June 11, 2026
We study dimension-free Markov–Bernstein inequalities for polynomials with respect to product probability measures. In the Gaussian case, for \(p\ge4\), we prove that \[\|\nabla f\|_{L^p(\gamma^n)} \le C(p)d^{\frac{1}{2}+\theta_p} \|f\|_{L^p(\gamma^n)}\] for every polynomial \(f\) of degree at most \(d\), where \(\theta_p\le \frac{2}{3p}\) and \(\theta_p=0\) whenever \(p\) is an even integer. Thus, for even integer exponents, we establish the sharp dependence on the degree conjectured by Eskenazis–Ivanisvili. For general \(p\ge4\), the estimate improves upon their dimension-free inequality.
We also obtain dimension-free Markov–Bernstein inequalities with sharp dependence on the degree for even integer exponents beyond the Gaussian setting. We first prove such estimates for the uniform distribution on the unit cube and then extend them to products of absolutely continuous measures with unimodal densities. Finally, we treat products of one-dimensional Freud measures with densities proportional to \(e^{-|t|^{2m}}\).
Markov–Bernstein inequalities estimate norms of derivatives of algebraic polynomials in terms of norms of the polynomials themselves. In the one-dimensional case, the classical Markov inequality on an interval asserts that \[\|f'\|_{L^\infty([-1,1])} \le d^2\|f\|_{L^\infty([-1,1])}\] for every algebraic polynomial \(f\) of degree at most \(d\). Bernstein’s inequality refines this estimate by introducing a weight which compensates for the endpoint effect: \[\|(1-x^2)^{1/2}f'\|_{L^\infty([-1,1])} \le d\|f\|_{L^\infty([-1,1])}.\] The factors \(d^2\) and \(d\) in these inequalities are sharp, as can already be seen from the Chebyshev polynomials. We refer to [1], [2] for detailed discussions of classical Markov and Bernstein inequalities. Integral versions of such inequalities were studied, for instance, in [3], [4], while weighted variants were developed in [5]–[8]. Besides their intrinsic interest, these inequalities play an important role in many areas of analysis, especially in approximation theory, see, for example, [9] and [10].
Multivariate analogues of the classical Markov–Bernstein inequalities, with the derivative replaced by the Euclidean norm of the gradient, have also been extensively studied in both the uniform norm and integral \(L^p\) norms; see, for instance, [11]–[15] and [16]–[21], respectively. A recent line of work of Kroó, Dai, and Prymak [22]–[26] demonstrates the importance of multivariate Markov–Bernstein inequalities in the construction of optimal polynomial meshes and in Marcinkiewicz-type discretization of \(L^p\) norms.
Most of the results cited above belong to the classical finite-dimensional framework, where the ambient dimension and the underlying domain are fixed, and the constants may depend on both the dimension and the geometry of the domain. In many modern problems at the interface of high-dimensional analysis and probability, a different point of view is more natural. The ambient dimension may be arbitrarily large, and one seeks estimates with constants independent of the dimension. In the Gaussian setting, such dimension-free results for polynomials are especially natural, since they can be transferred to abstract Wiener spaces and, in particular, to finite sums of multiple stochastic integrals, see [27].
Dimension-free estimates also arise naturally in questions concerning anti-concentration of polynomials [28], regularity of distributions of polynomials in log-concave random vectors [29], total variation distance estimates between such distributions [30], and oscillatory integrals with polynomial phases [31]. In these problems, one often needs to control the non-degeneracy of polynomial distributions, which is typically measured in terms of the variance of the polynomial with respect to the underlying measure. Markov–Bernstein-type inequalities are well suited for this purpose, since, when applied iteratively, they relate the leading coefficients of a polynomial to its \(L^2\) norm, and hence to its variance after centering. Related dimension-free coefficient estimates were recently studied by Glazer–Mikulincer [32], [33].
From this dimension-free viewpoint, integral Markov–Bernstein inequalities for the full gradient seem to have been investigated much less systematically and, to the best of our knowledge, are known only in a few specific settings. Namely, Kroó–Szabados [34] obtained \(L^2\) Markov–Bernstein inequalities for products of Gaussian and Gamma distributions, while general \(L^p\) Gaussian Markov–Bernstein inequalities were studied by Eskenazis–Ivanisvili [35].
The main goal of this paper is to study dimension-free integral \(L^p\) Markov–Bernstein inequalities with sharp dependence on the degree \(d\). For general high-dimensional measures, such estimates are not available for the whole class of algebraic polynomials, even in the log-concave setting, as already shown by the normalized Lebesgue measure on the isotropic Euclidean ball [32]. We therefore focus on product measures and prove sharp dimension-free estimates for even integer values of \(p\). In the Gaussian case, we also obtain improved estimates for all \(p\ge4\).
Let \(\gamma\) denote the standard Gaussian measure on \(\mathbb{R}\). Classical results of Freud [5], [6] imply that, for \(p\in[1,\infty)\), \[\|f'\|_{L^p(\gamma)} \le C(p)\sqrt d\,\|f\|_{L^p(\gamma)}\] for every polynomial \(f\) of degree at most \(d\).
Motivated by this one-dimensional estimate, Eskenazis and Ivanisvili [35] formulated the following multidimensional conjecture.
Conjecture 1. For every \(p\in[1,\infty)\), there exists a constant \(C(p)>0\) such that, for every \(n,d\in\mathbb{N}\), one has \[\|\nabla f\|_{L^p(\gamma^n)} \le C(p)\sqrt d\,\|f\|_{L^p(\gamma^n)} \quad \forall f\in \mathcal{P}_d(\mathbb{R}^n).\]
Here \(\gamma^n\) denotes the \(n\)-fold product of the one-dimensional measure \(\gamma\), and \(\mathcal{P}_d(\mathbb{R}^n)\) denotes the space of all algebraic polynomials of degree at most \(d\) in \(n\) variables.
For \(p=2\), the standard Hermite polynomial expansion implies \[\label{eq-MB-2} \|\nabla f\|_{L^2(\gamma^n)} \le \sqrt d\,\|f\|_{L^2(\gamma^n)} \quad \forall f\in \mathcal{P}_d(\mathbb{R}^n).\tag{1}\]
For \(p\in[1,2)\cup(2,\infty)\), Eskenazis and Ivanisvili [35] proved the following weaker estimate: \[\label{eq-EI-est} \|\nabla f\|_{L^p(\gamma^n)} \le C(p)d^{ \frac{1}{2}+\frac{1}{\pi}\arctan\bigl(\frac{|p-2|}{2\sqrt{p-1}}\bigr)} \|f\|_{L^p(\gamma^n)} \quad \forall f\in\mathcal{P}_d(\mathbb{R}^n).\tag{2}\]
Our first result improves upon this estimate for \(p\ge4\) and confirms Conjecture 1 for even integer exponents.
Theorem 2. For every \(p\ge4\), there is a constant \(C(p)>0\) such that, for every \(n,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^p(\gamma^n)} \le C(p)d^{\frac{1}{2}+\theta_p}\|f\|_{L^p(\gamma^n)},\] where \[\theta_p:= \frac{1}{\pi}\arctan\biggl( \frac{p-2\lfloor p/2\rfloor}{2\sqrt{\lfloor p/2\rfloor(p-\lfloor p/2\rfloor)}} \biggr)\le \frac{2}{3p}.\] In particular, if \(p\) is an even integer, then \(\theta_p=0\) and one may take \(C(p)=3\sqrt p\).
The proof is carried out in two main steps. First, an integration by parts argument gives \[\|\nabla f\|_{L^p(\gamma^n)}^p \le C(p)\|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \Bigl( \|\nabla f\|_{L^p(\gamma^n)}^p + \int_{\mathbb{R}^n} |\nabla f|^{p-2}\|D^2f\|_{\rm HS}^2 \,d\gamma^n \Bigr)^{1/2},\] see Corollary 2. The last integral should be thought of as a multidimensional analogue of the one-dimensional expression \(|g|^{p-2}|g'|^2\). When \(p=2k\) is an even integer, this expression can be written as \(k^{-2}|(g^k)'|^2\) and can therefore be controlled by the \(L^2\) Markov–Bernstein inequality 1 , since \(g^k\) is again a polynomial. This reduction is formalized in Lemma 2. Applying Hölder’s inequality, we then obtain \[\|\nabla f\|_{L^{2k}(\gamma^n)}^{2k} \le C(k)d^{1/2} \|f\|_{L^{2k}(\gamma^n)} \|\nabla f\|_{L^{2k}(\gamma^n)}^{2k-1},\] which gives the sharp bound for even integer exponents. To obtain the improved estimate for all \(p\ge4\), we use the Eskenazis–Ivanisvili estimate 2 with exponent \(p/k\), \(k:=\lfloor p/2\rfloor\), in place of the \(L^2\) estimate 1 . This approach, with some technical adjustments, is then applied to the unit cube and to Freud-type densities.
