$ matrix $$B = (b_{ij})$$ is negative in the sense that all its eigenvalues have negative real parts, and it gives the drift of $$\mathcal L$$.

The semigroup is formally $$\mathcal H_t=e^{t\mathcal L}$$, $${t> 0}$$, but to write it more explicitly we first introduce the positive definite, symmetric matrices

(1.1)

and the normalized Gaussian measures $$\gamma _t$$ in $$\mathbb {R}^n$$, with $$t\in (0,+\infty ]$$, having density

\begin{aligned} y\mapsto (2\pi )^{-\frac{n}{2}} (\text {det} \, Q_t)^{-\frac{1}{2} } \exp \left( {-\frac{1}{2} \langle Q_t^{-1}y,y\rangle }\right) \end{aligned}

with respect to Lebesgue measure. Then for functions f in the space of bounded continuous functions in $$\mathbb {R}^n$$ one has

\begin{aligned} \mathcal H_t f(x)= \int f(e^{tB}x-y)\,d\gamma _t (y)\,, \quad x\in \mathbb {R}^n\,, \end{aligned}
(1.2)

a formula due to Kolmogorov. The measure $$\gamma _\infty$$ is invariant under the action of $$\mathcal H_t$$; it will be our basic measure, replacing Lebesgue measure.

We remark that $$\big ( \mathcal H_t \big )_{t> 0}$$ is the transition semigroup of the stochastic process

\begin{aligned}\chi (x,t)= e^{tB}+ \int _0^t e^{(t-s)B} \, dW(s),\end{aligned}

where W is a Brownian motion in $$\mathbb {R}^n$$ with covariance Q.

We are interested in the maximal operator defined as

\begin{aligned} \mathcal H_* f(x)= \sup _{t> 0} \big | \mathcal H_t f(x) \big |. \end{aligned}

Under the above assumptions on Q and B, our main result is the following.

### Theorem 1.1

The Ornstein–Uhlenbeck maximal operator $$\mathcal H_*$$ is of weak type (1, 1) with respect to the invariant measure $$\gamma _\infty$$, with an operator quasinorm that depends only on the dimension and the matrices Q and B.

In other words, the inequality

\begin{aligned} \gamma _\infty \{x\in \mathbb {R}^n : \mathcal H_*f(x)> \alpha \} \le \frac{C}{\alpha }\,\Vert f\Vert _{L^1( \gamma _\infty )},\qquad \alpha >0, \end{aligned}
(1.3)

holds for all functions $$f\in L^1 (\gamma _\infty )$$, with $$C=C(n,Q,B)$$.

For large values of the time parameter, we also obtain a refinement of this result. Indeed, we prove in Proposition 6.1 that

\begin{aligned} \gamma _\infty \left\{ x\in \mathbb {R}^n : \sup _{t> 1}|\mathcal H_t f(x)| > \alpha \right\} \le \frac{C}{ \alpha \ {\sqrt{\log \alpha }}}\, \end{aligned}
(1.4)

for large $$\alpha >0$$ and all normalized functions $$f\in L^1 (\gamma _\infty )$$. Here $$C=C(n,Q,B)$$, and this estimate is shown to be sharp. It cannot be extended to $$\mathcal H_*$$, since the maximal operator corresponding to small values of t only satisfies the ordinary weak type inequality. This sharpening is not surprising, in the light of some recent results for the standard case $$Q=I$$ and $$B=-I$$ by Lehec [8]. He proved the following conjecture, proposed by Ball, Barthe, Bednorz, Oleszkiewicz and Wolff [2]:

For each fixed $$t>0$$, there exists a function $$\psi _t=\psi _t(\alpha )$$, with $$\displaystyle \lim _{\alpha \rightarrow +\infty } \psi _t (\alpha )=0$$, satisfying

\begin{aligned} \gamma _\infty \{x\in \mathbb {R}^n : |\mathcal H_t f(x)| > \alpha \} \le \frac{\psi _t (\alpha )}{\alpha }\, \end{aligned}

for all large $$\alpha >0$$ and all $$f\in L^1(\gamma _\infty )$$ such that $$\Vert f\Vert _{L^1( \gamma _\infty )}=1$$. Lehec proved this conjecture with $$\psi _t (\alpha )={C(t)}/{\sqrt{\log \alpha }}$$ independent of the dimension, and this $$\psi _t$$ is sharp. Our estimates depend strongly on the dimension n, but on the other hand we estimate the supremum over large t.

The history of $$\mathcal H_*$$ is quite long and started with the first attempts to prove $$L^p$$ estimates. When $$\big ( \mathcal H_t \big )_{t> 0}\,$$ is symmetric, i.e., when each operator $$\mathcal H_t$$ is self-adjoint on $$L^2 (\gamma _\infty )$$, then $$\mathcal H_*$$ is bounded on $$L^p (\gamma _\infty )$$ for $$1<p\le \infty$$, as a consequence of the general Littlewood–Paley–Stein theory for symmetric semigroups of contractions on $$L^p$$ spaces [16,   Ch. III].

It is easy to see that the maximal operator is unbounded on $$L^1 (\gamma _\infty )$$. This led, about fifty years ago, to the study of the weak type (1, 1) of $$\mathcal H_*$$ with respect to $$\gamma _\infty$$. The first positive result is due to B. Muckenhoupt [13], who proved the estimate (1.3) in the one-dimensional case with $$Q=I$$ and $$B=-I$$. The analogous question in the higher-dimensional case was an open problem until 1983, when the third author [15] proved the weak type (1, 1) in any finite dimension. Other proofs are due to Menárguez, Pérez and Soria [11] (see also [10, 14]) and to Garcìa-Cuerva, Mauceri, Meda, Sjögren and Torrea [7]. Moreover, a different proof of the weak type (1, 1) of $$\mathcal H_*$$, based on a covering lemma halfway between covering results by Besicovitch and Wiener, was given by Aimar, Forzani and Scotto [1]. A nice overview of the literature may be found in [17,   Ch.4].

In [4] the present authors recently considered a normal Ornstein–Uhlenbeck semigroup in $$\mathbb {R}^n$$, that is, we assumed that $$\mathcal H_t$$ is for each $$t> 0$$ a normal operator on $$L^2 (\gamma _\infty )$$. Under this extra assumption, we proved that the associated maximal operator is of weak type (1, 1) with respect to the invariant measure $$\gamma _\infty$$. This extends earlier work in the non-symmetric framework by Mauceri and Noselli [9], who proved that if $$Q=I$$ and $$B=\lambda (R-I)$$ for some positive $$\lambda$$ and a real skew-symmetric matrix R generating a periodic group, then the maximal operator $$\mathcal H_*$$ is of weak type (1, 1).

In Theorem 1.1 we go beyond the hypothesis of normality. The proof has a geometric core and relies on the ad hoc technique developed by the third author in [15]. It is worth noticing that, while the proof in [4] required an analysis of the special case when $$Q=I$$ and $$B=(-\lambda _1, \ldots , -\lambda _n)$$, with $$\lambda _j>0$$ for $$j=1, \ldots , n$$, and then the application of factorization results, we apply here directly, avoiding many intermediate steps, the "forbidden zones" technique introduced in [15].

Since the maximal operator $$\mathcal H_*$$ is trivially bounded from $$L^{\infty }$$ to $$L^{\infty }$$, we obtain by interpolation the following corollary.

### Corollary 1.2

The Ornstein–Uhlenbeck maximal operator $$\mathcal H_*$$ is bounded on $$L^p (\gamma _\infty )$$ for all $$p>1$$.

This result improves Theorem 4.2 in [9], where the $$L^p$$ boundedness of $$\mathcal H_*$$ is proved for all $$p>1$$ in the normal framework, under the additional assumption that the infinitesimal generator of $$\big ( \mathcal H_t \big )_{t> 0}\,$$ is a sectorial operator of angle less than $$\pi /2$$.

In this paper we focus our attention on the Ornstein–Uhlenbeck semigroup in $$\mathbb {R}^n$$. In view of possible applications to stochastic analysis and to SPDE’s, it would be very interesting to investigate the case of the infinite-dimensional Ornstein-Uhlenbeck maximal operator as well (see [3, 6, 18] for an introduction to the infinite-dimensional setting). The Riesz transforms associated to a general Ornstein–Uhlenbeck semigroup in $$\mathbb {R}^n$$ have been studied in the authors’ paper [5].

The scheme of the paper is as follows. In Sect. 2 we introduce the Mehler kernel $$K_t(x,u)$$, that is, the integral kernel of $$\mathcal H_t$$. Some estimates for the norm and the determinant of $$Q_t$$ and related matrices are provided in Sect. 3. As a consequence, we obtain bounds for the Mehler kernel. In Sect. 4 we consider the relevant geometric features of the problem, and introduce in Sect. 4.1 a system of polar-like coordinates. We also express Lebesgue measure in terms of these coordinates. Sections 5, 6, 7 and 8 are devoted to the proof of Theorem 1.1. First, Sect. 5 introduces some preliminary simplifications of the proof; in particular, we restrict the variable x to an ellipsoidal annulus. In Sect. 6 we consider the supremum in the definition of the maximal operator taken only over $$t> 1$$ and prove the sharp estimate (1.4). Section 7 is devoted to the case of small t under an additional local condition. Finally, in Sect. 8 we treat the remaining case and conclude the proof of Theorem 1.1, by proving the estimate (1.3) for small t under a global assumption.

In the following, we use the “variable constant convention”, according to which the symbols $$c>0$$ and $$C<\infty$$ will denote constants which are not necessarily equal at different occurrences. They all depend only on the dimension and on Q and B. For any two nonnegative quantities a and b we write $$a\lesssim b$$ instead of $$a \le C b$$ and $$a \gtrsim b$$ instead of $$a \ge c b$$. The symbol $$a\simeq b$$ means that both $$a\lesssim b$$ and $$a \gtrsim b$$ hold.

