doi:10.1112/S0025579312001064
L1-SMOOTHING FOR THE ORNSTEIN–UHLENBECK SEMIGROUP
K. BALL, F. BARTHE, W. BEDNORZ, K. OLESZKIEWICZANDP. WOLFF
Abstract. Given a probability density, we estimate the rate of decay of the measure of the level sets of its evolutes by the Ornstein–Uhlenbeck semigroup.
The rate is faster than what follows from the preservation of mass and Markov’s inequality.
§1. Introduction. Let N> 1. For t > 0, consider the probability measure µt =1 − e−t
2 δ−1+1 + e−t 2 δ1.
We write simply µ for µ∞=(δ−1+δ1)/2. On the (multiplicative) group {−1, 1}N, we consider the semigroup of operators(Tt)t>0defined for functions
f : {−1, 1}N → R by
Tt f = f ∗µtN. In other words,
Tt f(x) =Z
f(x · y)Kt(y) dµN(y) where Kt(y) =
N
Y
i =1
(1 + e−tyi).
For A ⊂ {1, . . . , N}, we define wA: {−1, 1}N → R by wA(y) = Qi ∈ A yi, with the convention thatw∅=1. This family, known as the Walsh system, forms an orthonormal basis of L2({−1, 1}N, µN). Expanding the product in the definition of the kernel Kt, one readily checks that TtwA=e−tcard(A)wA.
The above formulations show that Ts◦Tt=Ts+tand that Tt is self-adjoint on L2and preserves positivity and integrals (with respect toµN). As a consequence, Ttis a contraction from Lp=Lp({−1, 1}N, µN) into itself: kTt f kp6 k f kpfor p> 1. Actually, the hypercontractive estimate of Bonami [2]and Beckner [1] tells us more: if 1< p < q < +∞ and e2t > (q − 1)/( p − 1), then
kTtf kq6 k f kp.
Hence the semigroup improves the integrability of functions in Lp, provided p> 1. A challenging problem is to understand the improving effects of Tt
on functions f ∈ L1. In the paper [5], Talagrand asks the following question:
Received 31 January 2012, published online 2 August 2012.
MSC (2010): 46N30, 60E15 (primary).
The research of the third, fourth and fifth authors was partially supported by the Polish MNiSzW grants 1 PO3A 012 29 and N N201 397437.
for t> 0, is there a function ψt : [1, +∞) → (0, +∞) with limu→+∞ψt(u) = +∞such that for every N > 1, every function f on {−1, 1}N with k f k16 1 and all u> 1,
µN {x : |Tt f(x)| > u} 6 1
uψt(u) ? (1)
This would be a strong improvement on the following simple consequence of Markov’s inequality and the contractivity property:
µN {x : |Tt f(x)| > u} 6kTtf k1
u 6kf k1 u .
Talagrand actually asks a more specific question withψt(u) = c(t)plog(u), and he observes that one cannot expect a faster rate in u. Question (1) is still open;
only in some special cases is an affirmative answer known (see the final section).
Its difficulty is essentially due to the lack of convexity of the tail condition.
Nevertheless, the paper [5] contains a similar result for the averaged operator M :=R1
0 Ttdt: there exists K such that for all N and u> 1, µN {x : |M f(x)| > uk f k1} 6 Klog log u
ulog u ·
The goal of this note is to study the analogue of Question (1) in Gauss space.
§2. Gaussian setting. Let n> 1. We work on Rn with its canonical Euclidean structure(h· , ·i, | · |). Denote by γnthe standard Gaussian probability measure on Rn:
γn(dx) = e−|x |2/2 d x (2π)n/2·
Let G be a standard Gaussian random vector, with distributionγn. Let f : Rn→ R be measurable. Then the Ornstein–Uhlenbeck semigroup (Ut)t>0is defined by
Utf(x) = E f e−tx +p
1 − e−2t G
= Z
Rn
f e−tx +p
1 − e−2t y e−|y|2/2 d y (2π)n/2
=(1 − e−2t)−n/2Z
Rn
f(z) e−|z−e−tx |2/(2(1−e−2t)) d z (2π)n/2
=(1 − e−2t)−n/2e|x |2/2 Z
Rn
f(z) e−[e−2t/(2(1−e−2t))]|z−etx |2dγn(z), when f is non-negative or belongs to L1(γn). The operators Ut preserve positivity and mean. They are self-adjoint on L2(γn). By Nelson’s hypercontractivity theorem [3], Ut is a contraction from Lp(γn) to Lq(γn) provided 1< p 6 q and (p − 1)e2t > q − 1. It is natural to ask the analogue of Question (1) for Ut: does there exist a functionψt with limu→+∞ψt(u) = +∞
such that for all n and all non-negative orγn-integrable functions f : Rn→ R, γn {x : |Utf(x)| > uk f kL1(γn)} 6 1
uψt(u) ? (2)
This inequality would actually follow from Talagrand’s conjecture on the discrete cube. Indeed, if f : Rn→ R is continuous and bounded, consider the function g : {−1, 1}nk→ R defined by
g (xi, j)i6n, j6k = f x1,1+ · · · +√ x1,k
k , . . . , xn,1+ · · · +√ xn,k
k
.