Remark 3. While it is natural to try to interpolate between the sharp estimates obtained for arbitrarily large even values of \(p\), it is not clear how to carry out such an interpolation. A first natural attempt is to consider the linear operator \[T_dg:=\nabla P_dg,\] where \(P_d\) is the \(L^2(\gamma^n)\)-orthogonal projection onto \(\mathcal{P}_d(\mathbb{R}^n)\). This naive approach does not seem to give the desired result. Indeed, the sharp estimate for polynomials in \(L^{2k}(\gamma^n)\) gives only \[\|T_dg\|_{L^{2k}(\gamma^n)} \le C(k)\sqrt d\,\|P_dg\|_{L^{2k}(\gamma^n)}.\] Thus, interpolation through this operator would require a bound for \(P_d\) on \(L^{2k}(\gamma^n)\) that is uniform in \(d\). Such a bound is not available for \(k\ne1\) already in dimension one, since Hermite polynomial expansions are known to converge in \(L^p(\gamma)\) only in the case \(p=2\), see [36].
Sharp dimension-free Markov–Bernstein inequalities on the unit cube \([-1,1]^n\) in the uniform norm were obtained by Skalyga [37], [38]. However, the case of integral norms in this classical multivariate setting appears to be largely unexplored. To the best of our knowledge, even the \(L^2([-1,1]^n)\) inequality for the full gradient, with a dimension-free constant and sharp dependence on \(d\), has not been explicitly stated in the literature. We close this gap for even integer exponents \(p\) by proving the following Markov-type and weighted Bernstein-type inequalities.
Theorem 4. Let \(k,n,d\in\mathbb{N}\). Then, for every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\Bigl( \int_{[-1,1]^n} \Bigl( \sum_{j=1}^n(1-x_j^2)(\partial_{x_j}f)^2 \Bigr)^k\,dx \Bigr)^{\frac{1}{2k}} \le C_1(k)d\,\|f\|_{L^{2k}([-1,1]^n)}\] and \[\|\nabla f\|_{L^{2k}([-1,1]^n)} \le C_2(k)d^2\,\|f\|_{L^{2k}([-1,1]^n)}.\] One may take \(C_1(1)=\sqrt{2}\), \(C_2(1)=12\), and \(C_1(k)=46k^2\), \(C_2(k)=368k^3\) for \(k\ge2\).
The dependence on \(d\) is sharp already in dimension one.
Next, we transfer the cube estimate to products of probability measures with bounded unimodal densities.
Definition 1. We say that a density \(\varrho\) on \(\mathbb{R}\) is unimodal if there exists \(x_0\in\mathbb{R}\) such that \(\varrho\) is nondecreasing on \((-\infty,x_0]\) and nonincreasing on \([x_0,\infty)\).
In particular, all superlevel sets \(\{\varrho\ge t\}\) are intervals. We use this fact to represent a unimodal density as a mixture of uniform distributions on intervals. This reduces the Markov–Bernstein inequalities for products of unimodal densities to the case of products of intervals and gives the following extension of the cube result.
Theorem 5. Let \(k,n,d\in\mathbb{N}\) and let \(\mu=\mu_1\otimes\cdots\otimes\mu_n\), where each \(\mu_j\) has a bounded unimodal density \(\varrho_j\) on \(\mathbb{R}\). Set \[M:=\max_{1\le j\le n}\|\varrho_j\|_\infty.\] Then, for every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^{2k}(\mu)} \le C(k)Md^2 \|f\|_{L^{2k}(\mu)}.\] One may take \(C(1)=24\) and \(C(k)=736k^3\) for \(k\ge2\).
The dependence on \(d\) is sharp already for the cube.
Since log-concave densities are unimodal and an isotropic log-concave density on \(\mathbb{R}\) is bounded by \(1\), see [39], we obtain the following corollary.
Corollary 1. Let \(k,n,d\in\mathbb{N}\) and let \(\mu=\mu_1\otimes\cdots\otimes\mu_n\), where \(\mu_1,\ldots,\mu_n\) are isotropic log-concave probability measures on \(\mathbb{R}\). Then, for every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^{2k}(\mu)} \le C(k)d^2 \|f\|_{L^{2k}(\mu)}.\] One may take \(C(1)=24\) and \(C(k)=736k^3\) for \(k\ge2\).
The dependence on \(d\) in Theorem 5 and Corollary 1 is sharp for the whole classes of bounded unimodal densities and isotropic log-concave densities, respectively. For individual densities, however, this dependence may be improved, as illustrated by the Gaussian case.
In the last part of the paper, we extend the Gaussian estimates for even integer exponents \(p\) to products of probability measures with Freud-type densities, namely \[\nu_m(dt)=c_m e^{-|t|^{2m}}\,dt,\] where \(m\in\mathbb{N}\) and \(c_m\) is the normalizing constant. In dimension one, Markov–Bernstein inequalities for general Freud weights were obtained by Levin and Lubinsky [7]. In the particular case of the measures \(\nu_m\), for arbitrary \(p\in[1,\infty)\), this estimate asserts, see [40], that \[\label{Freud-MB} \|f'\|_{L^p(\nu_m)} \le C(m,p)d^{1-\frac{1}{2m}}\|f\|_{L^p(\nu_m)} \quad \forall f\in \mathcal{P}_d(\mathbb{R}).\tag{3}\] For even integer \(p\), we extend this result to the multidimensional product setting.
Theorem 6. Let \(m,k\in\mathbb{N}\). There exists a constant \(C(m,k)>0\) such that, for every \(n,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^{2k}(\nu_m^n)} \le C(m,k)d^{1-\frac{1}{2m}} \|f\|_{L^{2k}(\nu_m^n)}.\]
As mentioned above, \(\mathcal{P}_d(\mathbb{R}^n)\) denotes the space of all algebraic polynomials on \(\mathbb{R}^n\) of degree at most \(d\). We denote by \(C^\infty(\mathbb{R}^n)\) the space of all smooth functions on \(\mathbb{R}^n\), and by \(C^\infty(\mathbb{R}^n,\mathbb{R}^n)\) the space of all smooth vector fields on \(\mathbb{R}^n\). We denote by \(C^\infty_P(\mathbb{R}^n)\) the space of all smooth functions on \(\mathbb{R}^n\) such that the function itself and all its partial derivatives have at most polynomial growth. Similarly, \(C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\) denotes the space of all vector fields whose components belong to \(C^\infty_P(\mathbb{R}^n)\).
Throughout the paper, constants are denoted by \(C\) and may change from line to line. Their dependence on parameters is always indicated explicitly, for instance by writing \(C(p)\), \(C(k)\), or \(C(m,k)\). When several constants appear in the same argument, we distinguish them by subscripts, writing \(C_1,C_2\), etc. These subscripts are only labels and do not indicate any additional dependence.