By $$\mathbb {N}$$ we mean the set of all nonnegative integers. If A is an $$n\times n$$ matrix, we write $$\Vert A\Vert$$ for its operator norm on $$\mathbb {R}^n$$ with the Euclidean norm $$|\cdot |$$.

## 2 The Mehler kernel

For $$t>0$$, the difference

\begin{aligned} Q_\infty -Q_t =\int _t^{\infty }e^{sB}Qe^{sB^*}ds \end{aligned}
(2.1)

is a symmetric and strictly positive definite matrix. So is the matrix

\begin{aligned} Q_t ^{-1} - Q_\infty ^{-1} = Q_t^{-1}( Q_\infty -Q_t )Q_\infty ^{-1}, \end{aligned}
(2.2)

and we can define

\begin{aligned} D_t = (Q_t^{-1}-Q_\infty ^{-1} )^{-1} Q_t^{-1} e^{tB}\,,\qquad t>0. \end{aligned}
(2.3)

Then formula (1.2), the definition of the Gaussian measure and some elementary computations yield

\begin{aligned} \mathcal H_t f(x)&= (2\pi )^{-\frac{n}{2}} (\text {det} \, Q_t)^{-\frac{1}{2} } \int f(e^{tB}x-y) \exp \left[ {-\frac{1}{2} \langle Q_t^{-1}y,y\rangle }\right] dy\nonumber \\&= \Big ( \frac{\text {det} \, Q_\infty }{\text {det} \, Q_t} \Big )^{{1}/{2} } \exp \left[ {\frac{1}{2} \langle Q_t^{-1} e^{tB}x, D_t\, x-e^{tB}x\rangle }\right] \nonumber \\&\quad \times \int f(u) \, \exp \left[ {\frac{1}{2} \langle (Q_\infty ^{-1}- Q_t^{-1}) (u-D_t\, x)\,,\, u-D_t\, x\rangle }\right] d\gamma _\infty (u)\,, \end{aligned}
(2.4)

where we repeatedly used the fact that $$Q_\infty ^{-1}- Q_t^{-1}$$ is symmetric. We now express the matrix $$D_t$$ in various ways.

### Lemma 2.1

For all $$x\in \mathbb {R}^n$$ and $$t>0$$ we have

1. (i)

$$D_t = Q_\infty e^{-tB^*} Q_\infty ^{-1}$$;

2. (ii)

$$D_t = e^{tB} + Q_t e^{-tB^*}Q_\infty ^{-1}$$.

### Proof

1. (i)

The formulae (2.1) and (1.1) imply

\begin{aligned} Q_\infty - Q_t= e^{tB} Q_\infty e^{tB^*} \end{aligned}
(2.5)

(see also [12,   formula (2.1)]). From (2.3) and (2.2) it follows that

\begin{aligned} {D_t = Q_\infty ( Q_\infty -Q_t)^{-1}\, e^{tB},}\end{aligned}

and combining this with (2.5) we arrive at (i).

2. (ii)

Multiplying (2.5) by $$e^{-tB^*} Q_\infty ^{-1}$$ from the right, we obtain

\begin{aligned} Q_\infty e^{-tB^*}Q_\infty ^{-1}- Q_t e^{-tB^*}Q_\infty ^{-1} = e^{tB}, \end{aligned}

and (ii) now follows from (i).

$$\square$$

By means of (i) in this lemma, we can define $$D_t$$ for all $$t\in \mathbb {R}$$, and they will form a one-parameter group of matrices.

Now (ii) in Lemma 2.1 yields

\begin{aligned} \langle Q_t^{-1} e^{tB}x, D_t\, x-e^{tB}x\rangle = \langle Q_t^{-1} e^{tB}x, Q_t e^{-tB^*}Q_\infty ^{-1}x \rangle =\langle Q_\infty ^{-1}x ,x \rangle . \end{aligned}

Thus (2.4) may be rewritten as

\begin{aligned} \mathcal H_t f(x)&= \int K_t (x,u)\, f(u)\, d\gamma _\infty (u) \,, \end{aligned}

where $$K_t$$ denotes the Mehler kernel, given by

(2.6)

for $$x,u\in \mathbb {R}^n$$. Here we introduced the quadratic form

## 3 Some auxiliary results

In this section we collect some preliminary bounds, which will be essential for the sequel.

### Lemma 3.1

For $$s>0$$ and for all $$x\in \mathbb {R}^n$$ the matrices $$D_{s }$$ and $$D_{-s }= D_{s }^{-1}$$ satisfy

\begin{aligned} e^{cs}|x| \lesssim |D_s\, x| \lesssim e^{Cs} |x|, \end{aligned}

and

\begin{aligned} e^{-Cs}|x| \lesssim |D_{-s}\, x| \lesssim e^{-cs} |x|. \end{aligned}

This also holds with $$D_{s }$$ replaced by $$e^{-sB}$$ and $$e^{-sB^*}$$.

### Proof

We make a Jordan decomposition of $$B^*$$, thus writing it as the sum of a complex diagonal matrix and a triangular, nilpotent matrix, which commute with each other. This leads to expressions for $$e^{-sB^*}$$ and $$e^{sB^*}$$, and since $$B^*$$ like B has only eigenvalues with negative real parts, we see that

\begin{aligned} \Vert e^{-sB^*} \Vert \lesssim e^{Cs} \qquad \text { and } \qquad \Vert e^{sB^*} \Vert \lesssim e^{-cs}. \end{aligned}
(3.1)

From (i) in Lemma 2.1, we now get the claimed upper estimates for $$D_{\pm s}$$. To prove the lower estimate for $$D_s$$, we write

\begin{aligned} |x| = |D_{-s}\, D_s\, x| \lesssim e^{-cs} |D_s\, x|. \end{aligned}

The other parts of the lemma are completely analogous. $$\square$$

In the following lemma, we collect estimates of some basic quantities related to the matrices $$Q_t$$.

### Lemma 3.2

For all $$t>0$$ we have

1. (i)

$$\det { \, Q_t} \simeq (\min (1,t))^{n}$$;

2. (ii)

$$\Vert Q_t^{-1}\Vert \simeq (\min (1,t))^{-1}$$;

3. (iii)

$$\Vert Q_\infty -Q_t \Vert \lesssim e^{-ct}$$;

4. (iv)

$$\Vert Q_t^{-1}-Q_\infty ^{-1} \Vert \lesssim {t}^{-1}\,{e^{-ct}}$$;

5. (v)

$$\Vert \left( Q_t^{-1}-Q_\infty ^{-1} \right) ^{-1/2}\Vert \lesssim {t^{1/2}}\, e^{Ct}$$.

### Proof

(i) and (ii) Using (3.1), we see that for each $$t>0$$ and for all $$v\in \mathbb {R}^n$$

\begin{aligned} \langle Q_{t} v ,v\rangle&= \left\langle \int _0^t e^{sB} Q e^{sB^*}v \, ds,v\right\rangle =\int _0^t \langle Q^{1/2} e^{sB^*}v , Q^{1/2} e^{sB^*} v\rangle \, ds\\ {}&= \int _0^t \big | Q^{1/2} e^{sB^*}v \big |^2 \, ds\simeq \int _0^t \big | e^{sB^*}v \big |^2 \, ds\\ {}&\lesssim \int _0^t e^{-cs} \, ds \,|v|^2 \simeq \min (1,t)\, |v|^2. \end{aligned}

Since $$\Vert \left( e^{s B^*}\right) ^{-1}\Vert = \Vert e^{-s B^*}\Vert \lesssim e^{Cs}$$, there is also a lower estimate

\begin{aligned} \int _0^t&\big | e^{sB^*}v \big |^2 \, ds \gtrsim \int _0^t e^{-Cs} \, ds \, |v|^2 \simeq \min (1,t) |v|^2. \end{aligned}

Thus any eigenvalue of $$Q_t$$ has order of magnitude $$\min (1,t)$$, and (i) and (ii) follow.

(iii) From the definition of $$Q_t$$ and (3.1), we get

\begin{aligned} \Vert Q_\infty -Q_t \Vert =\left\| \int _t^\infty e^{sB}Qe^{sB^*}ds\right\| \lesssim e^{-ct}. \end{aligned}

(iv) Using now (ii) and (iii), we have

\begin{aligned} \Vert Q_t^{-1}-Q_\infty ^{-1}\Vert&=\Vert Q_t^{-1}(Q_\infty -Q_t) Q_\infty ^{-1} \Vert \lesssim \Vert Q_t^{-1}\Vert \, \Vert Q_\infty -Q_t\Vert \\&\lesssim (\min (1,t))^{-1}\, e^{-ct} \lesssim t^{-1}\, e^{-ct}. \end{aligned}

(v) Since $$\Vert A^{1/2}\Vert = \Vert A\Vert ^{1/2}$$ for any symmetric positive definite matrix A, we consider $$(Q_t^{-1}-Q_\infty ^{-1})^{-1}$$, which can be rewritten as

\begin{aligned} (Q_t^{-1}-Q_\infty ^{-1})^{-1}= (Q_\infty ^{-1}(Q_\infty -Q_t) Q_t^{-1})^{-1} =Q_t(Q_\infty -Q_t)^{-1}Q_\infty . \end{aligned}
(3.2)

