Ornstein-Uhlenbeck Pinball and the Poincaré Inequality in a Punctured Domain

Abstract

In this paper we study the Poincaré constant for the Gaussian measure restricted to $D={\mathbb R}^d - \mathbb {B}$ where $\mathbb {B}$ is the disjoint union of bounded open sets. We will mainly look at the case where the obstacles are Euclidean balls B(x _i, r _i) with radii r _i, or hypercubes with vertices of length 2r _i, and d ≥ 2. This will explain the asymptotic behavior of a d-dimensional Ornstein-Uhlenbeck process in the presence of obstacles with elastic normal reflections (the Ornstein-Uhlenbeck pinball).

Appendices

Appendix 1: Existence and Uniqueness of the Process

The main (actually unique) result of this section is the following (recall that the notion of solution for a reflected system involves both X and the local times L, see e.g. [9, 24])

Theorem 1.6

Assume (1.1). Then the system (1.2) has a unique (non explosive) strong solution for any allowed starting point x. In addition $\mu _{\lambda ,\mathbb {X}}$ is the unique invariant (actually symmetric) probability measure.

The remainder of this section is devoted to the proof of this result.

In the sequel we shall denote by L the (formal) infinitesimal generator

$$\displaystyle \begin{aligned} L = \frac 12 \, \varDelta - \lambda \, \langle x,\nabla\rangle \, ,\end{aligned} $$

(1.39)

whose domain is some extension of the set of smooth functions f compactly supported in $\bar D$ such that for all i,

$$\displaystyle \begin{aligned}\frac{\partial \, f}{\partial n_i}(y)=0\end{aligned} $$

at any y such that |y − x _i| = r _i, where n _i denotes the normal vector field on the sphere of center x _i and radius r _i.

We shall denote by $\mathbb {D}(L)$ this core.

1.1.1 Finite Number of Obstacles

When N is finite, existence of a unique (strong) solution of (1.2) starting from any point (belonging to $\bar D$ for (1.2)), up to the explosion time, is standard (see e.g. [9] for references) at least when the boundary of the obstacles is smooth. That is why we have chosen to smooth the hypercubes when looking at this particular situation. The only point is to show that the explosion time is almost surely infinite.

To this end, define

$$\displaystyle \begin{aligned} d_N = \max_{i=1,\ldots,N} \, |x_i| \quad , \quad r=\max_{i=1,\ldots,N} \, r_i \, , \end{aligned} $$

(1.40)

and choose a smooth function h _N such that h _N ≥ 1 everywhere,

$$\displaystyle \begin{aligned} h_N(x) = 1 \mbox{ if } |x|< d_N+2r \quad , \quad h_N(x)= 1 + |x|{}^2 \mbox{ if } |x|> d_N+3r+1 \, . \end{aligned} $$

(1.41)

It is enough to remark that $h_N \in \mathbb {D}(L)$ and satisfies

$$\displaystyle \begin{aligned} Lh_N \leq - \, \lambda \, h_N \, \quad , \mathrm{ for } |x|>d_L=(d/2\lambda)^{\frac 12}\vee (d_N+3r+1) \, . \end{aligned} $$

(1.42)

h _N can thus play the role of a Lyapunov function for Hasminskii’s non explosion test.

We can thus define for any x in $\bar D$ the law P _t(x, dy) of the process at time t, X _t starting from x, as well as a Markov semi-group P _t acting on continuous and bounded functions. It is known that, for all t > 0,

$$\displaystyle \begin{aligned}P_t(x,dy)=p_t(x,y) \, dy\end{aligned}$$

where $p_t \in C^\infty (\bar {D})$ (see [9, 10]). Furthermore, the density p _t is everywhere positive. This is a consequence of (1.1) (which implies in particular that D is path connected) and standard tools about the support of the law of the whole process.

$\mu _{\lambda ,\mathbb {X}}$ is clearly a symmetric, hence invariant, probability measure. Uniqueness follows from the previous regularity and positivity as usual. Let us denote by q _t the density of the law of X _t w.r.t. $\mu _{\lambda ,\mathbb {X}}$ i.e.

$$\displaystyle \begin{aligned}q_t(x,y) = p_t(x,y) \, \frac{dx}{d\mu_{\lambda,\mathbb{X}}} \, .\end{aligned}$$

Application of the Chapman-Kolmogorov formula and standard regularization arguments yield

$$\displaystyle \begin{aligned} q_{2t}(x,x) = \int \, q_t(x,y) \, q_t(y,x) \, \mu_{\lambda,\mathbb{X}}(dy) = \int \, q_t^2(x,y) \, \mu_{\lambda,\mathbb{X}}(dy) \, , \end{aligned} $$

(1.43)

thanks to symmetry, i.e. $q_t \in {\mathbb L}^2(\mu _{\lambda ,\mathbb {X}})$.