For \(f\in C^\infty_P(\mathbb{R}^n)\), set \[Lf:=\Delta f-\langle x,\nabla f\rangle .\] For \(j\in\{1,\ldots,n\}\), define \[\delta_{x_j}f:=x_jf-\partial_{x_j}f .\] For a vector field \(u\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\), define \[\delta u:=\sum_{j=1}^n \delta_{x_j}u_j =\langle x,u\rangle-\operatorname{div}u .\] In particular, \[\delta\nabla f=-Lf.\]
For \(f,g\in C^\infty_P(\mathbb{R}^n)\), the Gaussian integration by parts formula gives \[\int_{\mathbb{R}^n}\partial_{x_j}f \, g\,d\gamma^n = \int_{\mathbb{R}^n} f\,\delta_{x_j}g\,d\gamma^n .\] Consequently, for \(f\in C^\infty_P(\mathbb{R}^n)\) and \(u\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\), \[\label{int-by-parts-1} \int_{\mathbb{R}^n}\langle \nabla f,u\rangle\,d\gamma^n = \int_{\mathbb{R}^n} f\,\delta u\,d\gamma^n .\tag{4}\]
We will also use the following consequence. Since \[\partial_{x_j}(\delta u) = u_j+\sum_{k=1}^n\delta_{x_k}\partial_{x_j}u_k ,\] we obtain \[\begin{align} \label{int-by-parts-2} \int_{\mathbb{R}^n}(\delta u)^2\,d\gamma^n &= \int_{\mathbb{R}^n} \sum_{j=1}^n u_j \Bigl( u_j+\sum_{k=1}^n\delta_{x_k}\partial_{x_j}u_k \Bigr)\,d\gamma^n \\&= \int_{\mathbb{R}^n}|u|^2\,d\gamma^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n \partial_{x_k}u_j\,\partial_{x_j}u_k\,d\gamma^n \notag \\ &\le \int_{\mathbb{R}^n}|u|^2\,d\gamma^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n (\partial_{x_k}u_j)^2\,d\gamma^n . \notag \end{align}\tag{5}\]
Lemma 1. Let \(n,d\in\mathbb{N}\), let \(r\ge q>0\), and let \(f\in\mathcal{P}_d(\mathbb{R}^n)\). Then \[\begin{align} \|\nabla f\|_{L^{r+2}(\gamma^n)}^{r+2} &\le 2\|f|\nabla f|^{r-q}\|_{L^2(\gamma^n)} \Bigl( \int_{\mathbb{R}^n} |\nabla f|^{2q+2} + |\nabla f|^{2q}\|D^2f\|_{\rm HS}^2 \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad+ r^2|\nabla f|^{2q-2}|D^2f\nabla f|^2\,d\gamma^n \Bigr)^{1/2}. \end{align}\]
Proof. Let \(\eta\in C^\infty(\mathbb{R})\) be a function such that \[\eta(t)=0 \quad \text{for } |t|\le 1, \quad \eta(t)=1 \quad \text{for } |t|\ge 2, \quad 0\le \eta(t)\le 1 \quad \text{for all } t\in\mathbb{R}.\] For \(\varepsilon>0\), set \(\eta_\varepsilon(t):=\eta(t/\varepsilon)\).
For \(a>0\), consider the vector field \[u_{a,\varepsilon} := |\nabla f|^a\eta_\varepsilon(|\nabla f|)\nabla f,\] where the expression is understood as zero on the set \(\{|\nabla f|\le\varepsilon\}\). With this convention, \(u_{a,\varepsilon}\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\).
We have \[\delta u_{a,\varepsilon} = -|\nabla f|^a\eta_\varepsilon(|\nabla f|)Lf - \bigl( a|\nabla f|^{a-2}\eta_\varepsilon(|\nabla f|) + |\nabla f|^{a-1}\eta_\varepsilon'(|\nabla f|) \bigr) \langle D^2f\nabla f,\nabla f\rangle .\] Therefore, \[\label{eq-delta-u} \delta u_{r,\varepsilon} = |\nabla f|^{r-q} \bigl( \delta u_{q,\varepsilon} - (r-q)|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) \langle D^2f\nabla f,\nabla f\rangle \bigr).\tag{6}\]
By Gaussian integration by parts 4 , \[\int_{\mathbb{R}^n} |\nabla f|^{r+2}I_{\{|\nabla f|\ge 2\varepsilon\}}\,d\gamma^n \le \int_{\mathbb{R}^n} |\nabla f|^{r+2}\eta_\varepsilon(|\nabla f|)\,d\gamma^n = \int_{\mathbb{R}^n} \langle \nabla f,u_{r,\varepsilon}\rangle\,d\gamma^n = \int_{\mathbb{R}^n} f\,\delta u_{r,\varepsilon}\,d\gamma^n .\] Using the identity 6 and applying the Cauchy–Schwarz inequality, we obtain \[\begin{align} &\int_{\mathbb{R}^n} |\nabla f|^{r+2}I_{\{|\nabla f|\ge 2\varepsilon\}}\,d\gamma^n \\ &\le \Bigl( \int_{\mathbb{R}^n} |f|^2|\nabla f|^{2r-2q}\,d\gamma^n \Bigr)^{1/2} \Bigl( \int_{\mathbb{R}^n} \bigl( \delta u_{q,\varepsilon} - (r-q)|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) \langle D^2f\nabla f,\nabla f\rangle \bigr)^2\,d\gamma^n \Bigr)^{1/2}. \end{align}\]
We estimate the second factor. Since \[\langle D^2f\nabla f,\nabla f\rangle^2 \le |\nabla f|^2|D^2f\nabla f|^2,\] we get \[\begin{align} &\int_{\mathbb{R}^n} \Bigl( \delta u_{q,\varepsilon} - (r-q)|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) \langle D^2f\nabla f,\nabla f\rangle \Bigr)^2\,d\gamma^n \\ &\le 2\int_{\mathbb{R}^n}(\delta u_{q,\varepsilon})^2\,d\gamma^n + 2(r-q)^2 \int_{\mathbb{R}^n} |\nabla f|^{2q-2}\eta_\varepsilon^2(|\nabla f|) |D^2f\nabla f|^2\,d\gamma^n . \end{align}\] By 5 , applied to \(u_{q,\varepsilon}\), \[\int_{\mathbb{R}^n}(\delta u_{q,\varepsilon})^2\,d\gamma^n \le \int_{\mathbb{R}^n}|u_{q,\varepsilon}|^2\,d\gamma^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n \bigl(\partial_{x_k}(u_{q,\varepsilon})_j\bigr)^2\,d\gamma^n .\] We have \[|u_{q,\varepsilon}|^2 = |\nabla f|^{2q+2}\eta_\varepsilon^2(|\nabla f|) \le |\nabla f|^{2q+2}\] and \[\partial_{x_k}(u_{q,\varepsilon})_j = |\nabla f|^q\eta_\varepsilon(|\nabla f|) \partial_{x_kx_j}^2f + \bigl( q|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) + |\nabla f|^{q-1}\eta_\varepsilon'(|\nabla f|) \bigr) \langle \nabla\partial_{x_k}f,\nabla f\rangle \partial_{x_j}f .\] Therefore, \[\sum_{j,k=1}^n \bigl(\partial_{x_k}(u_{q,\varepsilon})_j\bigr)^2 \le 2|\nabla f|^{2q}\|D^2f\|_{\rm HS}^2 + 2\bigl( q|\nabla f|^{q-1} + |\nabla f|^q|\eta_\varepsilon'(|\nabla f|)| \bigr)^2 |D^2f\nabla f|^2 .\] Since \(\eta_\varepsilon'(t)=\varepsilon^{-1}\eta'(t/\varepsilon)\) and \(\eta_\varepsilon'\) is supported on \(\{\varepsilon\le |t|\le 2\varepsilon\}\), we have \[|\nabla f|^q|\eta_\varepsilon'(|\nabla f|)| \le \frac{|\nabla f|^q}{\varepsilon}\|\eta'\|_\infty I_{\{\varepsilon\le|\nabla f|\le 2\varepsilon\}} \le 2\|\eta'\|_\infty |\nabla f|^{q-1} I_{\{\varepsilon\le|\nabla f|\le 2\varepsilon\}}.\]
Combining the preceding estimates, we obtain \[\begin{align} \int_{\mathbb{R}^n} |\nabla f|^{r+2}&I_{\{|\nabla f|\ge 2\varepsilon\}}\,d\gamma^n \\ &\le \|f|\nabla f|^{r-q}\|_{L^2(\gamma^n)} \Biggl( 2\int_{\mathbb{R}^n} |\nabla f|^{2q+2}\,d\gamma^n + 4\int_{\mathbb{R}^n} |\nabla f|^{2q}\|D^2f\|_{\rm HS}^2\,d\gamma^n \\ &+ \int_{\mathbb{R}^n} \Bigl( 4\bigl(q+2\|\eta'\|_\infty I_{\{\varepsilon\le|\nabla f|\le 2\varepsilon\}}\bigr)^2 + 2(r-q)^2 \Bigr) |\nabla f|^{2q-2}|D^2f\nabla f|^2\,d\gamma^n \Biggr)^{1/2}. \end{align}\] Letting \(\varepsilon\to0\) and using Lebesgue’s dominated convergence theorem, since \[4q^2+2(r-q)^2\le 4r^2\] for \(r\ge q>0\), we obtain the announced estimate. ◻
Corollary 2. Let \(n,d\in\mathbb{N}\), let \(p>2\), and let \(f\in\mathcal{P}_d(\mathbb{R}^n)\). Then \[\begin{align} \|\nabla f\|_{L^p(\gamma^n)}^p &\le 2\|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \Bigl( \int_{\mathbb{R}^n} |\nabla f|^p + |\nabla f|^{p-2}\|D^2f\|_{\rm HS}^2 \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad + (p-2)^2|\nabla f|^{p-4}|D^2f\nabla f|^2\,d\gamma^n \Bigr)^{1/2}. \end{align}\]
Proof. The estimate follows by taking \(r=p-2\) and \(q=p/2-1\) in Lemma 1. ◻
Lemma 2. Let \(k\in\mathbb{N}\). For every \(f\in C^\infty(\mathbb{R}^n)\) we have \[\sum_{1\le j_1,\ldots, j_k\le n } |\nabla (\partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f)|^2 = k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f \nabla f|^2.\]
Proof. Let \[u_j:=\partial_{x_j}f,\quad j=1,\ldots,n.\] Fix \(i\in\{1,\ldots,n\}\). For a multi-index \((j_1,\ldots,j_k)\in\{1, \ldots, n\}^k\), we set \[Q_{j_1,\ldots,j_k}:=u_{j_1}\cdot\ldots\cdot u_{j_k}.\] Then \[\partial_{x_i}Q_{j_1,\ldots,j_k} = \sum_{\ell=1}^k \partial_{x_i}u_{j_\ell} \prod_{\substack{r=1\\ r\ne \ell}}^k u_{j_r}.\] Therefore \[\sum_{1\le j_1,\ldots,j_k\le n} \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2 = \sum_{1\le j_1,\ldots,j_k\le n} \sum_{\ell,m=1}^k \partial_{x_i}u_{j_\ell}\partial_{x_i}u_{j_m} \prod_{\substack{r=1\\ r\ne \ell}}^k u_{j_r} \prod_{\substack{s=1\\ s\ne m}}^k u_{j_s}.\] We split the last sum into the diagonal part \(\ell=m\) and the off-diagonal part \(\ell\ne m\).