It follows from (2.5) that $$(Q_\infty - Q_t)^{-1}= e^{-tB^*} Q_\infty ^{-1}e^{-tB},$$

so that

\begin{aligned}\Vert (Q_\infty - Q_t)^{-1}\Vert \lesssim e^{Ct} \end{aligned}

as a consequence of (3.2). Inserting this and the simple estimate $$\Vert Q_t \Vert \lesssim t$$ in (3.2), we obtain $$\Vert (Q_t^{-1}-Q_\infty ^{-1})^{-1}\Vert \lesssim t e^{Ct}$$, and (v) follows. $$\square$$

### Proposition 3.3

For $$t\ge 1$$ and $$w\in \mathbb {R}^n$$, we have

\begin{aligned} \langle (Q_t^{-1}- Q_\infty ^{-1} ) D_t \, w,\, D_t \, w \rangle&\simeq |w|^2. \end{aligned}

### Proof

By (2.3) and Lemma 2.1 (i) we have

\begin{aligned} \langle (Q_t^{-1}- Q_\infty ^{-1} ) D_t \, w,\, D_t \, w \rangle&= \langle Q_t^{-1} e^{tB} w\,,\, Q_\infty e^{-tB^*} Q_\infty ^{-1} \,w \rangle \\&= \langle Q_\infty Q_t^{-1} e^{tB} w\,,\, e^{-tB^*} Q_\infty ^{-1} \,w \rangle . \end{aligned}

Since $$Q_\infty Q_t^{-1}=I+(Q_\infty -Q_t)Q_t^{-1}$$, this leads to

\begin{aligned}&\langle (Q_t^{-1}- Q_\infty ^{-1} ) D_t \, w,\, D_t \, w \rangle \\&\quad = \langle e^{tB} w\,,\, e^{-tB^*} Q_\infty ^{-1} \,w \rangle + \langle ( Q_\infty - Q_t) Q_t^{-1} e^{tB} w\,,\, e^{-tB^*} Q_\infty ^{-1} \,w \rangle \\&\quad = \langle Q_\infty ^{-1} w, w\rangle + \langle e^{-tB} ( Q_\infty - Q_t) Q_t^{-1} e^{tB} w\,,\, Q_\infty ^{-1} \,w \rangle . \end{aligned}

Here $$\langle Q_\infty ^{-1} w, w\rangle \simeq |w|^2$$. Using (2.1) and then the definition of $$Q_\infty$$, we observe that the last term can be written as

\begin{aligned}&\left\langle \int _t^\infty e^{(s-t)B}Qe^{(s-t)B^*} \,ds\,e^{tB^*} \, Q_t^{-1} e^{tB} w \,,\, Q_\infty ^{-1} \,w \right\rangle \nonumber \\&\quad = \big \langle Q_\infty \,e^{tB^*} \, Q_t^{-1} e^{tB} w\, \,,\, Q_\infty ^{-1} \,w \big \rangle \nonumber \\&\quad = \langle \,e^{tB^*} \, Q_t^{-1} e^{tB} w\, \,,\, \,w \rangle \nonumber \\&\quad = \, \, \big |Q_t^{-1/2} e^{tB} w \big |^2. \end{aligned}
(3.3)

Since $$\big |Q_t^{-1/2} e^{tB} w \big |^2 \lesssim |w|^2$$ for $$t\ge 1$$ by Lemmata 3.1 and 3.2 (ii), the proposition follows.$$\square$$

We finally give estimates of the kernel $$K_t$$, for small and large values of t. When $$t\le 1$$, one has $$\Vert (Q_t^{-1}-Q_\infty ^{-1})^{1/2}\Vert \simeq t^{-1/2}$$ and $$\Vert (Q_t^{-1}-Q_\infty ^{-1})^{-1/2}\Vert \simeq t^{1/2}$$, by (iv) and (v) in Lemma 3.2. Combined with (2.6), this implies

\begin{aligned} \frac{ e^{R( x)}}{t^{n/2}}\exp \left( -C\,\frac{|u-D_t \,x |^2}{t}\right) \lesssim K_t(x,u) \lesssim \frac{ e^{R( x)}}{t^{n/2}} \exp \left( -c\,\frac{|u-D_t\, x |^2}{t}\right) , \quad \; 0 < t\le 1.\nonumber \\ \end{aligned}
(3.4)

### Lemma 3.4

For $$t\ge 1$$ and $$x,u\in \mathbb {R}^n$$, we have

\begin{aligned} e^{R(x)} \exp \Big [ -C \big | D_{-t}\,u- x \big |^2 \Big ] \lesssim K_t (x,u)&\lesssim e^{R(x)} \exp \Big [ - c\big | D_{-t}\,u- x\big |^2 \Big ]. \end{aligned}
(3.5)

### Proof

This follows from (2.6), if we write $$u-D_t\, x= D_t (D_{-t}\,u- x)$$ and apply Proposition 3.3 with $$w=D_{-t}\,u- x$$. $$\square$$

## 4 Geometric aspects of the problem

### 4.1 A system of adapted polar coordinates

We first need a technical lemma.

### Lemma 4.1

For all x in $$\mathbb {R}^n$$ and $$s\in \mathbb {R}$$, we have

\begin{aligned}&\langle B^* Q_\infty ^{-1} x, x\rangle = - \frac{1}{2}\, |Q^{1/2}\, Q_\infty ^{-1} x|^2; \end{aligned}
(4.1)
\begin{aligned}&\frac{\partial }{\partial s} D_s\, x = - Q_\infty B^*\,Q_\infty ^{-1} D_s\, x = -Q_\infty e^{-sB^*}B^*Q_\infty ^{-1} x; \end{aligned}
(4.2)
\begin{aligned}&\frac{\partial }{\partial s} R\big ( D_s\, x \big ) =\frac{1}{2}\, \big | Q^{1/2} Q_\infty ^{-1} D_s\, x\big |^2 \simeq \big | D_s\, x\big |^2 . \end{aligned}
(4.3)

### Proof

To prove (4.1), we use the definition of $$Q_\infty$$ to write for any $$z\in \mathbb {R}^n$$

\begin{aligned} \langle B^* z, Q_\infty z\rangle&= \int _0^\infty \langle B^* z, e^{sB}\,Q\, e^{sB^*}z\rangle \,ds\\&= \int _0^\infty \langle e^{sB^*}\, B^* z, \, Q\, e^{sB^*} z \rangle \,ds\\&=\frac{1}{2}\,\int _0^\infty \frac{d}{ds}\langle e^{sB^*}\, z,\, Q\, e^{sB^*} z \rangle \, ds\\&= -\frac{1}{2}\, |Q^{1/2}\, z|^2. \end{aligned}

Setting $$z=Q_\infty ^{-1} x$$, we get (4.1).

Further, (4.2) easily follows if we observe that

\begin{aligned} \frac{\partial }{\partial s} D_s\, x = \frac{\partial }{\partial s} \left( Q_\infty e^{-sB^*} Q_\infty ^{-1} x \right) = - Q_\infty B^*\, Q_\infty ^{-1} Q_\infty e^{-sB^*} Q_\infty ^{-1} x =- Q_\infty B^*\,Q_\infty ^{-1} D_s\, x. \end{aligned}

Finally, we get by means of (4.2) and (4.1)

\begin{aligned} \frac{\partial }{\partial s} R\left( D_s\, x \right)&= \frac{1}{2} \frac{\partial }{\partial s} \langle Q_\infty ^{-1/2} D_s\, x, Q_\infty ^{-1/2} D_s\, x \rangle \\&=- \langle Q_\infty ^{-1/2} Q_\infty B^* Q_\infty ^{-1} D_s\, x, Q_\infty ^{-1/2} D_s\, x \rangle \\&= \frac{1}{2} \big | Q^{1/2} Q_\infty ^{-1}D_s\, x\big |^2\,, \end{aligned}

and (4.3) is verified. $$\square$$

We observe here that an integration of (4.2) leads to

\begin{aligned} |x- D_t \,x| \lesssim t\, |x|, \qquad 0\le t\le 1. \end{aligned}
(4.4)

Fix now $$\beta >0$$ and consider the ellipsoid

\begin{aligned} E_\beta =\{x\in \mathbb {R}^n:\, R(x)= \beta \} \,.\end{aligned}

As a consequence of (4.3), the map $$s\mapsto R(D_s z)$$ is strictly increasing for each $$0 \ne z\in \mathbb {R}^n$$. Hence any $$x\in \mathbb {R}^n,\, x\ne 0$$, can be written uniquely as

\begin{aligned} x=D_s\, {\tilde{x}} \,, \end{aligned}
(4.5)

for some $${\tilde{x}}\in E_\beta$$ and $$s\in \mathbb {R}$$. We consider s and $${\tilde{x}}$$ as the polar coordinates of x. Our estimates in what follows will be uniform in $$\beta$$.

Next, we shall write Lebesgue measure in terms of these polar coordinates. A normal vector to the surface $$E_\beta$$ at the point $${\tilde{x}} \in E_\beta$$ is $$\mathbf{{N}} ({\tilde{x}})=Q_\infty ^{-1}{\tilde{x}}$$, and the tangent hyperplane at $${\tilde{x}}$$ is $$\mathbf{{N}}({\tilde{x}})^\perp$$. For $$s > 0$$ the tangent hyperplane of the surface $$D_sE_\beta = \{D_s\, {\tilde{x}}: {\tilde{x}}\in E_\beta \}$$ at the point $$D_s\,{\tilde{x}}$$ is $$D_s(\mathbf{{N}}({\tilde{x}})^\perp )$$, and a normal to $$D_sE_\beta$$ at the same point is $$w = (D_s^{-1})^* (\mathbf{{N}}({\tilde{x}}))=D_{-s}^* Q_\infty ^{-1}{\tilde{x}} = Q_\infty ^{-1}e^{sB} {\tilde{x}}$$.