1.1.2 Infinite Number of Obstacles

We now consider the case of infinitely many obstacles, still satisfying the non overlapping condition (1.1), for the locally finite collection $\mathbb {X}$. We can thus construct the process up to exit times of an increasing sequence of relatively compact open subsets U _n, each of which containing only a finite number of (closed) obstacles, the remaining (closed) obstacles being included into $(\bar {U_n})^c$. The sequence T _n of exit times of U _n is thus growing to the explosion time, but now it is much more difficult to control this limit.

A standard way is to use Dirichlet forms theory. Namely let us consider

$$\displaystyle \begin{aligned} \mathbb{E} (f) = \int \, |\nabla f|{}^2 \, d\mu_{\lambda,\mathbb{X}} \end{aligned} $$

(1.44)

defined for f which are smooth, bounded with bounded derivatives.

Our goal is to show that $\mathbb {E}$ is a conservative local Dirichlet form, so that one can associate to $\mathbb {E}$ a stationary Hunt process (Y _t)_t≥0 which is a non exploding diffusion process. This process coincides with X up to the exit time of U _n for all n, provided X ₀ has distribution $\mu _{\lambda ,\mathbb {X}}$ (exit time can be equal to 0). But, since Y _t − Y ₀ is an additive functional of finite energy, it can be decomposed (Lyons-Zheng decomposition) for 0 ≤ t ≤ T into

$$\displaystyle \begin{aligned}Y_t - Y_0 = M_t + RM^T_t\end{aligned}$$

where M _. (resp. $RM^T_.$) is a forward (resp. backward) ${\mathbb L}^2$ martingale with brackets 〈M〉_t = 〈RM ^T〉_t = t, hence are Brownian motions. It follows that for any K > 0,

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathbb P}\left( \sup_{t\in[0;T]}|Y_t| \geq K \right) &\displaystyle \leq&\displaystyle {\mathbb P}\left( \sup_{t\in[0;T]} |Y_t-Y_0| \geq \frac{K}{2} \text{ or } |Y_0| \geq \frac{K}{2} \right) \\ &\displaystyle \leq&\displaystyle {\mathbb P}\left( \sup_{t\in[0;T]} |M_t| \geq \frac K4 \right) + {\mathbb P}\left( \sup_{t\in[0;T]} |RM^T_t| \geq \frac K4 \right) \\ &\displaystyle &\displaystyle + {\mathbb P} \left(|Y_0| \geq \frac{K}{2} \right) \end{array} \end{aligned} $$

and Doob’s inequality allows us to conclude that the latter upper bound goes to 0 as K goes to infinity. It follows that the supremum of the exit times of the balls B(0, K) is almost surely infinite, hence so does the supremum of the T _n’s, implying that Y  and X coincide up to any time and that X does not explode, when the initial law is $\mu _{\lambda ,\mathbb {X}}$.

Standard arguments (see [19]) imply that there is no explosion starting from quasi every point x (i.e. all x’s not belonging to some polar set $\mathbb {N}$, recall that here polar sets coincide with sets of zero capacity), though here we only need that this property holds for $\mu _{\lambda ,\mathbb {X}}$ almost all x’s, which is an immediate consequence of disintegration of the measure.

Now let x be some point in D, and choose a small ball B(x, ε) ⊂ D. If ${\mathbb P}_y$ denotes the law of X starting from y as usual, we have for all z ∈ B(x, ε),

$$\displaystyle \begin{aligned}{\mathbb P}_z (\sup_n T_n< +\infty) = \int_{|y-x|=\varepsilon} \, {\mathbb P}_y(\sup_n T_n< +\infty) \, \eta_z(dy)\end{aligned}$$

where η _z denotes the ${\mathbb P}_z$ law of X _τ with τ the exit time of B(x, ε) (that τ is almost surely finite is well known and actually follows from the arguments below).

Up to the exit time of B(x, ε), X is just an Ornstein-Uhlenbeck process, so that its law is equivalent to the one of the Brownian motion. For Brownian motion, it is well known that τ is a.s. finite, that the exit measure (starting from z) is simply the harmonic measure (related to z) on the sphere S(x, ε), hence is equivalent to the surface measure σ _x. Thus the same properties hold true for our Ornstein-Uhlenbeck process.