By the central limit theorem, as k goes to infinity, the distribution of g underµnk tends to that of f underγn, while the distribution of Ttgunderµnk tends to that of Ut f underγn(see, e.g., [1]). This allows us to pass from (1) for g to (2) for f . The above argument uses boundedness and continuity. These assumptions can be removed by using a classical truncation argument and the semigroup property:
Ut f = Ut/2Ut/2f where Ut/2f is automatically continuous. We omit the details.
To conclude this introduction, let us provide evidence that the functionsψt(u) in (2) cannot grow faster thanplog u. We will do this for n = 1, which implies the general case (by choosing functions depending on only one variable). We have shown that
Utf(x) =Z
R
Qt(x, z) f (z) dγ1(z), (3) where
Qt(x, z) = (1 − e−2t)−1/2exp 1 2
x2− (z − etx)2 e2t−1
. We are going to choose specific functions f > 0 with R
f dγ1=1 for which Utf can be explicitly computed. Note that (3) allows us to extend the definition of Ut to (non-negative) measures ν with R ϕ dν < +∞, where ϕ(t) = e−t2/2/√
2π is the Gaussian density. The simplest choice is then to take normalized Dirac masses eδy:=ϕ(y)−1δy as test measures. Obviously, R ϕ deδy=1 and Uteδy=Qt(· , y). Actually, by the semigroup property,
Qt(· , y) = Ut/2Ut/2eδy=Ut/2Qt/2(· , y),
where x 7→ Qt/2(x, y) is a non-negative function with unit Gaussian integral.
Hence,
{Qt(· , y) : y ∈ R} ⊂n
Ut/2f : f > 0 and Z
f dγ1=1o.
Fix t> 0 and let u > (1 − e−2t)−1/2. Then, using Qt(x, y) = Qt(y, x) and settingv = u√
1 − e−2t, one readily obtains that
x : Qt(x, y) > u = (
x :exp 1 2
y2−(x − ety)2 e2t −1
> v )
= ety −
q(e2t −1)(y2−2 logv)+,
ety +
q(e2t −1)(y2−2 logv)+.
For the particular choice y = y0:=etp2 logv, one gets {x : Qt(x, y0) > u} =p
2 logv, (2e2t−1)p2 log v . Since for 0< a < b,
γ1((a, b)) > 1
√2π Z b
a
s
be−s2/2ds =e−a2/2−e−b2/2 b
√2π ,
we can deduce that
γ1 {x : Qt(x, y0) > u} > 1 2√
π(2e2t −1)plog v
1
v − 1
v(2e2t−1)2
.
Combining the above observations yields lim inf
u→+∞up
log u supnγ1({x : Ut/2f(x) > u}) : f > 0 and Z
f dγ1=1o> 0.
Henceψt(u) in (2) cannot grow faster thanplog u.
Using the same one-dimensional test functions and similar calculations, one can check that for t> 0, the image under Ut of the unit ball B1= {f ∈ L1(γn) : kf k16 1} is not uniformly integrable, that is:
lim inf
c→+∞ sup
f ∈B1
Z
|Utf |1|Utf |>cdγn> 0.
Consequently, Ut:L1(γn) → Lφ(γn) is not continuous when φ is a Young function with limt →+∞φ(t)/t = +∞. Next, we turn to positive results.
§3. Main results. In the rest of this section B(a, r) denotes the closed ball of center a and radius r , while
C(a, r1, r2) = {x ∈ Rn:r16 |x − a| 6 r2}. We start with an easy inclusion of the upper level-sets of Ut f.
LEMMA1. Let f : Rn→ R+ be such that R
f dγn=1. Then for all t, u > 0,
{x ∈ Rn:Ut f(x) > u} ⊂ B 0,q
2 log u + n log(1 − e−2t)+c
. Proof. As already explained,
Utf(x) = (1 − e−2t)−n/2ex2/2 Z
Rn
f(z)e−[e−2t/(2(1−e−2t))](z−etx)2dγn(z).
Consequently,
Utf(x) 6 (1 − e−2t)−n/2ex2/2 Z
f dγn.
Our normalization hypothesis then implies that
{x : Ut f(x) > u} ⊂ {x : |x|2> 2 log u + n log(1 − e−2t)}. 2 The probability measure of complements of balls appearing in the above lemma can be estimated thanks to the following classical fact.