The scalar product of w and the tangent of the curve $$s \mapsto D_s\,{\tilde{x}}$$ at the point $$D_s\,{\tilde{x}}$$ is, because of (4.2) and (4.1),

\begin{aligned}&\left\langle \frac{\partial }{ \partial s} D_s\, {\tilde{x}}, w \right\rangle \nonumber \\&\quad = -\langle Q_\infty e^{-sB^*}B^* Q_\infty ^{-1} {\tilde{x}},\, Q_\infty ^{-1}e^{sB} {\tilde{x}}\rangle = - \langle B^* Q_\infty ^{-1} {\tilde{x}}, {\tilde{x}}\rangle = \frac{1}{2}\, |Q^{1/2}\, Q_\infty ^{-1} {\tilde{x}}|^2 > 0. \end{aligned}
(4.6)

Thus the curve $$s \mapsto D_s\,{\tilde{x}}$$ is transversal to each surface $$D_sE_\beta$$. Let $$dS_s$$ denote the area measure of $$D_sE_\beta$$. Then Lebesgue measure is given in terms of our polar coordinates by

\begin{aligned} dx= H&(s, {\tilde{x}}) \, dS_s(D_s\, {\tilde{x}})\,ds, \end{aligned}
(4.7)

where

\begin{aligned} H(s,{\tilde{x}})= \left\langle \frac{\partial }{ \partial s} D_s\, {\tilde{x}}, \frac{w}{|w|}\right\rangle = \frac{ |Q^{1/2}\, Q_\infty ^{-1} {\tilde{x}} |^2 }{2\ | Q_\infty ^{-1} e^{sB} {\tilde{x}} |}. \end{aligned}

To see how $$dS_s$$ varies with s, we take a continuous function $$\varphi =\varphi ({\tilde{x}})$$ on $$E_\beta$$ and extend it to $${\mathbb {R}}^n \setminus \{0 \}$$ by writing $$\varphi (D_s\, {\tilde{x}}) =\varphi ({\tilde{x}})$$. For any $$t>0$$ and small $$\varepsilon >0$$, we define the shell

\begin{aligned} \Omega _{t,\varepsilon } = \{D_s\, {\tilde{x}}: t<s<t+ \varepsilon , \;{\tilde{x}} \in E_\beta \}. \end{aligned}

Then $$\Omega _{t,\varepsilon }$$ is the image under $$D_t$$ of $$\Omega _{0,\varepsilon }$$, and the Jacobian of this map is $$\det D_t = e^{-t{{\,\mathrm{tr}\,}}B}$$. Thus

\begin{aligned} \int _{\Omega _{t,\varepsilon }} \varphi (x)\,dx = e^{-t{{\,\mathrm{tr}\,}}B} \int _{\Omega _{0,\varepsilon }} \varphi (D_t\, x)\,dx, \end{aligned}

which we can rewrite as

\begin{aligned}&\int _{t<s<t+ \varepsilon } \int _{{\tilde{x}} \in E_\beta } \varphi ({\tilde{x}})\, H(s,{\tilde{x}})\, dS_s(D_s\, {\tilde{x}})\,ds \\&\quad =\, e^{-t{{\,\mathrm{tr}\,}}B}\, \int _{0<s< \varepsilon } \int _{{\tilde{x}} \in E_\beta } \varphi ({\tilde{x}})\, H(s,{\tilde{x}})\, dS_s(D_s\, {\tilde{x}})\,ds. \end{aligned}

Now we divide by $$\varepsilon$$ and let $$\varepsilon \rightarrow 0$$, getting

\begin{aligned} \int _{ E_\beta } \varphi ({\tilde{x}})\, H(t,{\tilde{x}})\, dS_t(D_t\, {\tilde{x}}) = e^{-t{{\,\mathrm{tr}\,}}B}\, \int _{ E_\beta } \varphi ({\tilde{x}})\, H(0,{\tilde{x}})\, dS_0( {\tilde{x}}). \end{aligned}

Since this holds for any $$\varphi$$, it follows that

\begin{aligned} dS_t(D_t\, {\tilde{x}}) = e^{-t{{\,\mathrm{tr}\,}}B}\, \frac{H(0,{\tilde{x}})}{H(t,{\tilde{x}})}\, dS_0( {\tilde{x}}). \end{aligned}

Together with (4.7), this implies the following result.

### Proposition 4.2

The Lebesgue measure in $${\mathbb {R}}^n$$ is given in terms of polar coordinates $$(t, {\tilde{x}})$$ by

\begin{aligned} dx = e^{-t{{\,\mathrm{tr}\,}}B}\, \frac{ |Q^{1/2}\, Q_\infty ^{-1} {\tilde{x}} |^2}{2\,| Q_\infty ^{-1} {\tilde{x}} |}\, dS_0( {\tilde{x}})\,dt\,. \end{aligned}

We also need estimates of the distance between two points in terms of the polar coordinates. The following result is a generalization of Lemma 4.2 in [4], and its proof is analogous.

### Lemma 4.3

Fix $$\beta > 0$$. Let $$x^{(0)},\; x^{(1)}\in \mathbb {R}^n \setminus \{ 0\}$$ and assume $$R(x^{(0)}) > \beta /2$$. Write

\begin{aligned} x^{(0)} = D_{ s^{(0)}} ({\tilde{x}}^{(0)} )\qquad \text { and } \qquad x^{(1)} = D_{s^{(1)}}({\tilde{x}}^{(1)}) \end{aligned}

with $$s^{(0)}$$, $$s^{(1)}\in \mathbb {R}$$ and $${\tilde{x}}^{(0)},\; {\tilde{x}}^{(1)} \in E_\beta$$.

1. (i)

Then

\begin{aligned} \big |x^{(0)} - x^{(1)}\big | \gtrsim c\, \big |{\tilde{x}}^{(0)} - {\tilde{x}}^{(1)}\big | .\end{aligned}
(4.8)
2. (ii)

If also $$s^{(1)} \ge 0$$, then

\begin{aligned} \big |x^{(0)} - x^{(1)}\big | \gtrsim c\,\sqrt{\beta }\,|s^{(0)} -s^{(1)}|. \end{aligned}
(4.9)

### Proof

Let $$\Gamma : [0,1] \rightarrow \mathbb {R}^n \setminus \{0\}$$ be a differentiable curve with $$\Gamma (0) =x^{(0)}$$ and $$\Gamma (1) =x^{(1)}$$. It suffices to bound the length of any such curve from below by the right-hand sides of (4.8) and (4.9).

For each $$\tau \in [0,1]$$, we write

\begin{aligned} \Gamma (\tau ) = D_{s{(\tau )}}\,{\tilde{x}}{(\tau )} , \end{aligned}

with $${\tilde{x}}(\tau ) \in E_\beta$$ and $${\tilde{x}}{(i)}= \tilde{x}^{(i)}$$, $$s{(i)}= s^{(i)}$$ for $$i=0,1$$. Thus

\begin{aligned} \Gamma '(\tau )&= -s'(\tau ) \,\frac{\partial }{\partial s}\, {D_{s}}_{\big |s=s{(\tau )}} \,{\tilde{x}}{(\tau )} + D_{s{(\tau )}}\, {\tilde{x}}' (\tau ). \end{aligned}

The group property of $$D_s$$ implies that

\begin{aligned} \frac{\partial }{\partial s} {D_{s}}_{\big |s=s{(\tau )}}= {D_{s(\tau )}} \frac{\partial }{\partial s} {D_{s}}_{\big |s=0}, \end{aligned}

and so

\begin{aligned} \Gamma '(\tau )&= {D_{s(\tau )}} \, v, \end{aligned}

with

\begin{aligned} v&= -s'(\tau ) \,\frac{\partial }{\partial s}\, {D_{s}}_{\big |s=0} \,{\tilde{x}}{(\tau )} + {\tilde{x}}' (\tau ). \end{aligned}

The vector $${\tilde{x}}' (\tau )$$ is tangent to $$E_\beta$$ and thus orthogonal to $$\mathbf{{N} }({\tilde{x}})$$. Then (4.6) (with $$s=0$$) implies that the angle between $$\frac{\partial }{\partial s} {D_{s}}_{\big |s=0} \,{\tilde{x}} (\tau )$$ and $${\tilde{x}}'(\tau )$$ is larger than some positive constant. It follows that

\begin{aligned} |v|^2 \gtrsim |s'(\tau )|^2\, \Big | \frac{\partial }{\partial s}\, {D_{s}}_{\big |s=0} \,{\tilde{x}}{(\tau )} \Big |^2 + \big | {\tilde{x}}' (\tau )\big |^2 \gtrsim |s'(\tau )|^2\, \beta + \big | {\tilde{x}}' (\tau )\big |^2, \end{aligned}
(4.10)

where we also used the fact that, by (4.2),

\begin{aligned} \Big | \frac{\partial }{\partial s}\, {D_{s}}_{\big |s=0} \,\tilde{x}{(\tau )} \Big |\simeq |{\tilde{x}}(\tau )| \simeq \sqrt{\beta }. \end{aligned}

Since

\begin{aligned} |v|&= \big |{D_{-s(\tau )}}\, \Gamma '(\tau )\big | \le \big \Vert {D_{-s(\tau )}} \big \Vert \, \big | \Gamma '(\tau )\big | \lesssim e^{-C\min (s(\tau ),0)} \big | \Gamma '(\tau )\big | \end{aligned}

because of Lemma 3.1, we obtain from (4.10)

\begin{aligned} \big | \Gamma '(\tau )\big |&\gtrsim e^{C \min (s(\tau ), 0)} \ \big (\sqrt{\beta }\, |s'(\tau )| + \big | {\tilde{x}}' (\tau )\big |\big ). \end{aligned}
(4.11)

Next, we derive a lower bound for s(0); assume first that $$s{(0)} < 0$$. The assumption $$R(x^{(0)}) > \beta /2$$ implies, together with Lemma 3.1,

\begin{aligned} \beta /2 \le R(D_{s(0)}\, {\tilde{x}}^{(0)})&\lesssim \big | D_{ s{(0)}}\,\tilde{x}^{(0)} \big |^2\lesssim e^{c\, s(0)} \big | {\tilde{x}}^{(0)} \big |^2 \simeq e^{c\, s(0)}\beta .\end{aligned}

It follows that

\begin{aligned} s{(0)} > -{\tilde{s}}, \end{aligned}

for some $${\tilde{s}}$$ with $$0<{\tilde{s}}<C$$, and this obviously holds also without the assumption $$s(0)<0$$.