It follows that η _z is equivalent to the surface measure σ _x on the sphere S(x, ε), so that η _z and η _x are equivalent.

(One can see e.g. [11] theorem 4.18 for much more sophisticated situations).

Choose $z \notin \mathbb {N}$. The previous formula shows that for η _z almost all y ∈ S(x, ε), ${\mathbb P}_y(\sup _n T_n< +\infty )=0$, so that the same holds η _x almost surely and finally ${\mathbb P}_x(\sup _n T_n< +\infty )=0$, showing that no explosion occurs starting from any point.

It remains to show that $\mathbb {E}$ is a conservative and local Dirichlet form. To this end introduce the truncated form

$$\displaystyle \begin{aligned} \mathbb{E}_n (f) = \frac{1}{\mu_{\lambda,\mathbb{X}}(U_n)} \, \int_{U_n} \, |\nabla f|{}^2 \, d\mu_{\lambda,\mathbb{X}} \end{aligned} $$

(1.45)

corresponding to the reflected O-U process in U _n with hard obstacles. It is not difficult to see that we can build the open sets U _n in such a way that ∂U _n is smooth. It thus follows that $\mathbb {E}_n$ is a conservative and local Dirichlet form, to which is associated a non-exploding process X ⁿ. The same reasoning as before shows that we can start from any point x ∈ U _n. We use the superscript n for the Markov law corresponding to $\mathbb {E}_n$

Let τ _K be the exit time from the ball B(0, K) and let n _K be such that for n ≥ n _K, B(0, K) ⊂ U _n. All processes X ⁿ (n ≥ n _K), starting from x ∈ B(0, K), coincide up to time τ _K (and coincide with the possibly exploding X). Now choose some initial measure ν(dy) = u(y)dy where u is bounded and has compact support included in B(0, R). Then ν is absolutely continuous with respect to $\mu ^n_{\lambda ,r}$ and one can find some constant C(K, ν) such that

$$\displaystyle \begin{aligned}\parallel \frac{d\nu}{d\mu_{\lambda,\mathbb{X}}^n}\parallel_\infty \, \leq \, C(K,\nu) \quad \mbox{ for all } n\geq n_K \, .\end{aligned}$$

For any T > 0, it yields, using the Lyons-Zheng decomposition as before

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathbb P}_\nu\left( \sup_{t\in[0;T]}|X^n_t| \geq K \right) &\displaystyle \leq&\displaystyle C(K,\nu) \, {\mathbb P}_{\mu^n_{\lambda,r}}\left( \sup_{t\in[0;T]} |X^n_t-X^n_0| \geq \frac{K}{2} \text{ or } |X^n_0| \geq \frac{K}{2} \right) \\ &\displaystyle \leq&\displaystyle C(K,\nu) \, \left({\mathbb P}_{\mu^n_{\lambda,r}} \left( \sup_{t\in[0;T]} |M^n_t| \geq \frac K4 \right) \right. \\ &\displaystyle &\displaystyle \left. + {\mathbb P}_{\mu^n_{\lambda,r}}\left( \sup_{t\in[0;T]} |RM^{T,n}_t| \geq \frac K4 \right)\right) \\ &\displaystyle &\displaystyle + \, C(K,\nu) \, {\mathbb P}_{\mu^n_{\lambda,r}} \left(|X^n_0| \geq \frac{K}{2} \right) \\ &\displaystyle \leq&\displaystyle C(K,\nu) \, \left(C_1 \, e^{- \, C_2 \, K^2/T} + \mu_{\lambda,\mathbb{X}}^n(B^c(0,K/2))\right) \\ &\displaystyle \leq&\displaystyle C(K,\nu) \, \left(C_1 \, e^{- \, C_2 \, K^2/T} + \, \frac{\mu_{\lambda,\mathbb{X}}(B^c(0,K/2))}{\mu_{\lambda,\mathbb{X}}(U_n)} \right) \end{array} \end{aligned} $$

for well chosen universal constants C ₁, C ₂. It immediately follows that ${\mathbb P}_\nu (\tau _K \leq T)$ (here we consider the process X) goes to 0 as K goes to + ∞, so that the process starting from ν does not explode. This is of course sufficient for our purpose, since conservativeness follows by choosing a sequence ν _j converging to $\mu _{\lambda ,\mathbb {X}}$.

Remark 1.3

Once the non explosion is proven, standard arguments show that the process is Feller. Hence compact sets are closed petite sets in the terminology of [17, 18]. We refer to the latter reference for a complete discussion.

Appendix 2: Useful Estimates for Exponential Moments of Hitting Times

In this section we shall recall some estimates of exponential moments of hitting times for some special linear processes. Denotes by S(r) the first exit time of the symmetric interval [−r, r] for a one dimensional process.