LEMMA2. For all n ∈ N∗ there exists a constant cn such that for all u>
√ 2n,
γn B(0, u)c 6 cnun−2e−u2/2.
Actually, when n6 2 this is valid for all u > 0. Also, one may take c1=√ 2/π.
Proof. Polar integration gives
γn B(0, u)c =(2π)−n/2·nvoln(B(0, 1))Z +∞
u
rn−1e−r2/2dr.
For u2> 2n − 4, the map r 7→ rn−2e−r2/4is non-increasing on(u, ∞). Thus we may bound the last integral as
Z ∞ u
rn−2e−r2/4·re−r2/4dr6 Z ∞
u
un−2e−u2/4·re−r2/4dr =2un−2e−u2/2,
and this proves the lemma. 2
Combining the previous statements gives a satisfactory estimate in dimension one, which improves on the Markov estimate γn(Utf > u) 6 min(1, 1/u) if f is non-negative with integral 1.
PROPOSITION3. Let f : R → R+ be integrable. Then for all t> 0 and v > 1,
γ1
x : Ut f(x) > v
R f dγ1
√
1 − e−2t
6 1
vpπ log v·
In higher dimensions, the above reasoning gives a weaker estimate than Markov’s inequality. However, a more precise approach allows us to get a slightly weaker decay for the level sets of Ut f. Our main result is stated next. It contains a dimensional dependence that we have not been able to remove.
THEOREM4. Let n> 2 and t > 0. Then there exists a constant K (n, t) such that for all non-negative functions f defined on Rn withR
f dγn=1, we have that for all u> 10,
γn {x ∈ Rn: Ut f(x) > u} 6 K (n, t)log log u uplog u·
Proof. Note that it is enough to show the inequality for u larger than some number u0(n, t) > 10 depending only on n and t. We will just write that we choose u large enough, but an explicit value of u0(n, t) can be obtained from
our argument. Let us define
R1=R1(u, n, t) := 2 log u + n log(1 − e−2t)1/2+ , R2=R2(u, n) := 2 log u + (n − 1) log log u1/2. It is clear that for u large enough, R2>√
2n and also R2> R1> 0. So, by Lemma2,
γn B(0, R2)c 6 cne−R22/2R2n−2=cn
u
(2 log u + (n − 1) log log u)(n−2)/2 (log u)(n−1)/2
6 cn
u
((n + 1) log u)(n−2)/2
(log u)(n−1)/2 = c0n uplog u.
By Lemma1, {x : Utf(x) > u} is a subset of B(0, R1)c. Hence we may write, using Markov’s inequality on the annulus C(0, R1, R2) and the self-adjointness of Ut,
γn {x : Utf(x) > u}
6 γn {x : Ut f(x) > u} ∩ B(0, R2) + γn B(0, R2)c
=γn {x : Ut f(x) > u} ∩ B(0, R1)c∩B(0, R2) + γn B(0, R2)c 6
Z Ut f
u 1C(0,R1,R2)dγn+γn B(0, R2)c
= 1 u
Z
(Ut1C(0,R1,R2)) f dγn+γn B(0, R2)c
6 1
ukUt1C(0,R1,R2)k∞+ c0n uplog u· To prove the theorem, it remains to show that
kUt1C(0,R1,R2)k∞=O log log u plog u
.
First, note that for any set A ⊂ Rnand all x ∈ Rn, Ut1A(x) = E1A e−tx +p
1 − e−2t G
= P
G ∈ A −e−tx
√
1 − e−2t
=γn
A − e−tx
√
1 − e−2t
. Therefore
kUt1C(0,R1,R2)k∞= sup
a∈Rn
γn C(a,Re1,eR2), where eRi :=Ri/√
1 − e−2t. The main idea is that the shells described above can be covered by a thin slab and the complement of a large ball. Set
r = r(u) := 2(log log u)1/2;
then for u large enough, Lemma2yields
γn B(0, r)c 6 cne−r2/2rn−2=cn2n−2(log log u)(n−2)/2 (log u)2 6 c00n
log log u plog u . For an arbitrary point a ∈ Rn,
γn C(a,eR1,Re2) 6 γn C(a,eR1,eR2) ∩ B(0, r) + γn B(0, r)c 6 γn C(a,eR1,eR2) ∩ B(0, r) + c00n
log log u plog u . For u large enough, r<Re1, and Lemma5, which is stated below, ensures that C(a,eR1,Re2) ∩ B(0, r) is contained in a strip S of width
w :=eR2− q
eR12−r2.
By the product properties of the Gaussian measure, γn(S) coincides with the one-dimensional Gaussian measure of an interval of lengthw. Therefore it is no bigger thanw/√
2π 6 w. Hence γn C(a,eR1,eR2) ∩ B(0, r)
6 eR2− q
eR12−r2
=r 2 log u +(n − 1) log log u 1 − e−2t
−
s2 log u + n log(1 − e−2t)
1 − e−2t −4 log log u 6(n − 1 + 4(1 − e−2t)) log log u − n log(1 − e−2t)
√
1 − e−2tp2 log u +(n − 1) log log u 6 κ(n, t)log log u
plog u ,
where the last inequality is valid for u large enough. The proof of the theorem is
therefore complete. 2
LEMMA5. Let 0< r < ρ1< ρ2and a, b ∈ Rn; then the set C(a, ρ1, ρ2) ∩ B(b, r)
is contained in a strip of width at mostρ2−
qρ12−r2.
Proof. Assume that the intersection is not empty. Then without loss of generality, a = 0 and b = t e1 with t> 0. Let z be an arbitrary point in the intersection. Obviously, z16 |z| 6 ρ2. Next, since z ∈ B(b, r) and |z| > ρ1, one gets
r2> |z − te1|2= |z|2−2t z1+t2> ρ12−2t z1+t2.
Hence, by the arithmetic mean–geometric mean inequality, the first coordinate of z satisfies
z1> 1 2
ρ12−r2
t +t
!
>
qρ12−r2.
Summarizing, we have z ∈q
ρ12−r2, ρ2 × Rn−1. 2
§4. Product functions on the discrete cube.Finally, we provide an affirmative answer to Question (1) in the case of functions with product structure.
PROPOSITION6. Assume that functions f1, f2, . . . , fN: {−1, 1} → [0, ∞) satisfy R
fidµ = 1 for i = 1, 2, . . . , N. Let f = f1⊗ f2⊗ · · · ⊗ fN, i.e.
f(x) = Qi =1N fi(xi). Then for every t > 0 there exists a positive constant ct
such that for all u> 1,
µN {x : |Tt f(x)| > u} 6 ct
uplog u.
Proof. The above result is immediately implied by the following inequality.
PROPOSITION7 [4]. Let b> a > 0. Let X1, X2, . . . , XN be independent non-negative random variables such that E Xi =1 and a6 Xi6 b a.s. for i =1, 2, . . . , N. Then for every u > 1 we have
P
N
Y
i =1
Xi > u
6 Cu−1(1 + log u)−1/2,
where C is a positive constant which depends only on a and b.
Indeed,
Tt f = Tt f1⊗Tt f2⊗ · · · ⊗Tt fN,
where Ttfi : {−1, 1} → [1 − e−t, 1 + e−t]satisfyR Tt fi dµ = R fidµ = 1 for i =1, 2, . . . , N. Thus, the random variables X1, X2, . . . , XN defined on the probability space({−1, 1}N, µN) by Xi(x) = Tt fi(xi) satisfy the assumptions of Proposition7with a = 1 − e−t and b = 1 + e−t, while f =QN
i =1Xi. 2
References
1. W. Beckner, Inequalities in Fourier analysis. Ann. of Math. (2) 102(1) (1975), 159–182.
2. A. Bonami, Etude des coefficients de Fourier des fonctions de Lp(G). Ann. Inst. Fourier (Grenoble) 20(2) (1971), 335–402.
3. E. Nelson, The free Markoff field. J. Funct. Anal. 12 (1973), 211–227.
4. K. Oleszkiewicz, On Eaton’s property (in preparation).
5. M. Talagrand, A conjecture on convolution operators, and a non-Dunford–Pettis operator on L1. Israel J. Math.68(1) (1989), 82–88.
K. Ball,
Mathematics Institute, University of Warwick, Coventry, CV4 7AL, U.K.
E-mail: kmb120205@googlemail.com
F. Barthe,
Institut de Mathématiques de Toulouse (CNRS UMR 5219),
Université Paul Sabatier, 31062 Toulouse cedex 09, France
E-mail: barthe@math.univ-toulouse.fr W. Bednorz,
Institute of Mathematics, University of Warsaw, Banacha 2, 02-097 Warszawa, Poland
E-mail: W.Bednorz@mimuw.edu.pl
K. Oleszkiewicz, Institute of Mathematics, University of Warsaw, Banacha 2, 02-097 Warszawa, Poland
and
Institute of Mathematics, Polish Academy of Sciences,
´Sniadeckich 8, 00-956 Warszawa, Poland
E-mail: K.Oleszkiewicz@mimuw.edu.pl P. Wolff,
Institute of Mathematics, Polish Academy of Sciences,
´Sniadeckich 8, 00-956 Warszawa, Poland
E-mail: pwolff@mimuw.edu.pl