For the diagonal part, for each fixed \(\ell\) we get \[\sum_{1\le j_1,\ldots,j_k\le n} (\partial_{x_i}u_{j_\ell})^2 \prod_{\substack{r=1\\ r\ne \ell}}^k u_{j_r}^2 = \Bigl(\sum_{j=1}^n(\partial_{x_i}u_j)^2\Bigr) \Bigl(\sum_{j=1}^n u_j^2\Bigr)^{k-1}.\] Since there are \(k\) choices of \(\ell\), the diagonal contribution is \[k|\nabla f|^{2k-2}|\nabla\partial_{x_i}f|^2.\]
For the off-diagonal part, for each pair \(\ell\ne m\) we get \[\sum_{1\le j_1,\ldots,j_k\le n} \partial_{x_i}u_{j_\ell}\partial_{x_i}u_{j_m} u_{j_\ell}u_{j_m} \prod_{\substack{r=1\\ r\ne \ell,m}}^k u_{j_r}^2 = \Bigl(\sum_{j=1}^n \partial_{x_i}u_j u_j\Bigr)^2 \Bigl(\sum_{j=1}^n u_j^2\Bigr)^{k-2}.\] There are \(k(k-1)\) pairs \(\ell\ne m\). Hence the off-diagonal contribution is \[k(k-1)|\nabla f|^{2k-4} \langle \nabla\partial_{x_i}f,\nabla f\rangle^2.\]
Thus, for every \(i=1,\ldots,n\), \[\sum_{1\le j_1,\ldots,j_k\le n} \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2 = k|\nabla f|^{2k-2}|\nabla\partial_{x_i}f|^2 + k(k-1)|\nabla f|^{2k-4} \langle \nabla\partial_{x_i}f,\nabla f\rangle^2.\] Summing this identity over \(i=1,\ldots,n\), we obtain \[\sum_{1\le j_1,\ldots,j_k\le n} |\nabla Q_{j_1,\ldots,j_k}|^2 = k|\nabla f|^{2k-2} \|D^2f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4} |D^2f\nabla f|^2.\] This proves the lemma. ◻
Lemma 3. Let \(p\ge 2\) and let \(n, d\in\mathbb{N}\). Assume that there exists a constant \(M_{p,d}\) such that \[\|\nabla f\|_{L^p(\gamma^n)} \le M_{p,d}\|f\|_{L^p(\gamma^n)}\] for every \(f\in \mathcal{P}_d(\mathbb{R}^n)\). Then, for every \(m\in\mathbb{N}\) and for every \(g=(g_1,\ldots,g_m)\) with \(g_j\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\Bigl\| \Bigl(\sum_{j=1}^m |\nabla g_j|^2\Bigr)^{1/2} \Bigr\|_{L^p(\gamma^n)} \le \sqrt{p}\cdot M_{p,d} \Bigl\| \Bigl(\sum_{j=1}^m |g_j|^2\Bigr)^{1/2} \Bigr\|_{L^p(\gamma^n)}.\]
Proof. Let \(\varepsilon_1,\ldots,\varepsilon_m\) be independent Rademacher random variables, that is, \[\mathbb{P}(\varepsilon_i=1) = \mathbb{P}(\varepsilon_i=-1) = \frac{1}{2} .\] For each choice of signs \(\varepsilon=(\varepsilon_1,\ldots,\varepsilon_m)\), define \[f_\varepsilon(x):=\sum_{j=1}^m\varepsilon_j g_j(x).\] Then \(f_\varepsilon\in\mathcal{P}_d(\mathbb{R}^n)\), and therefore, by the assumed scalar estimate, \[\|\nabla f_\varepsilon\|_{L^p(\gamma^n)}^p \le M_{p,d}^p\|f_\varepsilon\|_{L^p(\gamma^n)}^p.\] Averaging over the signs and applying Hölder’s inequality, we obtain \[\begin{align} &\int_{\mathbb{R}^n} \Bigl(\sum_{j=1}^m|\nabla g_j|^2 \Bigr)^{p/2}\,d\gamma^n = \int_{\mathbb{R}^n} \Bigl(\mathbb{E}_\varepsilon\Bigl| \sum_{j=1}^m\varepsilon_j\nabla g_j \Bigr|^2\Bigr)^{p/2} \, d\gamma^n = \int_{\mathbb{R}^n} \mathbb{E}_\varepsilon \Bigl| \sum_{j=1}^m\varepsilon_j\nabla g_j \Bigr|^p\,d\gamma^n \\ &\le \mathbb{E}_\varepsilon \int_{\mathbb{R}^n} \Bigl| \sum_{j=1}^m\varepsilon_j\nabla g_j \Bigr|^p\,d\gamma^n \le M_{p,d}^p \mathbb{E}_\varepsilon \int_{\mathbb{R}^n} \Bigl| \sum_{j=1}^m\varepsilon_j g_j \Bigr|^p\,d\gamma^n = M_{p,d}^p \int_{\mathbb{R}^n} \mathbb{E}_\varepsilon\Bigl| \sum_{j=1}^m\varepsilon_j g_j \Bigr|^p\,d\gamma^n. \end{align}\] By Kahane–Khintchine inequality for Rademacher sums (see, for instance, [41] and [42] for the sharp constant), we have \[\mathbb{E}_\varepsilon \Bigl| \sum_{j=1}^m\varepsilon_j g_j \Bigr|^p \le p^{p/2} \Bigl(\sum_{j=1}^m|g_j|^2\Bigr)^{p/2}.\] Therefore, \[\int_{\mathbb{R}^n} \Bigl(\sum_{j=1}^m|\nabla g_j|^2 \Bigr)^{p/2}\,d\gamma^n \le p^{p/2}\cdot M_{p,d}^p \int_{\mathbb{R}^n} \Bigl(\sum_{j=1}^m|g_j|^2\Bigr)^{p/2}\,d\gamma^n.\] Taking the \(p\)-th root yields the announced estimate. ◻
Let \(k:=\lfloor p/2\rfloor\).