For the linear Brownian motion it is well known, (see [27] Exercise 3.10) that

$$\displaystyle \begin{aligned}E_0\left(e^{\theta \, S(r)}\right)= \frac{1}{\cos{}(r \, \sqrt{2\theta})}<+\infty\end{aligned}$$

if and only if

$$\displaystyle \begin{aligned}\theta \leq \frac{\pi^2}{8 \, r^2} \, .\end{aligned}$$

Surprisingly enough (at least for us) a precise description of the Laplace transform of S(r) for the O-U process is very recent: it was first obtained in [21]. A simpler proof is contained in [22] Theorem 3.1. The result reads as follows

Theorem 1.7 (See [21, 22])

If S(r) denotes the exit time from [−r, r] of a linear O-U process with drift − λx (λ > 0), then for θ ≥ 0,

$$\displaystyle \begin{aligned}E_0\left(e^{- \, \theta \, S(r)}\right) = \frac{1}{{}_1F_1\left(\frac{\theta}{2\lambda}\, , \, \frac 12 \, , \, \lambda \, r^2\right)} \, ,\end{aligned} $$

where ₁ F ₁ denotes the confluent hypergeometric function.

The function ₁F ₁ is also denoted by Φ (in [21] for instance) or by M in [1] (where it is called Kummer function) and is defined by

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} {}_1F_1(a,b,z) = \sum_{k=0}^{+\infty} \, \frac{(a)_k}{(b)_k} \, \frac{z^k}{k!} \quad \mbox{ where } \quad (a)_k = a(a+1)\ldots(a+k-1) \, , \, (a)_0=1 \, . \end{array} \end{aligned} $$

(1.46)

In our case, $b=\frac {1}{2}$, so that ₁F ₁ is an analytic function, as a function of both z and θ. It follows that $\theta \mapsto E_0\left (e^{- \, \theta \, S(r)}\right )$ can be extended, by analytic continuation, to θ < 0 as long as λr ² is not a zero of ${ }_1F_1{(\frac {\theta }{2\lambda } \, , \, \frac {1}{2},.)}$.

The zeros of the confluent hypergeometric function are difficult to study. Here we are looking for the first negative real zero. For − 1 < a < 0, b > 0, it is known (and easy to see) that there exists only one such zero, denoted here by u. Indeed ₁F ₁(a, b, 0) = 1 and all terms in the expansion (1.46) are negative for z > 0 except the first one, implying that the function is decaying to −∞ as z → +∞. However, an exact or an approximate expression for u are unknown (see the partial results of Slater in [1, 28], or in [20]). Our situation however is simpler than the general one, and we shall obtain a rough but sufficient bound.

First, comparing with the Brownian motion, we know that for all λ > 0 we must have

$$\displaystyle \begin{aligned}\frac{- \theta}{\lambda} \leq \frac{\pi^2}{8 (r\sqrt{\lambda})^2} \, .\end{aligned} $$

So, if λ r ² > π ²∕8 and − θ∕2λ ≥ 1∕2, the Laplace transform (or the exponential moment) is infinite. We may thus assume that − θ∕2λ < 1∕2.

Hence, for ${ }_1F_1\left (\frac {\theta }{2\lambda }\, , \, \frac 12 \, , \, \lambda \, r^2\right )$ to be negative it is enough that

$$\displaystyle \begin{aligned} \begin{array}{rcl} 1 &\displaystyle <&\displaystyle \frac{-\theta}{\lambda} \, \left((\lambda \, r^2) + \sum_{k=2}^{+\infty} \, \frac{(1+\frac{\theta}{2 \lambda})(2+\frac{\theta}{2 \lambda}) \ldots(k-1+\frac{\theta}{2 \lambda})}{(1+\frac 12)(2+\frac 12) \ldots (k-1+\frac 12)} \, \frac{(\lambda \, r^2)^k}{k!}\right) \\ &\displaystyle <&\displaystyle \frac{-\theta}{ \lambda} \, \, \left(\sum_{k=1}^{+\infty} \, \frac{(\lambda \, r^2)^k}{k!}\right) \, , \end{array} \end{aligned} $$

i.e.

$$\displaystyle \begin{aligned} \mbox{ as soon as } \quad \beta = - \theta \, > \, \frac{\lambda}{e^{\lambda \, r^2} -1} \quad \mbox{ then } \quad {\mathbb E}_{0}\left(e^{\beta \, S(r)}\right) = + \infty \, . \end{aligned} $$

(1.47)

So there is a drastically different behavior between both processes.

Finally we shall also need estimates for a general CIR process or generalized squared radial Ornstein-Uhlenbeck process, i.e. the solution of

$$\displaystyle \begin{aligned}dU_t=2\sqrt{U_t} dB_t + (\delta+2\beta \, U_t) \, dt\end{aligned}$$

when β > 0 and δ > 0. According to [21] Theorem 3, for θ > 0,

$$\displaystyle \begin{aligned} E_0\left(e^{- \, \theta \, S(u)}\right) = \frac{e^{\beta \, u}}{{}_1F_1\left(\frac{(\theta+\beta \delta)}{2\beta}\, , \, \frac \delta 2 \, , \, \beta \, u\right)} \, . \end{aligned} $$

(1.48)

It follows that for 0 < θ < β δ, $E_0\left (e^{\theta \, S(u)}\right )<+ \infty$.

Appendix 3: The Case N = 1: Another Estimate for a General y Using Decomposition of Variance

A very usual method to deal with dimension controls is the decomposition of variance. This method can be used here in order to transfer the results of Proposition 1.1 to a non centered obstacle. Though the results are non optimal in many directions, the method contains some interesting features.

In this section for simplicity we will first assume that λ = 1, and second that d ≥ 3. Recall that we are looking here at the case of an unique spherical obstacle B(y, r), so that we simply denote by μ _d,r the restricted Gaussian measure $\mu _{\lambda ,\mathbb {X}}$. Since we will use an induction procedure on the dimension d we explicitly make it appear in the notation.

Using rotation invariance we may also assume that y = (a, 0) for some $a\in {\mathbb R}^+$, 0 being the null vector of $\mathbb R^{d-1}$. So, writing $x=(u,\bar x) \in {\mathbb R}\times {\mathbb R}^{d-1}$,

$$\displaystyle \begin{aligned}\mu_{d,r}(du,d\bar x)= \nu^0_{d-1,R(u)}(d\bar x) \, \mu_1(du) \, ,\end{aligned}$$

where $\nu ^0_{d-1,R(u)}(d\bar x)$ is the d − 1 dimensional Gaussian measure restricted to B ^c(0, R(u)) as in Sect. 1.2.1 with $R(u)=\sqrt {\left (\left (r^2 - (u-a)^2\right )_+\right )}$ and μ ₁ is the first marginal of μ _d,r given by

$$\displaystyle \begin{aligned}\mu_1(du) = \frac{\gamma_{d-1}(B^c(0,R(u)))}{\gamma_d(B^c(y,r))} \, \, \gamma_1(du) \, ,\end{aligned}$$

γ _n denoting the n dimensional Gaussian measure $c_n \, e^{-|x|{ }^2} \, dx$.

The standard decomposition of variance tells us that for a nice f,

$$\displaystyle \begin{aligned} \operatorname{\mathrm{Var}}_{\mu_{d,r}}(f) = \int \, \left(\operatorname{\mathrm{Var}}_{\nu^0_{d-1,R(u)}}(f)\right) \, \mu_1(du) + \operatorname{\mathrm{Var}}_{\mu_1}(\bar f) \, , \end{aligned} $$

(1.49)

where

$$\displaystyle \begin{aligned}\bar f(u) = \int \, f(u,\bar x) \, \nu^0_{d-1,R(u)}(d \bar x) \, .\end{aligned}$$

According to Proposition 1.1, on one hand, it holds for all u,

$$\displaystyle \begin{aligned} \operatorname{\mathrm{Var}}_{\nu^0_{d-1,R(u)}}(f) \leq \left(1+\frac{(r^2 - (u-a)^2)_+}{d-1}\right) \, \int \, |\nabla_{\bar x} f|{}^2 \, d\nu^0_{d-1,R(u)} \, , \end{aligned} $$

(1.50)

so that

$$\displaystyle \begin{aligned} \int \, \left(\operatorname{\mathrm{Var}}_{\nu^0_{d-1,R(u)}}(f)\right) \, \mu_1(du) \leq \left(1+ \frac{r^2}{d-1}\right) \, \int \, |\nabla_{\bar x} f|{}^2 \, d\mu_{d,r} \, . \end{aligned} $$

(1.51)

On the other hand, μ ₁ is a logarithmically bounded perturbation of γ ₁ hence satisfies some Poincaré inequality so that

$$\displaystyle \begin{aligned} \operatorname{\mathrm{Var}}_{\mu_1}(\bar f) \leq C_1 \, \int \, \left|\frac{d\bar f}{du}\right|{}^2 \, d\mu_1 \, . \end{aligned} $$

(1.52)

So we have first to get a correct bound for C ₁, second to understand what $\frac {d\bar f}{du}$ is.

1.1.1 A Bound for C ₁

Since μ ₁ is defined on the real line, upper and lower bounds for C ₁ may be obtained by using Muckenhoupt bounds (see [3] Theorem 6.2.2). Unfortunately we were not able to obtain the corresponding explicit expression in our situation as μ ₁ is not sufficiently explicitly given to use Muckenhoupt criterion. So we shall give various upper bounds using other tools.

The usual Holley-Stroock perturbation argument combined with the Poincaré inequality for γ ₁ imply that

$$\displaystyle \begin{aligned} C_1 & \leq \frac 12 \, \frac{\sup_u \, \{\gamma_{d-1}(B^c(0,R(u)))\}}{\inf_u \, \{\gamma_{d-1}(B^c(0,R(u)))\}} \leq \frac 12 \, \frac{\int_0^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho}{\int_r^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho} \\ &= \frac 12 \, \left(1+ \frac{\int_0^r \, \rho^{d-2} \, e^{-\rho^2} \, d\rho}{\int_r^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho}\right) \, . \end{aligned} $$

(1.53)

Using the first inequality and the usual lower bound for the denominator, it follows that

$$\displaystyle \begin{aligned} \mbox{for all }\ r>0, \quad C_1 \leq \pi^{(d-2)/2} \, \frac{e^{r^2}}{r^{d-3}} \, .\end{aligned}$$

The function $\rho \mapsto \rho ^{d-2} \, e^{-\rho ^2}$ increases up to its maximal value which is attained for ρ ² = (d − 2)∕2 and then decreases to 0. It follows, using the second form of the inequality (1.53) that

• if $r\leq \sqrt {\frac {d-2}{2}}$ we have $C_1 \leq \frac 12 +r^2$, while

• if $r\geq \sqrt {\frac {d-2}{2}}$ we have

  $$\displaystyle \begin{aligned}C_1 \leq \frac 12 \, + \left(\frac{d-2}{2}\right)^{\frac{d-2}{2}} \, e^{- \frac{d-2}{2}} \, \frac{e^{r^2}}{r^{d-4}} \, .\end{aligned}$$

These bounds are quite bad for large r’s but do not depend on y.

Why is it bad? First for a = 0 (corresponding to the situation of Sect. 1.2.1) we know that $C_1 \leq 1 + \frac {r^2}{d}$ according to Proposition 1.1 applied to functions depending on x ₁. Actually the calculations we have done in the proof of Proposition 1.1, are unchanged for f(z) = z ₁, so that it is immediately seen that $C_1 \geq \max (\frac 12,\frac {r^2}{d})$.

Intuitively the case a = 0 is the worst one, though we have no proof of this. We can nevertheless give some hints.

The natural generator associated to μ ₁ is

The additional drift term behaves badly for a ≤ u ≤ a + r, since in this case it is larger than − u, while for u ≤ a it is smaller. In stochastic terms it means that one can compare the induced process with the Ornstein-Uhlenbeck process except possibly for a ≤ u ≤ a + r. In analytic terms let us look for a Lyapunov function for L ₁. As for the O-U generator the simplest one is g(u) = u↦u ² for which

Remember that a ≥ r so that $-au \leq - \, \frac 12 \, u^2$. It follows

$$\displaystyle \begin{aligned} \mbox{ provided}\ a\geq r, \quad L_1 g \leq 2 - 2g \, . \end{aligned} $$

(1.54)

For $|u|\geq \sqrt {2}$ we then have L ₁g(u) ≤− g(u), so that g is a Lyapunov function outside the interval $[-\sqrt {2}, \sqrt {2}]$ and the restriction of μ ₁ to this interval coincides (up to the constants) with the Gaussian law γ ₁ hence satisfies a Poincaré inequality with constant $\frac 12$ on this interval. According to the results in [5] we recalled in the previous section, we thus have that C ₁ is bounded above by some universal constant c.

We may gather our results

Lemma 1.6

The following upper bound holds for C ₁ :

• (1) (small obstacle) if $r\leq \sqrt {\frac {d-2}{2}}$ we have $C_1 \leq \frac 12 +r^2$ ,

• (2) (far obstacle) if $|y|>r + \sqrt {2}$ , C ₁ ≤ c for some universal constant c,

• (3) (centered obstacle) if y = 0, $C_1 \leq 1 + \frac {r^2}{d}$ ,

• (4) in all other cases, there exists c(d) such that $C_1\leq \, c(d) \, \frac {e^{r^2}}{r^{d-3}} \, .$

We conjecture that actually C ₁ ≤ C(1 + r ²) for some universal constant C.

Remark 1.4

In a recent work [25], the authors obtain a much better upper bound in case (4) (in fact a constant) when the origin belongs to the boundary of the ball and d = 3. ♢

1.1.2 Controlling $\frac {d \bar f}{du}$

It remains to understand what $\frac {d\bar f}{du}$ is and to compute the integral of its square against μ ₁.

Recall that

$$\displaystyle \begin{aligned}\bar f(u) = \int \, f(u,\bar x) \, \nu^0_{d-1,R(u)}(d \bar x) \, .\end{aligned}$$

Hence

where dθ is the non-normalized surface measure on the unit sphere $\mathbb S^{d-2}$ and c(d) the normalization constant for the Gaussian measure. Hence, for |u − a|≠ r we have

Notice that if f only depends on u, $\bar f=f$ so that

$$\displaystyle \begin{aligned}\frac{d}{du} \bar f(u) = \frac{\partial f}{\partial x_1}(u) = \int \, \frac{\partial f}{\partial x_1}(u) \, \nu^0_{d-1,R(u)}(d \bar x) \, ,\end{aligned}$$

and thus the sum of the two remaining terms is equal to 0. Hence in computing the sum of the two last terms, we may replace f by $f - \int f(u,\bar x) \, \nu ^0_{d-1,R(u)}(d \bar x)$ or if one prefers, we may assume that the latter $\int f(u,\bar x) \, \nu ^0_{d-1,R(u)}(d \bar x)$ vanishes. Observe that this change will not affect the gradient in the $\bar x$ direction.

Assuming this, the second term becomes

We thus have using a scale change

Our goal is to control the last term using the gradient of f. One good way to do it is to use the Green-Riemann formula, in a well adapted form. Indeed, let V  be a vector field written as

$$\displaystyle \begin{aligned} V(\bar x) = - \, \frac{\varphi(|\bar x|)}{|\bar x|{}^{d-1}} \, \, \bar x \quad \mbox{ where } \varphi(R(u))=R^{d-2}(u) \, . \end{aligned} $$

(1.55)

This choice is motivated by the fact that the divergence, $\nabla . (\bar x/|\bar x|{ }^{d-1}) = 0$ on the whole ${\mathbb R}^{d-1}-\{0\}$.

Of course in what follows we may assume that R(u) > 0, so that all calculations make sense. The Green-Riemann formula tells us that, denoting $g_u(\bar x)= f(u, \bar x)$, for some well chosen φ

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{\mathbb S^{d-2}(R(u))} \, f(u, \, \theta) \, d\theta &\displaystyle =&\displaystyle \int_{\mathbb S^{d-2}(R(u))} \, g_u \, \langle V,(- \bar x/|\bar x|)\rangle \, d\theta \end{array} \end{aligned} $$

Now we choose $\varphi (s)=R^{d-2}(u) \, e^{R^2(u)} \, e^{-s^2}$ and recall that . We have finally obtained

We will bound the above quantities for each fixed u. To control the first one we use Cauchy-Schwarz inequality, while for the second one we use first Cauchy-Schwarz inequality yielding a term containing $\int f^2(u,\bar x) \, \nu ^0_{d-1,R(u)}(d\bar x)$ and then the Poincaré inequality for $\nu ^0_{d-1,R(u)}$ we obtained in Proposition 1.1, since $\int \, f(u,\bar x) \, \nu ^0_{d-1,R(u)}(d \bar x) \, = \, 0$. This yields

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\frac{d\bar f}{du}\right| &\displaystyle \leq&\displaystyle \, \int \, \left|\frac{\partial f}{\partial x_1}\right| \, \nu^0_{d-1,R(u)}(d \bar x) \, \\ &\displaystyle &\displaystyle + \, 2 \, \left(\int |\nabla_{\bar x}f|{}^2 \, \nu^0_{d-1,R(u)}(d \bar x)\right)^{\frac 12} \, (A_1(u)+ 2 \, A_2(u)) \end{array} \end{aligned} $$

where

and

It follows

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int \left|\frac{d\bar f}{du}\right|{}^2 \, d\mu_1 &\displaystyle \leq&\displaystyle \, 2 \, \int \, \left|\frac{\partial f}{\partial x_1}\right|{}^2 \, \mu_{d,r}(du,d \bar x) \, \\ &\displaystyle &\displaystyle + \, 4 \, \sup_u \, \left(A_1(u)+ 2 \, A_2(u)\right)^2 \, \int |\nabla_{\bar x}f|{}^2 \, \mu_{d,r}(du,d \bar x) \, . \end{array} \end{aligned} $$

It remains to study the final supremum.