Assume now that $$s(\tau ) > -{\tilde{s}}-1$$ for all $$\tau \in [0,1]$$. Then (4.11) implies

\begin{aligned} \big | \Gamma '(\tau )\big | \gtrsim \sqrt{\beta }\, |s'(\tau )| \end{aligned}

and

\begin{aligned} \big |\Gamma '(\tau )\big | \gtrsim |{\tilde{x}}'(\tau )|. \end{aligned}

Integrating these estimates with respect to $$\tau$$ in [0, 1], we immediately see that one can control the length of $$\Gamma$$ from below by the right-hand sides of (4.8) and (4.9).

If instead $$s(\tau ) \le -{\tilde{s}}-1$$ for some $$\tau \in [0,1]$$, we can proceed as in the proof of Lemma 4.2 in [4]. More precisely, since the image s([0, 1]) contains the interval $$[-{\tilde{s}}-1, \max (s(0), s(1))]$$, we can find a closed subinterval I of [0, 1] whose image s(I) is exactly the interval $$[-{\tilde{s}}-1, \max (s(0), s(1))]$$. Thus we may use (4.11) to control the length of $$\Gamma$$ by

\begin{aligned} \int _0^1 \big | \Gamma '(\tau )\big |\, d\tau \ge \int _I \big |\Gamma '(\tau )\big |\, d\tau \gtrsim \sqrt{\beta }\, \int _I |s'(\tau )|\,d\tau \ge \sqrt{\beta }\, \big (\max \, (s(0), s(1))\; + \;{\tilde{s}}+1\big ). \end{aligned}

Here

\begin{aligned} \sqrt{\beta }\, \big (\max \, (s(0), s(1))\; + \;{\tilde{s}}+1\big )&\gtrsim \sqrt{\beta }\gtrsim \text {diam}\, E_\beta \ge \big |{\tilde{x}}^{(0)} - {\tilde{x}}^{(1)}\big |, \end{aligned}

and (4.8) follows. Under the additional hypothesis $$s(1)\ge 0$$ of (ii), we have

\begin{aligned}{\tilde{s}}\ge \max \, (-s(0), -s(1))=-\min \,(s(0), s(1)).\end{aligned}

Then

\begin{aligned} \sqrt{\beta }\, \big (\max \, (s(0), s(1))\; + \;{\tilde{s}}+1\big )&\gtrsim \sqrt{\beta }\, \big (\max \, (s(0), s(1))-\min \,(s(0), s(1))\big )\\&= \sqrt{\beta }\, |s(0)-s(1)|, \end{aligned}

and (4.9) follows. $$\square$$

### 4.2 The Gaussian measure of a tube

We fix a large $$\beta > 0$$. Define for $$x^{(1)}\in E_\beta$$ and $$a>0$$ the set

\begin{aligned} \Omega = \left\{ x \in E_\beta : \left| x - x^{(1)}\right| < a \right\} . \end{aligned}

This is a spherical cap of the ellipsoid $$E_\beta$$, centered at $$x^{(1)}$$. Observe that $$|x| \simeq \sqrt{\beta }$$ for $$x \in \Omega$$, and that the area of $$\Omega$$ is $$|\Omega |\simeq \min \,( a^{n-1},\beta ^{(n-1)/2})$$. Then consider the tube

\begin{aligned} Z = \{D_s\, {\tilde{x}} :s\ge 0, \;{\tilde{x}} \in \Omega \}. \end{aligned}
(4.12)

### Lemma 4.4

There exists a constant C such that $$\beta > C$$ implies that the Gaussian measure of the tube Z fulfills

\begin{aligned} \gamma _\infty (Z)\lesssim \frac{a^{n-1}}{\sqrt{ \beta }}\, e^{-\beta }. \end{aligned}

### Proof

Proposition 4.2 yields, since $$H(0,{\tilde{x}}) \simeq |{\tilde{x}}|\simeq \sqrt{\beta }$$,

\begin{aligned} \gamma _\infty (Z) \simeq \int _0^\infty e^{-s {{\,\mathrm{tr}\,}}B} \ e^{-R(D_s\, {\tilde{x}})}\, \int _{\Omega } {H(0,{\tilde{x}})} \, dS ({\tilde{x}})\,ds \lesssim \sqrt{\beta }\,a^{n-1} \int _0^\infty e^{-s {{\,\mathrm{tr}\,}}B} e^{-R(D_s\, {\tilde{x}})}\,ds. \end{aligned}

By (4.3) we have

\begin{aligned} R ({D_s\, {\tilde{x}}})- R({\tilde{x}}) \simeq \int _0^s \big | D_{s'} \,{\tilde{x}}\big |^2 ds' \gtrsim s | {\tilde{x}}|^2\simeq s\beta , \end{aligned}

which implies

\begin{aligned} \gamma _\infty (Z) \lesssim \sqrt{\beta } \ a^{n-1}\, e^{-\beta } \int _0^\infty e^{-s {{\,\mathrm{tr}\,}}B} \, e^{-cs\beta }\, \ ds .\end{aligned}

Assuming $$\beta$$ large enough, one has $$c\beta > -2{{\,\mathrm{tr}\,}}B$$, and then the last integral is finite and no larger than $$C/\beta$$. The lemma follows. $$\square$$

## 5 Simplifications

In this section, we introduce some preliminary simplifications and reductions for the proof of (1.3), i.e., of Theorem 1.1.

1. (1)

We may assume that f is nonnegative and normalized in the sense that

\begin{aligned} \Vert f\Vert _{L^1( \gamma _\infty )}=1, \end{aligned}

since this involves no loss of generality.

2. (2)

We may assume that $$\alpha$$ is large, $$\alpha > C$$, since otherwise (1.3) and (1.4) are trivial.

3. (3)

In many cases, we may restrict x in (1.3) and (1.4) to the ellipsoidal annulus

\begin{aligned} {\mathcal E_\alpha }=\left\{ x \in \mathbb {R}^n:\, \frac{1}{2} \log \alpha \le R(x) \le 2 \log \alpha \, \right\} . \end{aligned}

To begin with, we can always forget the unbounded component of the complement of $$\mathcal E_\alpha$$, since

\begin{aligned}&\gamma _\infty \{ x\in \mathbb {R}^n : \,R( x)> 2 \log \alpha \}\nonumber \\ {}&\quad \lesssim \int _{ R(x)> 2 \log \alpha } \exp (-R(x) )\, dx\, \lesssim (\log \alpha )^{(n-2)/2}\,\exp ( {- 2 \log \alpha }) \lesssim \frac{1}{\alpha }. \end{aligned}
(5.1)
4. (4)

When $$t>1$$, we may forget also the inner region where $$R(x)<\frac{1}{2} \log \alpha$$. Indeed, from (3.5) we get, if $$(x,u)\in \mathbb {R}^n\times \mathbb {R}^n$$ with $$R(x) < \frac{1}{2} \log \alpha$$,

\begin{aligned} K_t (x,u) \lesssim \, e^{R(x)}< \sqrt{\alpha } < \alpha , \end{aligned}

since $$\alpha$$ is large. In other words, for any $$(x,u)\in \mathbb {R}^n\times \mathbb {R}^n$$

\begin{aligned} R(x) < \frac{1}{2} \log \alpha \; \qquad \Rightarrow \;\qquad K_t (x,u) \lesssim \alpha , \end{aligned}
(5.2)

for all $$t> 1$$.

Replacing $$\alpha$$ by $$C\alpha$$ for some C, we see from (3) and (4) that we can assume $$x \in {\mathcal E_\alpha }$$ in the proof of (1.3) and (1.4), when the supremum in the maximal operator is taken only over $$t> 1$$.

Before introducing the last simplification, we need to define a global region

\begin{aligned} G&=\left\{ (x,u)\in \mathbb {R}^n\times \mathbb {R}^n\,: \,|x-u|> \frac{1}{1+|x|} \right\} \end{aligned}

and a local region

\begin{aligned} L&=\left\{ (x,u)\in \mathbb {R}^n\times \mathbb {R}^n\,: \,|x-u|\le \frac{1}{1+|x|} \right\} . \end{aligned}

Notice that the definition of G and L does not depend on Q and B.

1. (5)

When $$t\le 1$$ and $$(x,u) \in G$$, we shall see that (5.2) is still valid, and it is again enough to consider $$x\in {\mathcal E_\alpha }$$.

To prove this, we need a lemma which will also be useful later.

### Lemma 5.1

If $$(x,u)\in G$$ and $$0<t\le 1$$, then

\begin{aligned} \frac{1}{(1+|x|)^2} \lesssim t^2 |x|^2 + | u-D_{t}\, x|^2. \end{aligned}

### Proof

From the definition of G and (4.4) we get

\begin{aligned} \frac{1}{1+|x|}&\le |x- u| \le |x-D_t\, x| +|D_t\, x -u| \lesssim t | x |+|u-D_{t}\,\, x|. \end{aligned}

The lemma follows. $$\square$$

To verify now (5.2) in the global region with $$t\le 1$$, we recall from (3.4) that

\begin{aligned} K_t (x,u) \lesssim \frac{e^{R(x)}}{ t^{n/2}}\, \exp \left( -c\,\frac{| u-D_t \,x|^2}{t} \right) . \end{aligned}

It follows from Lemma 5.1 that

\begin{aligned} t^2 \gtrsim \frac{1}{(1+|x|)^4}&\qquad \text {or} \qquad \frac{|u-D_{t}\, x|^2}{t} \gtrsim \frac{1}{(1+|x|)^2t}. \end{aligned}
(5.3)

The first inequality here implies that

\begin{aligned} K_t (x,u) \lesssim e^{R(x)}\, (1+| x|)^n \lesssim e^{2R(x)}, \end{aligned}

and (5.2) follows. If the second inequality of (5.3) holds, we have

\begin{aligned} K_t (x,u) \lesssim \frac{e^{R(x)}}{t^{n/2}} \exp \left( -\frac{c}{(1+|x|)^2 t} \right) \lesssim e^{R(x)}\, (1+|x|)^n, \end{aligned}

and we get the same estimate. Thus (5.2) is verified.

Finally, let

\begin{aligned} \mathcal H_*^{G} f(x)=\sup _{0 < t\le 1} \left| \int K_t (x,u)\,\chi _{{G}}(x,u)\,f(u) \, d\gamma _\infty (u) \, \right| \,, \end{aligned}

and

\begin{aligned} \mathcal H_*^{L} f(x)=\sup _{0 < t\le 1} \left| \int K_t (x,u)\,\chi _{{L}}(x,u)\,f(u) \, d\gamma _\infty (u) \, \right| \,. \end{aligned}

## 6 The case of large t

In this section, we consider the supremum in the definition of the maximal operator taken only over $$t> 1$$, and we prove (1.4).

### Proposition 6.1

For all functions $$f\in L^1 (\gamma _\infty )$$ such that $$\Vert f\Vert _{L^1( \gamma _\infty )}=1$$,

\begin{aligned} \gamma _\infty \left\{ x : \sup _{t> 1}|\mathcal H_t f(x)|> \alpha \right\} \lesssim \frac{1}{\alpha \sqrt{\log \alpha }}, \quad \quad \alpha >2. \end{aligned}
(6.1)

In particular, the maximal operator

\begin{aligned} \sup _{t> 1}|\mathcal H_t f(x)| \end{aligned}

is of weak type (1, 1) with respect to the invariant measure $$\gamma _\infty$$.

### Proof

We can assume that $$f\ge 0$$. Looking at the arguments in Sect. 5, items (3) and (4), we see that it suffices to consider points $$x\in {\mathcal E_\alpha }$$. For both x and u we use the coordinates introduced in (4.5) with $$\beta =\log \alpha$$, that is,

\begin{aligned} x=D_{s}\,{\tilde{x}}, \qquad u=D_{ s'}\, {\tilde{u}}, \end{aligned}

where $${\tilde{x}}, {\tilde{u}} \in E_{\log \alpha }$$ and $$s,s' \in \mathbb {R}$$.

From (3.5) we have

\begin{aligned} K_t (x,u)&\lesssim \exp (R( x))\exp \big ( - c\, \big | D_{-t}\,u-x\big |^2 \big ) \end{aligned}

for $$t> 1$$ and $$x,u\in \mathbb {R}^n$$. Since $$x \in {\mathcal E_\alpha }$$ and $$D_{-t}\,u= D_{s'-t}\, \tilde{u},$$ we can apply Lemma 4.3 (i), getting

\begin{aligned} \big | D_{-t}\,u-x \big | \gtrsim \big | {\tilde{x}}-{\tilde{u}}\big |, \end{aligned}

so that

\begin{aligned} \int K_t (x,u) f(u)\,d \gamma _\infty (u)&\lesssim \exp \big ( R(D_s\, {\tilde{x}}) \big ) \int \exp \big ( - c\, \big | {\tilde{x}}-{\tilde{u}}\big |^2 \big )\,f(u)\,d \gamma _\infty (u). \end{aligned}

In view of (4.3), the right-hand side here is strictly increasing in s, and therefore the inequality

\begin{aligned} \exp \big ( R(D_s\, {\tilde{x}}) \big ) \int \exp \big ( - c\, \big | {\tilde{x}}-{\tilde{u}}\big |^2 \big )\,f(u)\,d \gamma _\infty (u) > \alpha \end{aligned}
(6.2)

holds if and only if $$s> s_\alpha ({\tilde{x}})$$ for some function $${\tilde{x}}\mapsto s_\alpha ({\tilde{x}})$$, with equality for $$s=s_\alpha ({\tilde{x}})$$. Since $$\alpha >2$$ and $$\Vert f\Vert _{L^1 (\gamma _\infty )}=1$$, it follows that $$s_\alpha ({\tilde{x}})>0$$.

For some C, the set of points $$x\in {\mathcal E_\alpha }$$ where the supremum in (6.1) is larger than $$C\alpha$$ is contained in the set $$\mathcal A(\alpha )$$ of points $$D_s\, {\tilde{x}}\in {\mathcal E_\alpha }$$ fulfilling (6.2). We use Proposition 4.2 to estimate the $$\gamma _\infty$$ measure of $$\mathcal A(\alpha )$$. Observe that $$H(0,{\tilde{x}}) \simeq |{\tilde{x}}|\simeq \sqrt{\log \alpha }$$ and that $$D_s\, {\tilde{x}}\in {\mathcal E_\alpha }$$ implies $$s\lesssim 1$$, so that also $$e^{-s{{\,\mathrm{tr}\,}}B} \lesssim 1$$. We get

\begin{aligned} \gamma _\infty (\mathcal A(\alpha ) )&= \int _{\mathcal A(\alpha )\cap {\mathcal E_\alpha }} e^{-R(x)} dx \\&\lesssim { \sqrt{\log \alpha }} \int _{E_{\log \alpha }} \int _{s_\alpha ({\tilde{x}})}^{C} e^{-R(D_s\, {\tilde{x}})} \,ds\, dS({\tilde{x}}) \\&\lesssim { \sqrt{\log \alpha }} \int _{E_{\log \alpha }} \int _{s_\alpha ({\tilde{x}})}^{+\infty } \exp \left( - {R( D_{s_\alpha ({\tilde{x}})} \,{\tilde{x}}) -c\log \alpha \, (s-s_\alpha ({\tilde{x}})}) \right) \,ds\, dS({\tilde{x}}) , \end{aligned}

where the last inequality follows from (4.3), since $$|D_s\, {\tilde{x}}|^2\gtrsim |{\tilde{x}}|^2\simeq \log \alpha .$$ Integrating in s, we obtain

\begin{aligned} \gamma _\infty (\mathcal A (\alpha ) )&\lesssim \frac{1}{\sqrt{\log \alpha }}\int _{E_{\log \alpha }} \exp \big (-R( D_{{s_\alpha ({\tilde{x}})}} \, {\tilde{x}}) \big ) \,dS({\tilde{x}}) . \end{aligned}

Now combine this estimate with the case of equality in (6.2) and change the order of integration, to get

\begin{aligned} \gamma _\infty (\mathcal A (\alpha ) )&\lesssim \frac{1}{\alpha \sqrt{\log \alpha }} \int \int _{E_{\log \alpha }} \exp \big ( - c\, \big | {\tilde{x}}-{\tilde{u}}\big |^2 \big ) \,dS(\tilde{x})\,f(u)\,d \gamma _\infty (u)\\&\lesssim \frac{1}{\alpha \sqrt{\log \alpha }} \int f(u)\,d \gamma _\infty (u)\,, \end{aligned}

which proves Proposition 6.1. $$\square$$

Finally, we show that the factor $$1/\sqrt{\log \alpha }$$ in (6.1) is sharp.

### Proposition 6.2

For any $$t> 1$$ and any large $$\alpha$$, there exists a function f normalized in $$L^1 (\gamma _\infty )$$ and such that

\begin{aligned} \gamma _\infty \left\{ x :|\mathcal H_t f(x)| > \alpha \right\} \simeq \frac{1}{\alpha \sqrt{\log \alpha }}\,. \end{aligned}

### Proof

Take a point z with $$R(z)=\log \,\alpha$$, and let f be (an approximation of) a Dirac measure at the point $$u=D_t z$$. Then, as a consequence of (3.5), $$K_t (x,u) \simeq \exp (R( x))$$ when x is in the ball $$B(D_{-t}\,u,1)=B(z,1)$$. We then have $$\mathcal H_t f(x)=K_t (x,u)\gtrsim \alpha$$ in the set $$\mathcal B=\{x\in B(z,1)\!: R(x)>R(z) \}$$, whose measure is

\begin{aligned} \gamma _\infty \, (\mathcal B) \simeq e^{-R(z)}\, \frac{1}{\sqrt{R(z)}} = \frac{1}{\alpha \,\sqrt{\log \,\alpha }}. \end{aligned}

$$\square$$

## 7 The local case for small t

### Proposition 7.1

If $$(x,u) \in L$$ and $$0<t\le 1$$, then

\begin{aligned} \big |K_{t} (x,u)\big |\lesssim \,\frac{ \exp \big ( R(x) \big ) }{ t^{n/2} }\, \exp { \left( -c\, \frac{| u-x|^2}{ t} \, \right) } \,. \end{aligned}

### Proof

In view of (3.4), it is enough to show that

\begin{aligned} \frac{|u-D_t\, x|^2}{t} \ge \frac{|u-x|^2}{t} -C. \end{aligned}
(7.1)

We write

\begin{aligned}&|u-D_t\, x|^2 = |u-x +x-D_t \,x|^2 = |u-x|^2 +2 \langle u-x, x-D_t\, x\rangle + |x-D_t\, x|^2\\&\quad \ge |u-x|^2 -2 |u-x|\,|x-D_t\, x|. \end{aligned}

By (4.4),

\begin{aligned} |u-x|\, |x-D_t\, x| \lesssim |u-x|\,t\, |x|\le t \end{aligned}

since $$(x,u) \in L$$, and (7.1) follows. $$\square$$

### Proposition 7.2

The maximal operator $$\mathcal H_*^{L}$$ is of weak type (1, 1) with respect to the invariant measure $$\gamma _\infty$$.