First suppose that \(p\) is an even integer. Then \(p=2k\). Since \[\partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f \in \mathcal{P}_{k(d-1)}(\mathbb{R}^n)\] for any choice of \(j_1,\ldots,j_k\in\{1,\ldots,n\}\), Lemma 2 and 1 give \[\begin{align} &\int_{\mathbb{R}^n} k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2\,d\gamma^n \\ &= \sum_{1\le j_1,\ldots,j_k\le n} \int_{\mathbb{R}^n} \bigl| \nabla( \partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f) \bigr|^2\,d\gamma^n \\ &\le k(d-1) \sum_{1\le j_1,\ldots,j_k\le n} \int_{\mathbb{R}^n} |\partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f|^2\,d\gamma^n \\ &= k(d-1)\int_{\mathbb{R}^n}|\nabla f|^{2k}\,d\gamma^n. \end{align}\] Therefore, \[\label{eq-HS-est} \int_{\mathbb{R}^n} |\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + (k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2\,d\gamma^n \le (d-1)\int_{\mathbb{R}^n}|\nabla f|^{2k}\,d\gamma^n.\tag{7}\] Applying Corollary 2, we obtain \[\begin{align} \|\nabla f\|_{L^p(\gamma^n)}^p &\le 2\|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \Bigl( \int_{\mathbb{R}^n} |\nabla f|^p + |\nabla f|^{2k-2}\|D^2f\|_{\rm HS}^2 \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad+ 4(k-1)^2|\nabla f|^{2k-4}|D^2f\nabla f|^2 \,d\gamma^n \Bigr)^{1/2}. \end{align}\] By 7 , \[\int_{\mathbb{R}^n} |\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + 4(k-1)^2|\nabla f|^{2k-4}|D^2f\nabla f|^2\,d\gamma^n \le 4(k-1)(d-1)\int_{\mathbb{R}^n}|\nabla f|^{2k}\,d\gamma^n.\] Since \[2\sqrt{1+2(p-2)(d-1)} \le 2\sqrt{2(p-2)d} \le 3\sqrt{pd},\] it follows that \[\|\nabla f\|_{L^p(\gamma^n)}^p \le 3\sqrt{pd}\, \|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \|\nabla f\|_{L^p(\gamma^n)}^{p/2}.\] By Hölder’s inequality, \[\label{eq-holder} \|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \le \|f\|_{L^p(\gamma^n)} \|\nabla f\|_{L^p(\gamma^n)}^{p/2-1}.\tag{8}\] Hence \[\|\nabla f\|_{L^p(\gamma^n)}^p \le 3\sqrt{pd}\, \|f\|_{L^p(\gamma^n)} \|\nabla f\|_{L^p(\gamma^n)}^{p-1}.\] If \(\|\nabla f\|_{L^p(\gamma^n)}=0\), the conclusion is immediate. Otherwise, dividing by \(\|\nabla f\|_{L^p(\gamma^n)}^{p-1}\), we obtain the assertion for even integers \(p\).
It remains to consider the case where \(p>4\) is not an even integer. Then \[2k<p<2k+2, \quad p/k\in(2,3).\] Applying Lemma 3 to the estimate 2 with the exponent \(p/k\), we get \[\Bigl\| \Bigl(\sum_{j=1}^m |\nabla g_j|^2\Bigr)^{1/2} \Bigr\|_{L^{p/k}(\gamma^n)} \le C_1(p)d^{\frac{1}{2}+\theta_p} \Bigl\| \Bigl(\sum_{j=1}^m |g_j|^2\Bigr)^{1/2} \Bigr\|_{L^{p/k}(\gamma^n)}\] for every vector \(g=(g_1,\ldots,g_m)\) whose components belong to \(\mathcal{P}_{k(d-1)}(\mathbb{R}^n)\). Here we used that \[\frac{1}{\pi}\arctan\Bigl( \frac{p/k-2}{2\sqrt{p/k-1}} \Bigr) = \frac{1}{\pi}\arctan\Bigl( \frac{p-2k}{2\sqrt{k(p-k)}} \Bigr) = \theta_p.\] We apply this estimate to the vector whose components are \[\partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f, \quad 1\le j_1,\ldots,j_k\le n.\] Using Lemma 2, we obtain \[\begin{align} &\Bigl\| k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2 \Bigr\|_{L^{p/(2k)}(\gamma^n)}^{1/2} \\ &= \Bigl\| \Bigl( \sum_{1\le j_1,\ldots,j_k\le n} \bigl| \nabla( \partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f) \bigr|^2 \Bigr)^{1/2} \Bigr\|_{L^{p/k}(\gamma^n)} \\ &\le C_1(p)d^{\frac{1}{2}+\theta_p} \Bigl\| \Bigl( \sum_{1\le j_1,\ldots,j_k\le n} |\partial_{x_{j_1}}f\cdot\ldots\cdot \partial_{x_{j_k}}f|^2 \Bigr)^{1/2} \Bigr\|_{L^{p/k}(\gamma^n)} \\ &= C_1(p)d^{\frac{1}{2}+\theta_p} \||\nabla f|^k\|_{L^{p/k}(\gamma^n)} = C_1(p)d^{\frac{1}{2}+\theta_p} \|\nabla f\|_{L^p(\gamma^n)}^k. \end{align}\] Hence \[\Bigl\| k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2 \Bigr\|_{L^{p/(2k)}(\gamma^n)} \le C_2(p) d^{1+2\theta_p} \|\nabla f\|_{L^p(\gamma^n)}^{2k}.\] Since \(k\ge2\), \[\begin{align} &|\nabla f|^{p-2}\|D^2f\|_{\rm HS}^2 + (p-2)^2|\nabla f|^{p-4}|D^2f\nabla f|^2 \\ &\le C_3(p)|\nabla f|^{p-2k} \Bigl( k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2 \Bigr). \end{align}\] Therefore, by Hölder’s inequality, \[\begin{align} &\int_{\mathbb{R}^n} |\nabla f|^{p-2}\|D^2f\|_{\rm HS}^2 + (p-2)^2|\nabla f|^{p-4}|D^2f\nabla f|^2 \,d\gamma^n \\ &\le C_3(p) \bigl\||\nabla f|^{p-2k}\bigr\|_{L^{p/(p-2k)}(\gamma^n)} \Bigl\| k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2 + k(k-1)|\nabla f|^{2k-4}|D^2f\nabla f|^2 \Bigr\|_{L^{p/(2k)}(\gamma^n)} \\ &\le C_4(p)d^{1+2\theta_p} \|\nabla f\|_{L^p(\gamma^n)}^{p-2k} \|\nabla f\|_{L^p(\gamma^n)}^{2k} = C_4(p)d^{1+2\theta_p} \|\nabla f\|_{L^p(\gamma^n)}^p. \end{align}\] Substituting this estimate into Corollary 2, gives \[\|\nabla f\|_{L^p(\gamma^n)}^p \le C_5(p)d^{\frac{1}{2}+\theta_p} \|f|\nabla f|^{p/2-1}\|_{L^2(\gamma^n)} \|\nabla f\|_{L^p(\gamma^n)}^{p/2}.\] By 8 , \[\|\nabla f\|_{L^p(\gamma^n)}^p \le C_5(p)d^{\frac{1}{2}+\theta_p} \|f\|_{L^p(\gamma^n)} \|\nabla f\|_{L^p(\gamma^n)}^{p-1}.\] If \(\|\nabla f\|_{L^p(\gamma^n)}=0\), the conclusion is immediate. Otherwise, dividing by \(\|\nabla f\|_{L^p(\gamma^n)}^{p-1}\) proves the estimate for non-even \(p>4\).
Finally, we estimate the error term \(\theta_p\) for \(p\ge 4\). We write \[p=2k+\alpha,\quad 0\le \alpha<2, \quad k:=\lfloor p/2\rfloor.\] For \(p\ge4\) one has \(k\ge2\), and \[\theta_p = \frac{1}{\pi} \arctan\Bigl( \frac{\alpha}{2\sqrt{k(k+\alpha)}} \Bigr).\] Our goal is to prove that \(p\theta_p\le 2/3\). For fixed \(k\), the function \[\alpha\mapsto (2k+\alpha) \arctan\Bigl( \frac{\alpha}{2\sqrt{k(k+\alpha)}} \Bigr) = (2k+\alpha) \arctan\Bigl( \frac{1}{2\sqrt{k/\alpha(k/\alpha+1)}} \Bigr)\] is increasing on \([0,2]\). Therefore, \[p\theta_p \le \frac{2(k+1)}{\pi} \arctan\Bigl( \frac{1}{\sqrt{k(k+2)}} \Bigr) =\frac{2(k+1)}{\pi} \arcsin\frac{1}{k+1}.\] The function \[x\mapsto x\arcsin\frac{1}{x}\] is decreasing on \((1,\infty)\), since \[\frac{d}{dx}\biggl(x\arcsin\frac{1}{x}\biggr) = \arcsin\frac{1}{x}-\frac{1}{\sqrt{x^2-1}}<0.\] Thus, as \(k+1\ge3>1\), \[(k+1)\arcsin\frac{1}{k+1} \le 3\arcsin\frac{1}{3}.\] Finally, \[3\arcsin\frac{1}{3}<\frac{\pi}{3}.\] Indeed, if \(a=\arcsin(1/3)\), then \[\sin(3a)=3\sin a-4\sin^3a = 1-\frac{4}{27} = \frac{23}{27} < \frac{\sqrt3}{2} = \sin\frac{\pi}{3},\] and \(0<3a<\pi/2\). Therefore \[p\theta_p \le \frac{2}{\pi}\cdot \frac{\pi}{3} = \frac{2}{3},\] which gives the claimed estimate. 0◻
For two vector fields \(u,v\in C^\infty(\mathbb{R}^n,\mathbb{R}^n)\), let \[\langle u,v\rangle_0 := \sum_{j=1}^n(1-x_j^2)u_jv_j\] and let \[|u|_0^2:=\langle u,u\rangle_0.\] For \(r>0\), we set, on \([-1,1]^n\), \[|u|_0^r:=\bigl(|u|_0^2\bigr)^{r/2}.\]
For \(f\in C^\infty(\mathbb{R}^n)\), set \[\delta_{0,x_j}f := 2x_jf-(1-x_j^2)\partial_{x_j}f.\] For a vector field \(u=(u_1,\ldots,u_n)\in C^\infty(\mathbb{R}^n,\mathbb{R}^n)\), set \[\delta_0u:=\sum_{j=1}^n\delta_{0,x_j}u_j.\]
For \(f\in C^\infty(\mathbb{R}^n)\) and \(u\in C^\infty(\mathbb{R}^n,\mathbb{R}^n)\), we have \[\label{jacobi-int-by-parts-1} \int_{[-1,1]^n} (1-x_j^2)u_j\partial_{x_j}f\,dx = \int_{[-1,1]^n} f\delta_{0,x_j}u_j\,dx.\tag{9}\] Therefore, \[\label{jacobi-int-by-parts} \int_{[-1,1]^n}\langle \nabla f,u\rangle_0\,dx = \int_{[-1,1]^n}f\,\delta_0u\,dx.\tag{10}\]
Let \[L_0 f := \sum_{j=1}^n \Bigl( (1-x_j^2)\partial_{x_jx_j}^2f - 2x_j\partial_{x_j}f \Bigr).\] In particular, \[\delta_0\nabla f=-L_0f.\]
Lemma 4. Let \(D\in \mathbb{N}\) and let \(Q\in \mathcal{P}_D(\mathbb{R})\) be such that \(Q(t)\ge 0\) \(\forall t\in[-1, 1]\). Then \[\int_{-1}^1 Q(t)\,dt \le 16D^2 \int_{-1}^1(1-t^2)Q(t)\,dt .\]
Proof. By the standard Nikolskii inequality (see, for example, [9]), we have \[\max_{t\in[-1,1]} Q(t) \le 2D^2 \int_{-1}^1 Q(t)\,dt .\] Then, for any \(\delta\in(0,1)\), \[\int_{\{1-\delta\le |t|\le 1\}}Q(t)\,dt \le 2\delta \max_{t\in[-1,1]}Q(t) \le 4\delta D^2\int_{-1}^1Q(t)\,dt.\] Let \[\delta:=\frac{1}{8D^2}.\] Then \[\int_{\{|t|\le 1-\delta\}}Q(t)\,dt \ge \frac{1}{2}\int_{-1}^1Q(t)\,dt.\] Therefore \[\int_{-1}^1(1-t^2)Q(t)\,dt \ge \delta\int_{\{|t|\le 1-\delta\}}Q(t)\,dt \ge \frac{\delta}{2}\int_{-1}^1Q(t)\,dt.\] Since \(\delta=1/(8D^2)\), this gives \[\int_{-1}^1Q(t)\,dt \le 16D^2 \int_{-1}^1(1-t^2)Q(t)\,dt.\] This is the announced estimate. ◻
Lemma 5. For every \(n,k,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^{2k}([-1,1]^n)} \le 8kd\, \||\nabla f|_0\|_{L^{2k}([-1,1]^n)}.\]
Proof. We first record a one-dimensional consequence of Lemma 4. Let \(D,r,m\in\mathbb{N}\cup\{0\}\) and let \(Q\in\mathcal{P}_D(\mathbb{R})\). Then \[\label{eq-cor} \int_{-1}^1(1-t^2)^r|Q(t)|^2\,dt \le 64^m(D+r+m)^{2m} \int_{-1}^1(1-t^2)^{r+m}|Q(t)|^2\,dt.\tag{11}\] Indeed, for \(m=0\), this estimate is just the identity. For \(m\ge 1\), applying Lemma 4 to the non-negative polynomial \[(1-t^2)^r|Q(t)|^2,\] whose degree is at most \(2(D+r)\), gives \[\int_{-1}^1(1-t^2)^r|Q(t)|^2\,dt \le 64(D+r+1)^2 \int_{-1}^1(1-t^2)^{r+1}|Q(t)|^2\,dt.\] The shift \(+1\) covers the case \(D=r=0\). Iterating this estimate \(m\) times gives 11 .
For \((j_1,\ldots,j_k)\in\{1,\ldots,n\}^k\), set \[Q_{j_1,\ldots,j_k}:=\partial_{x_{j_1}}f\cdot\ldots\cdot\partial_{x_{j_k}}f\in \mathcal{P}_{k(d-1)}(\mathbb{R}^n).\] For \(j\in\{1,\ldots,n\}\), let \(m_j=m_j((j_1,\ldots,j_k))\) be the number of occurrences of \(j\) in \((j_1,\ldots,j_k)\). Then \[\sum_{j=1}^n m_j=k\] and \[\prod_{\ell=1}^k(1-x_{j_\ell}^2) = \prod_{j=1}^n(1-x_j^2)^{m_j}.\] Applying 11 successively in every variable gives \[\begin{align} \int_{[-1,1]^n}|Q_{j_1,\ldots,j_k}|^2\,dx &\le \prod_{j=1}^n 64^{m_j}\bigl(k(d-1)+m_j\bigr)^{2m_j} \int_{[-1,1]^n} \prod_{j=1}^n(1-x_j^2)^{m_j}|Q_{j_1,\ldots,j_k}|^2\,dx \\ &\le 64^k(kd)^{2k} \int_{[-1,1]^n} \prod_{\ell=1}^k(1-x_{j_\ell}^2)|Q_{j_1,\ldots,j_k}|^2\,dx \\ &= 64^k(kd)^{2k} \int_{[-1,1]^n} (1-x_{j_1}^2)(\partial_{x_{j_1}}f)^2 \cdot\ldots\cdot (1-x_{j_k}^2)(\partial_{x_{j_k}}f)^2\,dx. \end{align}\] Now summing over all \((j_1,\ldots,j_k)\), and using \[|\nabla f|^{2k} = \sum_{1\le j_1,\ldots,j_k\le n} \bigl(\partial_{x_{j_1}}f\cdot\ldots\cdot\partial_{x_{j_k}}f\bigr)^2\] and \[|\nabla f|_0^{2k} = \sum_{1\le j_1,\ldots,j_k\le n} (1-x_{j_1}^2)(\partial_{x_{j_1}}f)^2 \cdot\ldots\cdot (1-x_{j_k}^2)(\partial_{x_{j_k}}f)^2,\] we obtain \[\int_{[-1,1]^n}|\nabla f|^{2k}\,dx \le (8kd)^{2k} \int_{[-1,1]^n}|\nabla f|_0^{2k}\,dx.\] Taking the power \(1/(2k)\) gives the claimed estimate. ◻
Lemma 6. Let \(u\in C^\infty(\mathbb{R}^n, \mathbb{R}^n)\). Then \[\int_{[-1, 1]^n}(\delta_0 u)^2\,dx \le 2 \int_{[-1, 1]^n} |u|_0^2\,dx + \int_{[-1, 1]^n} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_j}u_k)^2\,dx.\]
Proof. Applying the integration by parts formula 10 with \(f=\delta_0 u\in C^\infty(\mathbb{R}^n)\) and the vector field \(u\), we get \[\int_{[-1,1]^n}(\delta_0 u)^2\,dx = \sum_{j,k=1}^n \int_{[-1,1]^n} (1-x_j^2)u_j \partial_{x_j}(\delta_{0, x_k}u_k) \,dx.\]