Recalling that $\nu ^0_{d-1,R(u)}$ is the (normalized) Gaussian measure restricted to $|\bar x|\geq R(u)$, we see that what we have to do is to get upper bounds for quantities like

$$\displaystyle \begin{aligned}\int_R^{+\infty} \, \rho^{-k} \, e^{-\rho^2} \, d\rho\end{aligned}$$

for k = d − 2 or k = d − 4, and a lower bound for

$$\displaystyle \begin{aligned}\int_R^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho \, .\end{aligned}$$

It is easily seen that for $k \in \mathbb N$,

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho^{-k} \, e^{-\rho^2} \, d\rho &\leq \, \frac{1}{R^{k+1}} \, \int_R^{+\infty} \, \rho \, e^{-\rho^2} \, d\rho \, \leq \, \frac{e^{-R^2}}{2 \, R^{1+k}} \quad \mbox{ if}\ R\geq 1, \end{aligned} $$

(1.56)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho^{-k} \, e^{-\rho^2} \, d\rho &\leq \frac{1}{2e} + \, \frac{1}{(k-1) R^{k-1}} \quad \mbox{ if}\ R\leq 1, k\geq 2, \end{aligned} $$

(1.57)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho^{-1} \, e^{-\rho^2} \, d\rho &\leq \frac{1}{2e} + \, \ln(1/R) \quad \mbox{ if }\ R\leq 1, \end{aligned} $$

(1.58)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, e^{-\rho^2} \, d\rho &\leq 1 \, + \, \frac{1}{2e} \quad \mbox{ if }\ R\leq 1, \end{aligned} $$

(1.59)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho \, e^{-\rho^2} \, d\rho &\leq \, \frac{1}{2} \quad \mbox{ if}\ R\leq 1, \end{aligned} $$

(1.60)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho &\geq \frac{R^{d-3}}{2 \, e^{R^2}} \quad \mbox{ if}\ R\geq 1, \end{aligned} $$

(1.61)

$$\displaystyle \begin{aligned} \int_R^{+\infty} \, \rho^{d-2} \, e^{-\rho^2} \, d\rho &\geq \frac{1}{2 \, e} \quad \mbox{ if}\ R\leq 1. \end{aligned} $$

(1.62)

Applying the previous bounds first for k = d − 2 (corresponding to A ₁) and k = d − 4 (corresponding to A ₂), we obtain first for R(u) ≥ 1,

$$\displaystyle \begin{aligned} A_1^2(u) &\leq r^2 \, R^{-2}(u) \, \leq \, r^2 \, ,\\ A_2^2(u) &\leq r^2 \, \left(1+\frac{R^2(u)}{d-1}\right) \, \leq \, r^2 \, \left(1+\frac{r^2}{d-1}\right) \end{aligned} $$

while for R(u) ≤ 1,

$$\displaystyle \begin{aligned} A_1^2(u) &\leq (1+2e) \, r^2 \, , \\ A_2^2(u) &\leq 2 \, (1+2e) \, r^2 \, , \end{aligned} $$

the latter bounds being obtained after discussing according to the dimension d = 3, 4, 5 or larger than 6.

Gathering all these results together we may state

Theorem 1.8

Assume d ≥ 3. There exists a function C(r, d) such that, for all $y \in {\mathbb R}^d$ ,

$$\displaystyle \begin{aligned}C_P(1,y,r) \, \leq \, C(r,d) \, .\end{aligned}$$

Furthermore, there exists some universal constant c such that

$$\displaystyle \begin{aligned}C(r,d) \leq C_1(r) \, C_2(r) \, ,\end{aligned}$$

C ₁(r) being given in Lemma 1.6 and C ₂(r) satisfying

• 1. if $r\leq \sqrt {(d-1)/2}$ , C ₂(r) ≤ c (1 + r ²),

• 2. if r ≥ 1, $C_2(r) \leq c \, r^2\left (1+\frac {r^2}{d-1}\right )$.

Remark 1.5

The previous theorem is interesting as it shows that when N = 1, the Poincaré constant is bounded uniformly in y and it furnishes some tractable bounds.

The method suffers nevertheless two defaults. First it does not work for d = 2, in which case the conditioned measure does no more satisfy a Poincaré inequality. More important for our purpose, the method does not extend to more than one obstacle, unless the obstacles have a particular location.