### Proof

The proof is standard, since Proposition 7.1 implies

\begin{aligned} \mathcal H_*^{L} f(x)&\lesssim \sup _{0<t\le 1} \frac{\exp \big ( R(x)\big )}{ t^{n/2}} \int \exp \Big (-c\, \frac{| x-u|^2 }{ t} \, \Big ) \,\chi _{L}(x,u)\,f(u) \, d\gamma _\infty (u). \end{aligned}

The supremum here defines an operator of weak type (1, 1) with respect to Lebesgue measure in $$\mathbb {R}^n$$. From this the proposition follows, cf. [7,   Section 3]. $$\square$$

## 8 The global case for small t

In this section, we conclude the proof of Theorem 1.1.

### Proposition 8.1

The maximal operator $$\mathcal H_*^{G}$$ is of weak type (1, 1) with respect to the invariant measure $$\gamma _\infty$$.

### Proof

We take f and $$\alpha$$ as in items (1) and (2) of Sect. 5. Then item (5) tells us that we need only consider $$\mathcal H_*^{G} f(x)$$ for $$x\in \mathcal E_\alpha$$.

For $$m\in \mathbb {N}$$ and $$0<t\le 1$$, we introduce regions $$\mathcal S^{m}_t$$. If $$m>0$$, we let

\begin{aligned} \mathcal S^{m}_t&=\left\{ (x,u)\in G: 2^{m-1} \sqrt{t}< |u-D_t\, x |\le 2^{m} \sqrt{t} \,\right\} . \end{aligned}

If $$m=0$$, we replace the condition $$2^{m-1} \sqrt{t}< |u-D_t\, x |\le 2^{m} \sqrt{t}$$ by $$| u-D_t\, x | \le \sqrt{t}$$. Note that for any fixed $$t\in (0,1]$$ these sets form a partition of G.

In the set $${\mathcal S^{m}_t}$$ we have, because of (3.4),

\begin{aligned} K_t (x,u)&\lesssim \frac{ \exp (R(x)) }{ t^{n/2} } \exp \left( { -c{2^{2 m} } }\right) . \end{aligned}

Then setting

\begin{aligned} {\mathcal K}_t^{{m}} (x,u)&= \frac{ \exp (R(x)) }{ t^{n/2} } \,\chi _{\mathcal S^{m}_t}(x,u), \end{aligned}

one has, for all $$(x,u)\in G$$ and $$0<t<1$$,

\begin{aligned} K_t (x,u)\lesssim \sum _{m=0}^\infty \exp \left( { -c{2^{2 m} } }\right) {\mathcal K}_t^{{m}} (x,u) \,. \end{aligned}

Hence, it suffices to prove that for $$m = 0,1,\dots$$

\begin{aligned} \gamma _\infty \left\{ x\in {\mathcal E_\alpha }: \sup _{0<t\le 1} \int {\mathcal K}_t^{{m}} \! (x,u)\, f(u)\, d \gamma _\infty (u) \! >\alpha \right\} \lesssim \frac{2^{Cm}}{\alpha } , \end{aligned}
(8.1)

for large $$\alpha$$ and some C, since this will allow summing in m in the space $$L^{1,\infty }(\gamma _\infty )$$.

Fix $$m\in {\mathbb {N}}$$ and assume that $$(x,u) \in S_t^m$$ for some $$t \in (0,1]$$, so that $$|u-D_t\, x| \le 2^{m}\sqrt{t}$$. Then Lemma 5.1 leads to

\begin{aligned} 1&\lesssim (1+|x|)^4 t^2+(1+|x|)^2 \, 2^{2m}\,t \le ((1+|x |)^2 \, 2^{2m}\,t)^2+ (1+|x|)^2 \,2^{2m}\,t. \end{aligned}

Consequently, a point $$x\in \mathcal E_\alpha$$ satisfies

\begin{aligned} (1+|x|)^2 \, 2^{2m}\,t \gtrsim 1 \end{aligned}
(8.2)

as soon as there exists a point u with $$\mathcal K_t^{m}(x,u)\ne 0$$, and then $$t\ge \varepsilon >0$$ for some $$\varepsilon =\varepsilon (\alpha ,m) >0$$. Hence the supremum in (8.1) will be the same if taken only over $$\varepsilon \le t\le 1$$, and it follows that this supremum is a continuous function of $$x\in {\mathcal E_\alpha }$$.

To prove (8.1), the idea, which goes back to [15], is to construct a finite sequence of pairwise disjoint balls $$\big (\mathcal B^{(\ell )}\big )_{\ell =1}^{\ell _0}$$ in $$\mathbb {R}^n$$ and a finite sequence of sets $$\big (\mathcal Z^{(\ell )}\big )_{\ell =1}^{\ell _0}$$ in $$\mathbb {R}^n$$, called forbidden zones. These zones will together cover the level set in (8.1). We will then verify that

\begin{aligned} \left\{ x\in {\mathcal E_\alpha }: \sup _{\varepsilon \le t\le 1} \int {\mathcal K}_t^{{m}} (x,u)\,f(u)\,d \gamma _\infty (u)\, \ge \alpha \right\} \subset \bigcup _{\ell =1}^{\ell _0}\mathcal Z^{(\ell )}, \end{aligned}
(8.3)

that for each $$\ell$$

\begin{aligned}&\gamma _\infty (\mathcal Z^{(\ell )}) \lesssim \frac{2^{Cm}}{\alpha } \int _{\mathcal B^{(\ell )}}f(u) \,d \gamma _\infty (u), \end{aligned}
(8.4)

and that the $$\mathcal B^{(\ell )}$$ are pairwise disjoint. This would imply

\begin{aligned} \gamma _\infty \Big (\bigcup _{\ell =1}^{\ell _0} \mathcal Z^{(\ell )} \Big ) \lesssim \frac{2^{Cm}}{\alpha }\, \sum _{\ell =1}^{\ell _0} \int _{\mathcal B^{(\ell )}}f(u)\,d \gamma _\infty (u) \lesssim \frac{2^{Cm}}{\alpha } ,\end{aligned}

and thus also (8.1) and Proposition 8.1.

The sets $$\mathcal B^{(\ell )}$$ and $$\mathcal Z^{(\ell )}$$ will be introduced by means of a sequence of points $$x^{(\ell )}, \ell =1,\ldots , \ell _0$$, which we define by recursion. To start, we choose as $$x^{(1)}$$ a point where the quadratic form R(x) takes its minimal value in the compact set

\begin{aligned} \mathcal A_1 (\alpha )= \left\{ x\in {\mathcal E_\alpha } : \sup _{\varepsilon \le t\le 1} \int {\mathcal K}_t^{{m}} (x,u)\,f(u)\,d \gamma _\infty \ge \alpha \right\} . \end{aligned}

However, should this set be empty, (8.1) is immediate.

We now describe the recursion to construct $$x^{(\ell )}$$ for $$\ell \ge 2$$. Like $$x^{(1)}$$, these points will satisfy

\begin{aligned} \sup _{\varepsilon \le t\le 1} \int {\mathcal K}_t^{{m}} (x^{(\ell )},u)\,f(u)\,d \gamma _\infty \ge \alpha . \end{aligned}

Once an $$x^{(\ell )},\;\ell \ge 1$$, is defined, we can thus by continuity choose $$t_\ell \in [\varepsilon , 1]$$ such that

\begin{aligned} \int {\mathcal K}_{t_\ell }^{{m}} (x^{(\ell )},u)\, f(u)\,d \gamma _\infty \ge \alpha . \end{aligned}
(8.5)

Using this $$t_\ell$$, we associate with $$x^{(\ell )}$$ the tube

\begin{aligned} \mathcal Z^{(\ell )} = \left\{ D_{ s}\,\eta \in \mathbb {R}^n :\,s\ge 0,\; R{( \eta )}= R(x^{(\ell )}), \; | \eta - x^{(\ell )} |< A\, 2^{3m}\, \sqrt{t_{\ell }}\right\} , \end{aligned}

Here the constant $$A>0$$ is to be determined, depending only on n, Q and B.

All the $$x^{(\ell )}$$ will be minimizing points of R(x). To avoid having them too close to one another, we will not allow $$x^{(\ell )}$$ to be in any $$\mathcal Z^{(\ell ')}$$ with $$\ell ' < \ell$$. More precisely, assuming $$x^{(1)}, \dots , x^{(\ell )}$$ already defined, we will choose $$x^{(\ell +1)}$$ as a minimizing point of R(x) in the set

\begin{aligned} \mathcal A_{\ell +1} (\alpha )= \left\{ x\in {\mathcal E_\alpha } \setminus \bigcup _{\ell '=1}^{\ell } \mathcal Z^{(\ell ')}:\, \sup _{\varepsilon \le t\le 1} \int {\mathcal K}_t^{{m}} (x,u)\,f(u)\,d\gamma _\infty (u) \ge \alpha \right\} , \end{aligned}
(8.6)

provided this set is nonempty. But if $$\mathcal A_{\ell +1} (\alpha )$$ is empty, the process stops with $$\ell _0=\ell$$ and (8.3) follows. We will see that this actually occurs for some finite $$\ell$$.

Now assume that $$\mathcal A_{\ell +1} (\alpha ) \ne \emptyset$$. In order to assure that a minimizing point exists, we must verify that $$\mathcal A_{\ell +1} (\alpha )$$ is closed and thus compact, although the $$\mathcal Z^{(\ell ')}$$ are not open. To do so, observe that for $$1\le \ell ' \le \ell$$, the minimizing property of $$x^{(\ell ')}$$ means that there is no point x in $$\mathcal A_{\ell '} (\alpha )$$ with $$R(x) < R(x^{(\ell ')})$$. Thus we have the inclusions

\begin{aligned} \mathcal A_{\ell +1} (\alpha ) \subset \mathcal A_{\ell '} (\alpha ) \subset \left\{ x: R(x)\ge R(x^{(\ell ')})\right\} , \qquad 1\le \ell ' \le \ell . \end{aligned}

It follows that

\begin{aligned}&\mathcal A_{\ell +1} (\alpha ) = \mathcal A_{\ell +1} (\alpha )\,\cap \, \bigcap _{1\le \ell ' \le \ell } \{x: R(x)\ge R(x^{(\ell ')})\} \\&\quad =\bigcap _{\ell '=1}^{\ell } \left\{ x\in {\mathcal E_\alpha } \setminus \mathcal Z^{(\ell ')}: R(x)\ge R(x^{(\ell ')}), \; \sup _{\varepsilon \le t\le 1} \int {\mathcal K}_t^{{m}} (x,u)\,f(u)\,d\gamma _\infty (u) \ge \alpha \right\} . \end{aligned}

For each $$\ell ' = 1,\dots , \ell$$ we have

\begin{aligned}&\{x\in {\mathcal E_\alpha } \setminus \mathcal Z^{(\ell ')}: R(x)\ge R(x^{(\ell ')})\} \\&\quad = \left\{ D_s\,\eta \in \mathcal E_\alpha :\, s\ge 0, \,R(\eta ) = R(x^{(\ell ')}), \;|\eta - x^{(\ell ')}| \ge A2^{3m} \sqrt{t_{\ell '}} \right\} , \end{aligned}

and this set is closed. It follows that $$\mathcal A_{\ell +1} (\alpha )$$ is compact, and a minimizing point $$x^{(\ell +1)}$$ can be chosen. Thus the recursion is well defined.

We observe that (8.2) applies to $$t_\ell$$ and $$x^{(\ell )}$$, and $$|x^{(\ell )}|$$ is large, so

\begin{aligned} |x^{(\ell )}|^2\, 2^{2m}\, t_\ell \gtrsim 1. \end{aligned}
(8.7)

Further, we define balls

\begin{aligned} \mathcal B^{(\ell )}=&\{ u \in \mathbb {R}^n :\, | u -D_{t_\ell } \,x^{(\ell )} | \le 2^{m} \sqrt{t_\ell }\, \}\,. \end{aligned}

Because of the definitions of $$\mathcal K_t^m$$ and $${\mathcal S^{m}_t}$$, the inequality (8.5) implies

\begin{aligned} \alpha&\le \frac{\exp \left( { R(x^{(\ell )}) }\right) }{ t_\ell ^{n/2}} \int _{\mathcal B^{(\ell )} }f(u)\,d \gamma _\infty (u). \end{aligned}
(8.8)

It remains to verify the claimed properties of $$\mathcal B^{(\ell )}$$ and $$\mathcal Z^{(\ell )}$$. The arguments below follow the lines of the proof of Lemma 6.2 in [4], with only slight modifications.

### Lemma 8.2

The balls $$\mathcal B^{(\ell )}$$ are pairwise disjoint.

### Proof

Two balls $$\mathcal B^{(\ell )}$$ and $$\mathcal B^{(\ell ')}$$ with $$\ell <\ell '$$ will be disjoint if

\begin{aligned} \big | D_{t_{\ell }} \, x^{(\ell )}- D_{t_{\ell '}} \, x^{(\ell ')}\big | > 2^m (\sqrt{t_\ell }+ \sqrt{ t_{\ell '}}). \end{aligned}
(8.9)

By means of our polar coordinates with $$\beta =R(x^{(\ell )})$$, we write

\begin{aligned} x^{(\ell ')} = D_{s}\, {\tilde{x}}^{(\ell ')} \end{aligned}

for some $${\tilde{x}}^{(\ell ')}$$ with $$R({\tilde{x}}^{(\ell ')})= R(x^{(\ell )})$$ and some $$s \in \mathbb {R}$$. Note that $$s\ge 0$$, because $$R(x^{(\ell ')})\ge R( x^{(\ell )})$$. Since $$x^{(\ell ')}$$ does not belong to the forbidden zone $$\mathcal Z^{(\ell )}$$, we must have

\begin{aligned} | {\tilde{x}}^{(\ell ')}- x^{(\ell )} |\ge A 2^{3m} \sqrt{t_\ell }. \end{aligned}
(8.10)

We first assume that $${t_{\ell '}} \ge M\, 2^{4m} \, t_\ell$$, for some $$M=M(n, Q,B)\ge 2$$ to be chosen. Lemma 4.3 (ii) implies

\begin{aligned} \big | D_{t_{\ell }} \, x^{(\ell )}- D_{t_{\ell '}} \, x^{(\ell ')}\big | = \big | D_{t_{\ell }} \, x^{(\ell )}- D_{t_{\ell '}+s} \, {\tilde{x}}^{(\ell ')}\big | \gtrsim |x^{(\ell )}|\,( t_{\ell '}+s-t_\ell ) \gtrsim |x^{(\ell )}|\, { t_{\ell '}}, \end{aligned}

the last step by our assumption. Using again the assumption and then (8.7), we get

\begin{aligned} |x^{(\ell )}|\,{ t_{\ell '}}\gtrsim |x^{(\ell )}|\, \sqrt{M} \, 2^{2m} \sqrt{ t_{\ell }}\, \sqrt{ t_{\ell '}} \gtrsim \sqrt{M} \, 2^{m}\sqrt{ t_{\ell '}} \simeq \sqrt{M}\, 2^{m}\, ( \sqrt{ t_{\ell '}} +\sqrt{ t_{\ell }}). \end{aligned}

Fixing M suitably large, we obtain (8.9) from the last two formulae.

It remains to consider the case when $${t_{\ell '}} < M\, 2^{4m}\, t_\ell$$. Then

\begin{aligned} \sqrt{t_{\ell }} > \frac{2^{-2m-1}}{\sqrt{M}} ( \sqrt{t_{\ell '}}+\sqrt{t_\ell }). \end{aligned}

Applying this to (8.10), we obtain (8.9) by choosing A so that $$A/\sqrt{M}$$ is large enough. $$\square$$

We next verify that the sequence $$(x^{(\ell )})$$ is finite. For $$\ell <\ell '$$, we have (8.10), and Lemma 4.3 (i) implies

\begin{aligned} \big | x^{(\ell ')}- x^{(\ell )} \big |&\gtrsim A\, 2^{3m} \sqrt{t_\ell }. \end{aligned}

Since $$t_\ell \ge \varepsilon$$, we see that the distance $$\left| x^{(\ell ')}- x^{(\ell )} \right|$$ is bounded below by a positive constant. But all the $$x^{(\ell )}$$ are contained in the bounded set $${\mathcal E_\alpha }$$, so they are finite in number. Thus the set considered in (8.6) must be empty for some $$\ell$$, and the recursion stops. This implies (8.3).

We finally prove (8.4) . Observe that the forbidden zone $$\mathcal Z^{(\ell )}$$ is a tube as defined in (4.12), with $$a=A\, 2^{3m} \sqrt{t_\ell }$$ and $$\beta =R(x^{(\ell )})$$. This value of $$\beta$$ is large since $$x^{(\ell )} \in {\mathcal E_\alpha }$$, and thus we can apply Lemma 4.4 to obtain

\begin{aligned} \gamma _\infty ( \mathcal Z^{(\ell )})\lesssim \frac{\big ( A 2^{3m} \sqrt{t_\ell }\big )^{n-1}}{ \sqrt{ R(x^{(\ell )})}}\, \exp \left( { -{ R(x^{(\ell )}) }}\right) . \end{aligned}

We bound the exponential here by means of (8.8) and observe that $$R(x^{(\ell )}) \sim |x^{(\ell )}|^2$$, getting

\begin{aligned}\gamma _\infty (\mathcal Z^{(\ell )})\lesssim \frac{1}{\alpha |x^{(\ell )}| {\sqrt{t_\ell }}} \, (A 2^{3m})^{n-1} \, \int _{\mathcal B^{(\ell )} }\!f(u)\,d \gamma _\infty (u). \end{aligned}

As a consequence of (8.7), we obtain

\begin{aligned} \gamma _\infty (\mathcal Z^{(\ell )}) \lesssim \frac{2^{m}}{\alpha } \, \big (A 2^{3m}\big )^{n-1} \, \int _{\mathcal B^{(\ell )}}f(u)\,d \gamma _\infty (u)\, \lesssim \frac{2^{Cm}}{\alpha }\, \, \int _{\mathcal B^{(\ell )}}f(u)\,d \gamma _\infty (u), \end{aligned}

proving (8.4). This concludes the proof of Proposition 8.1. $$\square$$