When \(j\ne k\), we have \[\partial_{x_j}(\delta_{0, x_k}u_k)= \partial_{x_j} \bigl( 2x_ku_k - (1-x_k^2)\partial_{x_k}u_k \bigr) = 2x_k\partial_{x_j}u_k - (1-x_k^2)\partial_{x_k}\partial_{x_j}u_k =\delta_{0, x_k}(\partial_{x_j}u_k).\] Applying the integration by parts formula 9 in the variable \(x_k\), we obtain \[\begin{align} \int_{[-1,1]^n} (1-x_j^2)u_j \partial_{x_j}(\delta_{0, x_k}u_k) \,dx &= \int_{[-1,1]^n} (1-x_j^2)u_j\delta_{0, x_k}(\partial_{x_j}u_k) \,dx \\ &= \int_{[-1,1]^n} (1-x_j^2)(1-x_k^2) \partial_{x_k}u_j\,\partial_{x_j}u_k\,dx. \end{align}\]
Now let \(j=k\). Then \[\begin{align} & \partial_{x_j}(\delta_{0, x_j}u_j) =\partial_{x_j} \bigl( 2x_ju_j - (1-x_j^2)\partial_{x_j}u_j \bigr) \\ &= 2u_j + 4x_j\partial_{x_j}u_j - (1-x_j^2)\partial_{x_jx_j}^2u_j = 2u_j + 2x_j\partial_{x_j}u_j + \delta_{0, x_j}(\partial_{x_j}u_j). \end{align}\] Therefore, \[\begin{align} &\int_{[-1,1]^n} (1-x_j^2)u_j \partial_{x_j}(\delta_{0, x_j}u_j)\,dx = \int_{[-1,1]^n} (1-x_j^2)u_j\bigl( 2u_j + 2x_j\partial_{x_j}u_j + \delta_{0, x_j}(\partial_{x_j}u_j)\bigr) \,dx \\ &= 2 \int_{[-1,1]^n} (1-x_j^2)u_j^2\,dx + \int_{[-1,1]^n} (1-x_j^2)u_j \bigl( 2x_j\partial_{x_j}u_j + \delta_{0, x_j}(\partial_{x_j}u_j)\bigr)\,dx. \end{align}\] Applying the integration by parts formula 9 in the second integral gives \[\begin{align} &\int_{[-1,1]^n} (1-x_j^2)u_j \bigl( 2x_j\partial_{x_j}u_j + \delta_{0, x_j}(\partial_{x_j}u_j)\bigr)\,dx \\ &= \int_{[-1,1]^n} (1-x_j^2)u_j 2x_j\partial_{x_j}u_j +(1-x_j^2)\partial_{x_j}\bigl((1-x_j^2)u_j\bigr) \partial_{x_j}u_j \,dx \\ &= \int_{[-1,1]^n} (1-x_j^2)^2(\partial_{x_j}u_j)^2\,dx. \end{align}\] Combining the diagonal and off-diagonal terms, we obtain \[\begin{align} \int_{[-1,1]^n}(\delta_0 u)^2\,dx &= 2 \int_{[-1,1]^n} |u|_0^2\,dx + \int_{[-1,1]^n} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) \partial_{x_j}u_k\,\partial_{x_k}u_j\,dx. \end{align}\] By the Cauchy–Schwarz inequality, \[\begin{align} &\sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) \partial_{x_j}u_k\,\partial_{x_k}u_j \\ &= \sum_{j,k=1}^n \bigl((1-x_j^2)^{1/2}(1-x_k^2)^{1/2} \partial_{x_j}u_k\bigr) \bigl((1-x_j^2)^{1/2}(1-x_k^2)^{1/2} \partial_{x_k}u_j\bigr) \\ &\le \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2)(\partial_{x_j}u_k)^2. \end{align}\] Thus, \[\begin{align} \int_{[-1,1]^n}(\delta_0 u)^2\,dx &\le 2 \int_{[-1,1]^n} |u|_0^2\,dx+ \int_{[-1,1]^n} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_j}u_k)^2\,dx. \end{align}\] This is the announced estimate. ◻
Lemma 7. Let \(\alpha_1,\ldots,\alpha_n>-1\) and let \[\mu_{\alpha_j}(dt):=c_{\alpha_j}(1-t^2)^{\alpha_j}\,dt\] be the corresponding probability measures on \([-1,1]\). Set \[\mu_\alpha(dx) := \bigotimes_{j=1}^n \mu_{\alpha_j}(dx_j) = \prod_{j=1}^n c_{\alpha_j}(1-x_j^2)^{\alpha_j}\,dx\] and \[\alpha_*:=\max_{1\le j\le n}\alpha_j.\] Then, for every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), \[\int_{[-1,1]^n} |\nabla f|_0^2 \,d\mu_\alpha \le d(d+2\alpha_*+1) \int_{[-1,1]^n}f^2\,d\mu_\alpha .\]
Proof. For each \(j\in\{1,\ldots,n\}\), consider the one-dimensional Jacobi operator \[L_{\alpha_j}h(t) = (1-t^2)h''(t)-2(\alpha_j+1)t h'(t).\] An integration by parts gives, for polynomials \(h_1,h_2\), \[\int_{-1}^1(1-t^2)h_1'(t)h_2'(t)\,\mu_{\alpha_j}(dt) = -\int_{-1}^1 h_1 L_{\alpha_j}h_2\,d\mu_{\alpha_j}.\] Define \[L_\alpha f := \sum_{j=1}^n \bigl( (1-x_j^2)\partial_{x_jx_j}^2f - 2(\alpha_j+1)x_j\partial_{x_j}f \bigr).\] Applying the preceding integration by parts formula in each coordinate gives \[\int_{[-1,1]^n} |\nabla f|_0^2\,d\mu_\alpha = \int_{[-1,1]^n} \sum_{j=1}^n(1-x_j^2)(\partial_{x_j}f)^2\,d\mu_\alpha = -\int_{[-1,1]^n} f L_\alpha f\,d\mu_\alpha.\]
Let \(\{u_m^{(\alpha_j)}\colon m\in\mathbb{N}\cup\{0\}\}\) be the system obtained by normalizing the Jacobi polynomials \(P_m^{(\alpha_j,\alpha_j)}\) in \(L^2(\mu_{\alpha_j})\). Then \(\{u_m^{(\alpha_j)}\colon m\in\mathbb{N}\cup\{0\}\}\) is an orthonormal system in \(L^2(\mu_{\alpha_j})\). The Jacobi differential equation (see [43]) gives \[L_{\alpha_j}u_m^{(\alpha_j)} = -m(m+2\alpha_j+1)u_m^{(\alpha_j)}.\]
For a multi-index \({\boldsymbol{m}}=(m_1,\ldots,m_n)\), \(m_j\in\mathbb{N}\cup\{0\}\), set \[U_{\boldsymbol{m}}(x):= \prod_{j=1}^n u_{m_j}^{(\alpha_j)}(x_j), \quad |{\boldsymbol{m}}|:=m_1+\ldots+m_n.\] The system \(\{U_{\boldsymbol{m}}\colon |{\boldsymbol{m}}|\le d\}\) is an orthonormal basis of \(\mathcal{P}_d(\mathbb{R}^n)\) with respect to the inner product of \(L^2(\mu_\alpha)\). Moreover, \[L_\alpha U_{\boldsymbol{m}} = -\lambda_{\boldsymbol{m}}U_{\boldsymbol{m}},\] where \[\lambda_{\boldsymbol{m}} = \sum_{j=1}^n m_j(m_j+2\alpha_j+1).\]
Since \(f\in\mathcal{P}_d(\mathbb{R}^n)\), we may write \[f=\sum_{|{\boldsymbol{m}}|\le d}a_{\boldsymbol{m}}U_{\boldsymbol{m}}.\] Therefore \[\int_{[-1,1]^n}f^2\,d\mu_\alpha = \sum_{|{\boldsymbol{m}}|\le d}|a_{\boldsymbol{m}}|^2\] and \[-\int_{[-1,1]^n}fL_\alpha f\,d\mu_\alpha = \sum_{|{\boldsymbol{m}}|\le d}\lambda_{\boldsymbol{m}}|a_{\boldsymbol{m}}|^2.\]
Now we note that, for \(|{\boldsymbol{m}}|\le d\), one has \[\lambda_{\boldsymbol{m}} = \sum_{j=1}^n m_j(m_j+2\alpha_j+1) \le (d+2\alpha_*+1)\sum_{j=1}^n m_j = (d+2\alpha_*+1)|{\boldsymbol{m}}| \le d(d+2\alpha_*+1).\] This implies the announced estimate. ◻
Lemma 8. Let \(r\ge q\ge 1\) and let \(f\in\mathcal{P}_d(\mathbb{R}^n)\). Then \[\begin{align} &\int_{[-1,1]^n} |\nabla f|_0^{r+2}\,dx \le 2\bigl\|f|\nabla f|_0^{r-q}\bigr\|_{L^2([-1, 1]^n)} \biggl( \int_{[-1,1]^n}|\nabla f|_0^{2q+2}\,dx \\ &+ \int_{[-1,1]^n}|\nabla f|_0^{2q} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2\,dx + \frac{r^2}{4}\int_{[-1,1]^n} |\nabla f|_0^{2q-2} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2 \,dx \biggr)^{1/2}. \end{align}\]
Proof. The proof is parallel to the proof of the Gaussian counterpart, Lemma 1.
Let \(\eta\in C^\infty(\mathbb{R})\) be such that \[\eta(t)=0 \quad \text{for } t\le1, \quad \eta(t)=1 \quad \text{for } t\ge2, \quad 0\le \eta(t)\le 1 \quad \text{for all } t\in\mathbb{R}.\] For \(\varepsilon>0\), set \(\eta_\varepsilon(t):=\eta(t/\varepsilon)\).
For \(a\ge 1\), we consider the vector field \[u_{a,\varepsilon} := |\nabla f|_0^a\eta_\varepsilon(|\nabla f|_0^2)\nabla f.\] We understand the coefficient \[|\nabla f|_0^a\eta_\varepsilon(|\nabla f|_0^2)\] as the composition of the polynomial \(|\nabla f|_0^2\) with the smooth function on \(\mathbb{R}\) which equals \(s^{a/2}\eta_\varepsilon(s)\) for \(s\ge0\) and is zero for \(s<0\). Thus, \[u_{a,\varepsilon} := |\nabla f|_0^a\eta_\varepsilon(|\nabla f|_0^2)\nabla f\in C^\infty(\mathbb{R}^n,\mathbb{R}^n).\]
We have \[\begin{align} &\delta_0 u_{r,\varepsilon} = \sum_{j=1}^n \bigl(2x_ju_{r,\varepsilon,j} - (1-x_j^2)\partial_{x_j}u_{r,\varepsilon,j}\bigr) \\ &= -|\nabla f|_0^r\eta_\varepsilon(|\nabla f|_0^2)L_0 f - \Bigl( \frac{r}{2} |\nabla f|_0^{r-2}\eta_\varepsilon(|\nabla f|_0^2) + |\nabla f|_0^r\eta_\varepsilon'(|\nabla f|_0^2) \Bigr)\langle \nabla |\nabla f|_0^2, \nabla f\rangle_0. \end{align}\] Therefore, \[\begin{align} \delta_0 u_{r,\varepsilon} &= |\nabla f|_0^{r-q} \Bigl( \delta_0 u_{q,\varepsilon} - \frac{r-q}{2} |\nabla f|_0^{q-2}\eta_\varepsilon(|\nabla f|_0^2) \langle \nabla|\nabla f|_0^2, \nabla f\rangle_0 \Bigr). \end{align}\]
By the integration by parts formula 10 , \[\int_{[-1,1]^n} |\nabla f|_0^{r+2}I_{\{|\nabla f|_0^2\ge2\varepsilon\}}\,dx \le \int_{[-1,1]^n}\langle \nabla f, u_{r,\varepsilon}\rangle_0\,dx = \int_{[-1,1]^n}f\,\delta_0 u_{r,\varepsilon}\,dx.\] Hence, by the Cauchy–Schwarz inequality, \[\begin{align} \label{eq-CS-est} &\int_{[-1,1]^n} |\nabla f|_0^{r+2}I_{\{|\nabla f|_0^2\ge2\varepsilon\}}\,dx \\ &\le \bigl\|f|\nabla f|_0^{r-q}\bigr\|_{L^2} \biggl( \int_{[-1,1]^n} \Bigl( \delta_0 u_{q,\varepsilon} - \frac{r-q}{2} |\nabla f|_0^{q-2}\eta_\varepsilon(|\nabla f|_0^2) \langle \nabla|\nabla f|_0^2, \nabla f\rangle_0 \Bigr)^2 \,dx \biggr)^{1/2}.\notag \end{align}\tag{12}\] The last integral is bounded by \[\label{eq-lem-2-4-middle} 2\int_{[-1,1]^n}(\delta_0 u_{q,\varepsilon})^2\,dx + \frac{(r-q)^2}{2} \int_{[-1,1]^n} |\nabla f|_0^{2q-2} \bigl|\nabla|\nabla f|_0^2\bigr|_0^2 \,dx.\tag{13}\] Applying Lemma 6 to \(u_{q,\varepsilon}\) in the first term gives \[\int_{[-1,1]^n}(\delta_0 u_{q,\varepsilon})^2\,dx \le 2 \int_{[-1,1]^n}|u_{q,\varepsilon}|_0^2\,dx + \int_{[-1,1]^n} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_j}u_{q,\varepsilon,k})^2\,dx.\] By the definition of \(u_{q, \varepsilon}\), \[\int_{[-1,1]^n}|u_{q,\varepsilon}|_0^2\,dx \le \int_{[-1,1]^n}|\nabla f|_0^{2q+2}\,dx.\] Next, \[u_{q,\varepsilon,k} = |\nabla f|_0^q\eta_\varepsilon(|\nabla f|_0^2)\partial_{x_k}f,\] and \[\partial_{x_j}u_{q,\varepsilon,k} = |\nabla f|_0^q\eta_\varepsilon(|\nabla f|_0^2)\partial_{x_jx_k}^2f + \Bigl( \frac{q}{2} |\nabla f|_0^{q-2}\eta_\varepsilon(|\nabla f|_0^2) + |\nabla f|_0^q\eta_\varepsilon'(|\nabla f|_0^2) \Bigr) \partial_{x_j}\bigl(|\nabla f|^2_0\bigr)\,\partial_{x_k}f.\] On the cube \([-1, 1]^n\) we have \[\Bigl| \frac{q}{2} |\nabla f|_0^{q-2}\eta_\varepsilon(|\nabla f|_0^2) + |\nabla f|_0^q\eta_\varepsilon'(|\nabla f|_0^2) \Bigr| \le \Bigl(\frac{q}{2}+2\|\eta'\|_\infty I_{\{\varepsilon\le |\nabla f|_0^2\le 2\varepsilon\}}\Bigr) |\nabla f|_0^{q-2}.\] Therefore, on the cube \([-1, 1]^n\), \[\begin{align} &\sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_j}u_{q,\varepsilon,k})^2 \\ &\le 2|\nabla f|_0^{2q} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2 \\ &+ 2\Bigl(\frac{q}{2}+2\|\eta'\|_\infty I_{\{\varepsilon\le |\nabla f|_0^2\le 2\varepsilon\}}\Bigr)^2|\nabla f|_0^{2q-4} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) \bigl(\partial_{x_j}(|\nabla f|_0^2)\bigr)^2(\partial_{x_k}f)^2 \\ &= 2|\nabla f|_0^{2q} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2 + 2\Bigl(\frac{q}{2}+2\|\eta'\|_\infty I_{\{\varepsilon\le |\nabla f|_0^2\le 2\varepsilon\}}\Bigr)^2|\nabla f|_0^{2q-2} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2. \end{align}\] Thus, \[\begin{align} \int_{[-1,1]^n}(\delta_0 u_{q,\varepsilon})^2\,dx &\le 2\int_{[-1,1]^n}|\nabla f|_0^{2q+2}\,dx + 2\int_{[-1,1]^n}|\nabla f|_0^{2q} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2\,dx \\ &+ 2\int_{[-1,1]^n} \Bigl(\frac{q}{2}+2\|\eta'\|_\infty I_{\{\varepsilon\le |\nabla f|_0^2\le 2\varepsilon\}}\Bigr)^2|\nabla f|_0^{2q-2} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2 \,dx. \end{align}\]
Substituting this estimate into 13 , and using \(q\le r\), we obtain \[\begin{align} &\int_{[-1,1]^n} \Bigl( \delta_0 u_{q,\varepsilon} - \frac{r-q}{2} |\nabla f|_0^{q-2}\eta_\varepsilon(|\nabla f|_0^2) \langle \nabla|\nabla f|_0^2, \nabla f\rangle_0 \Bigr)^2 \,dx \\ &\le 4\biggl( \int_{[-1,1]^n}|\nabla f|_0^{2q+2}\,dx + \int_{[-1,1]^n}|\nabla f|_0^{2q} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2\,dx \\ &+\int_{[-1,1]^n} \Bigl(\frac{r}{2}+2\|\eta'\|_\infty I_{\{\varepsilon\le |\nabla f|_0^2\le 2\varepsilon\}}\Bigr)^2|\nabla f|_0^{2q-2} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2 \,dx \biggr). \end{align}\] Returning to 12 and letting \(\varepsilon\to0\), the dominated convergence theorem gives the claimed estimate. ◻
Lemma 9. