Dimension independent Bernstein–Markov inequalitiesin Gauss space

# Dimension independent Bernstein–Markov inequalities in Gauss space

Alexandros Eskenazis  and  Paata Ivanisvili Department of Mathematics, Princeton University Department of Mathematics, Princeton University; UC Irvine.
###### Abstract.

We obtain the following dimension independent Bernstein–Markov inequality in Gauss space: for each there exists a constant such that for any and all polynomials on we have

 ∥∇P∥Lp(Rk,dγk)≤Cp(degP)12+1πarctan(|p−2|2√p−1)∥P∥Lp(Rk,dγk),

where is the standard Gaussian measure on . We also show that under some mild growth assumptions on any function with we have

 ∫RkB(|LP(x)|)dγk(x)≤∫RkB(10(degP)αB|P(x)|)dγk(x)

where is the generator of the Ornstein–Uhlenbeck semigroup and

 αB=1+2πarctan⎛⎜⎝12 ⎷sups∈(0,∞){sB′′(s)B′(s)+B′(s)sB′′(s)}−2⎞⎟⎠.
This work was carried out under the auspices of the Simons Algorithms and Geometry (A&G) Think Tank.

2010 Mathematics Subject Classification. Primary: 41A17; Secondary: 41A63, 42C10, 28C20.

Key words. Gaussian measure, Bernstein-Markov inequality, Freud’s inequality, weighted approximation.

## 1. Introduction

Let be the standard Gaussian measure on , given by

 dγk(x)=e−|x|2/2√(2π)kdx,

where is the Euclidean length of . Here and throughout, we will denote by be the density of the Gaussian measure with respect to the Lebesgue measure on . For , define to be the space of those measurable functions on for which

 ∥f∥Lp(Rk,dγk):=(∫Rk|f|pdγk)1p<∞.

As usual, is defined by the condition . For convenience of notation, we will abbreviate as .

### 1.1. Freud’s inequality in high dimensions

In his seminal paper , Freud obtained the following weighted Bernstein–Markov type inequality on the real line.

###### Theorem 1 (Freud’s inequality, ).

There exists a universal constant such that for any and all polynomials on , we have

 (1) (∫R|P′(x)φ1(x)|pdx)1/p≤C√degP(∫R|P(x)φ1(x)|pdx)1/p.

After making a change of variables in (1), Freud’s inequality can be rewritten in terms of norms as

 (2) ∥P′∥Lp(dγ1)≤C√degPp∥P∥Lp(dγ1),

for all . Notice that (2) breaks down for as for every non-constant polynomial , nevertheless inequality (1) still persists.

After proving Theorem 1, Freud  extended his Gaussian estimates (1) to more general weights on the real line, nowadays known as Freud weights, where the function satisfies certain growth and convexity assumptions. In this case, the bound in (1) is replaced by a certain quantity which depends on the so-called Mhaskar–Rakhmanov–Saff numbers of the weight . Since the works [5, 6] of Freud, several different proofs of such one-dimensional weighted Bernstein–Markov inequalities have been found (see, e.g., [7, 22, 14, 15, 16, 13]), in part due to important implications of such estimates in approximation theory (see, e.g., [5, Theorem 2] and [6, Theorems 4.1 and 5.1]). We refer the reader to the beautiful survey  of Lubinsky for a detailed exposition of results on this subject.

From the point of view of high dimensional probability, it is natural to ask whether inequality (2) admits a dimension independent extension for the Gaussian measure on . Throughout the ensuing discussion, for a smooth function and , we will denote

 (3) ∥∇f∥Lp(dγk):=(∫Rk(k∑j=1(∂jf)2(x))p/2dγk(x))1/p.

We first notice (see also Section 5) that the case of Freud’s inequality (1) easily extends in all with a constant independent of the dimension.

###### Proposition 2.

There exists a universal constant such that for any , and all polynomials on we have

 (4) ∥φk∇P∥L∞(Rk)≤C√degP∥φkP∥L∞(Rk)

For finite values of , the following question naturally arises, in analogy to (2).

###### Question 3 (Bernstein–Markov inequality in Gauss space).

Is it true that for each there exists a constant such that for any integer , and all polynomials on the the dimension independent Gaussian Bernstein–Markov inequality

 (5) ∥∇P∥Lp(dγk)≤Cp√degP∥P∥Lp(dγk)

holds true?

###### Remark 4.

Using (2), it is straightforward to obtain (5) with a dimension dependent constant . Also, inequality (5) can easily be proven for (and ) by expanding in the Hermite basis and using orthogonality.

Before moving to our main result, we mention that an elegant argument of Maurey and Pisier from , implies a weakening of Question 3 with a (suboptimal) linear bound on .

###### Proposition 5.

There exists a universal constant such that for any , any and all polynomials on , we have

 (6) ∥∇P∥Lp(dγk)≤CdegP√p+1∥P∥Lp(dγk).

The main result of the present paper is that the linear bound on in (6) can be improved.

###### Theorem 6.

For each there exists a constant such that for any , and all polynomials on , we have

 (7) ∥∇P∥Lp(dγk)≤Cp(degP)12+1πarctan(|p−2|2√p−1)∥P∥Lp(dγk).

Notice that for each we have , therefore (7) is worse than (5) but improves upon (6). Also, notice that for , inequality (7) recovers (5). Our proof of Theorem 6, relies on a similar Bernstein–Markov type inequality for the generator of the Ornstein–Uhlenbeck semigroup (see Theorem 8 below) and Meyer’s dimension-free Riesz transform inequalities in Gauss space from .

### 1.2. Reverse Bernstein–Markov inequality in Gauss space

Our initial motivation to study Question 3 comes from a dual question that Mendel and Naor [19, Remark 5.5 (2)] asked on the Hamming cube. A positive answer to their question would, by standard considerations, imply its continuous counterpart in Gauss space, namely a reverse Bernstein–Markov inequality. To state the latter question precisely, let be the probabilists’ Hermite polynomial of degree on , i.e.,

 (8) Hm(s)=∫R(s+it)mdγ1(t).

For and a multiindex , where , we consider the multivariate Hermite polynomial on , given by

 (9) Hα(x)=k∏j=1Hαj(xj).

The family forms the orthonormal system on . Denote by and let be the generator of the Ornstein–Uhlenbeck semigroup. Then, one has

 LHα(x)=−|α|Hα(x).

for every multiindex . The operator should be understood as the Laplacian in Gauss space. Now consider any polynomial on which lives on frequencies greater than , i.e., of the form

 (10) P(x)=∑|α|≥dcαHα(x)
###### Question 7 (Mendel–Naor, ).

Is it true that for each there exists a constant such that for any , any , and all polynomials of the form (10) on , living on frequencies greater than , we have

 (11) ∥LP∥Lp(dγk)≥cpd∥P∥Lp(dγk).

In our upcoming manuscript , we show that for every there exists some such that for all polynomials which live on frequencies , i.e. are of the form

 P(x)=∑d≤|α|≤d+mcαHα(x),

we have

 (12) ∥LP∥Lp(dγk)≥cpdm∥P∥Lp(dγk).

For small values of , (12) improves upon previously known bounds in Question 7 which follow from works of Meyer [19, Lemma 5.4] and Mendel and Naor [19, Theorem 5.10] on the Hamming cube for this smaller subclass of polynomials. In particular, when , (12) positively answers a special case of Question 7. We refer to  for further results on reverse Bernstein–Markov inequalities along with extensions for vector-valued functions on the Hamming cube.

### 1.3. Bernstein–Markov inequality with respect to L

In order to prove Theorem 6, it will be convenient to first study the analogue of Question 3 for the “second derivative” , namely, is it true that for every polynomial on , we have

 (13) ∥LP∥Lp(dγk)≤CpdegP∥P∥Lp(dγk) ?

The best result that we could obtain in this direction is the following theorem.

###### Theorem 8.

For any integer , any , and any polynomial on , we have

 (14) ∥LP∥Lp(dγk)≤10(degP)1+2πarctan(|p−2|2√p−1)∥P∥Lp(dγk).

### 1.4. General function estimates

Our techniques for proving Theorem 8 allow us to replace -th powers in norms in (14) by an arbitrary convex increasing function in the spirit of the works [1, 2] of Arestov. We recall that Arestov’s theorem asserts that if is nondecreasing convex function on , and

 fn(t)=a02+n∑k=1(akcos(kt)+bksin(kt))

is a trigonometric polynomial of degree at most , then for every , the sharp inequality

 (15) ∫2π0Φ(|f(r)n(t)|)dt≤∫2π0Φ(nr|fn(t)|)dt

holds true. In fact, Arestov’s result holds true for a somewhat larger class of functions which contains for every (instead of just ), thus implying the usual Bernstein–Markov inequality for trigonometric polynomials.

One straightforward way to obtain an analog of (15) in Gauss space is to invoke the rotational invariance of the Gaussian measure. Indeed, we will shortly show the following estimates.

###### Theorem 9.

Let be an increasing convex function. For any , and all polynomials on we have

 (16) ∫Rk∫RΦ(|t∇P(x)|)dγ1(t)dγk(x)≤∫RkΦ((degP)|P(x)|)dγk(x);

and

 (17) ∫RkΦ(|LP(x)|)dγk(x)≤∫RkΦ((degP)2|P(x)|)dγk(x).

Our main result of this section is that under mild assumptions on , one can further improve (17).

###### Theorem 10.

For any and all polynomials on , we have

 (18) ∫RkB(|LP(x)|)dγk(x)≤∫RkB(10(degP)αB|P(x)|)dγk(x)

for any function with , such that for every

 max{|B(x+ε)|,B′(x+ε),B′′(x+ε)}

for each fixed and some . Here

 (19) αB:=1+2πarctan⎛⎜⎝12 ⎷sups∈(0,∞){sB′′(s)B′(s)+B′(s)sB′′(s)}−2⎞⎟⎠.

A straightforward optimization shows that Theorem 8 follows from Theorem 10 by considering , where .

The rest of the paper is structured as follows. In Section 2, we present the proof of Theorem 9 and its consequence, Proposition 5. In Section 3, we prove our main result, Theorem 10 from which we also deduce Theorem 8. Finally, Section 4 contains the deduction of Theorem 6 from Theorem 8 and Section 5 contains the proof of Proposition 2.

### Acknowledgments

We would like to thank Volodymyr Andriyevskyy for discussions that helped us to simplify the proof of Lemma 15.

## 2. Proof of Theorem 9

We first prove Theorem 9. The argument is inspired by an idea of Maurey and Pisier  and only uses the rotational invariance of the Gaussian measure and Arestov’s inequality (15).

Proof of Theorem 9. Let

 X=(X1,…,Xk)andY=(Y1,…,Yk)

be two independent multivariate standard Gaussian random variables on . Take any polynomial on of degree , and consider the trigonometric polynomial

 t(θ):=P(Xcos(θ)+Ysin(θ)).

Clearly . It follows from Arestov’s inequality (15) that

 (20) ∫2π0Φ(|t′(θ)|)dθ≤∫2π0Φ(n|t(θ)|)dθ.

Since , and the random variables

 Xcos(θ)+Ysin(θ)and−Xsin(θ)+Ycos(θ)

are also independent multivariate standard Gaussians, we see that after taking the expectation of (20) with respect to , we obtain

 2πEΦ(|∇P(X)||Y1|) =∫2π0EΦ(|∇P(Xcos(θ)+Ysin(θ))⋅(−Xsin(θ)+Ycos(θ))|) ≤∫2π0EΦ(n|P(Xcos(θ)+Ysin(θ))|)=2πEΦ(|P(X)|).

This finishes the proof of (16). To prove (17) notice that

 t′′(θ) =⟨HessP(Xcos(θ)+Ysin(θ))(−Xsin(θ)+Ycos(θ)),(−Xsin(θ)+Ycos(θ))⟩ −∇P(Xcos(θ)+Ysin(θ))⋅(Xcos(θ)+Ysin(θ)),

where denotes inner product in . Therefore following the same steps as before and using Arestov’s inequality (15), we obtain

 EΦ(|⟨HessP(X)Y,Y⟩−∇P(X)⋅X|)≤EΦ(n2|P(X)|).

Finally, it remains to use convexity of the map and Jensen’s inequality to get

 EYΦ(|⟨HessP(X)Y,Y⟩−∇P(X)⋅X|)≥Φ(|EY(⟨HessP(X)Y,Y⟩−∇P(X)⋅X))=Φ(|LP(X)|),

which completes the proof of (17).

Deriving Proposition 5 is now straightforward.

Proof of Proposition 5. It follows from the proof of Theorem 9 that (16) holds true under the sole assumption that Arestov’s inequality (15) holds. Thus, applying (16) to , where , we deduce that

 (21) (E|g|p)1/p∥∇P∥Lp(dγk)≤degP∥P∥Lp(dγk),

where is a standard Gaussian random variable. Inequality (6) then follows from (21) since for .

## 3. Proof of Theorem 10

In this section, we prove the main result of this paper, Theorem 10, and its consequence, the Bernstein–Markov inequality of Theorem 8.

### 3.1. Step 1. A general complex hypercontractivity

Any polynomial on admits a representation of the form

 (22) P(x)=∑|α|≤deg(P)cαHα(x),

for some coefficients . Next, given , we define the action of the second quantization operator (or Mehler transform) on as

 (23) TzP(x)=∑|α|≤deg(P)z|α|cαHα(x).

Clearly , and .

In what follows we will be working with a real-valued function (and sometimes we will further require ) such that

 (24) |R(x)|,|R′(x)|,|R′′(x)|≤C(1+xN)

for some constants and every . These assumptions are sufficient to avoid integrability issues.

###### Lemma 11.

Fix , and let be a real-valued function such that . Assume that

 (25) (1−|z|2)R′(x)|w|2+2xR′′(x)((Rw)2−(Rzw)2)≥0for allw∈C and x≥0.

Then, for all , and for all polynomials on we have

 (26) ∫RkR(|TzP(x)|2)dγk(x)≤∫RkR(|P(x)|2)dγk(x).
###### Remark 12.

We will see later (see Section 3.4) that the reverse implication also holds, i.e., inequality (26) implies (25) under the additional assumption that .

###### Proof.

Denote the scaled Gaussian measure on of variance by

 dγ(s)k(x)=1√(2πs)ke−|x|2/2sdx,s∈(0,1].

Take any polynomial . We will denote partial derivatives by lower indices, for example

 gxixj(x):=∂2∂xi∂xjg(x).

Fix a complex number satisfying (25), and consider the map

 (27) g(x,u,s):=∫Rk∫Rkg((u+iv)+z(x+iy))dγ(s)k(v)dγ(1−s)k(y),

where for we denote . An analysis done in [11, 10, 12] suggests that one should study the monotonicity of the following map,

 (28) [0,1]∋s⟼r(s)=∫Rk∫RkR(|g(x,u,s)|2)dγ(s)k(u)dγ(1−s)k(x).

First notice that if the map is increasing then (26) follows. Indeed, consider any polynomial on of the form and define . Then,

 r(0)=∫RkR(∣∣∫Rkg(u+iv)dγk(v)∣∣2)dγk(u)\lx@stackrel=∫RkR(|P(u)|2)dγk(u)

and, similarly, also

 r(1)=∫RkR(∣∣∫Rkg(z(x+iy))dγk(y)∣∣2)dγk(x)\lx@stackrel=∫RkR(|TzP(x)|2)dγk(x).

Therefore, (26) can be rewritten as .

Next, we show that the monotonicity of (28) follows from (25). Notice that for any function such that for some we have

 (29) dds(∫RkQ(x)dγ(s)k(x))=∫RkΔQ(x)2dγ(s)k(x).

Indeed by making change of variables we obtain . Therefore

 dds(∫RkQ(x)dγ(s)k(x))=dds(∫RkQ(y√s)2dγk(y))=∫Rk∇Q(y√s)⋅y2√sdγk(y)

and

 ∫RkΔQ(x)2dγ(s)k(x)=∫RkΔQ(y√s)2dγk(y).

Notice that if we denote then (29) simply means that . The latter follows from integration by parts. Therefore, we have

To compute the first term, one differentiation gives

 [∫RkR(|g(x,u,s)|2)dγ(1−s)k(x)]uj=∫RkR′(|g(x,u,s)|2)[|g(x,u,s)|2]ujdγ(1−s)k(x),

which implies that

 [∫RkR(|(x,u,s)|2)dγ(1−s)k(x)]ujuj= ∫RkR′′(|g(x,u,s)|2)([|g(x,u,s)|2]uj)2+R′(|g(x,u,s)|2)[|g(x,u,s)|2]ujujdγ(1−s)k(x).

For the second term, we have

 (∫RkR(|g(x,u,s)|2)dγ(1−s)k(x))s=−12∫RkΔxR(|g(x,u,s)|2)−2[R(|g(x,u,s)|2)]sdγ(1−s)k(x).

Thus we get that where

 ~r(s)=ΔuR(|g(x,u,s)|2)−ΔxR(|g(x,u,s)|2)+2[R(|g(x,u,s)|2)]s.

Now, compute

 g(x,u,s)uj=gj(x,u,s):=gj   and   g(x,u,s)ujuj=gjj(x,u,s):=gjj,

where is the -th partial derivative of and we denote

 gj(x,u,s)=∫Rk∫Rkgj((u+iv)+z(x+iy))dγ(s)k(v)dγ(1−s)k(y),

and similarly means that we first differentiate the polynomial twice in the -th coordinate and then we apply the flow (27). Similarly, we have

 g(x,u,s)xj=zgj,  g(x,u,s)xjxj=z2gj  and  g(x,u,s)s=z2−12k∑j=1gjj.

Next, further abusing the notation, we will denote . We have

 (|g(x,u,s)|2)uj=gj¯g+g¯gj=2R(|g|2gjg); (|g(x,u,s)|2)ujuj=gjj¯g+g¯gjj+2gj¯gj=2R(|g|2gjjg)+2|gj|2; (|g(x,u,s)|2)xj=zgj¯g+g¯z¯gj=2R(|g|2zgjg); (|g(x,u,s)|2)xjxj=2R(|g|2z2gjjg)+2|z|2|gj|2; (|g(x,u,s)|2)s=R⎛⎝|g|2(z2−1)∑kj=1gjjg⎞⎠.

Therefore we obtain

 ΔuR(|g(x,u,s)|2)−ΔxR(|g(x,u,s)|2)+2[R(|g(x,u,s)|2)]s= R′(|g(x,u,s)|2)(Δu(|g(x,u,s)|2)−Δx(|g(x,u,s)|2)+2(|g(x,u,s)|2)s)+ R′′(|g(x,u,s)|2)(k∑j=1(|g(x,u,s)|2)2uj−(|g(x,u,s)|2)2xj)= 2k∑j=1(1−|z|2)R′(|g|2)|gj|2+2R′′(|g|2)((Rgj¯g)2−(Rzgj¯g)2).

Notice that if then by (25) we have , and hence in this case. So assume that . Then, denoting and , we see that follows from

 (1−|z|2)R′(x)|w|2+2xR′′(x)((Rw)2−(Rzw)2)≥0,x≥0,w∈C.

The latter condition is exactly (25) and the proof is complete. ∎

###### Lemma 13.

Let , be such that , and satisfies (24) for each fixed . Take any , such that the inequality

 (30) (1−|z|2)R′(x)|w|2+2xR′′(x)((Rw)2−(Rzw)2)≥0

holds for all , and all . Then for all polynomials on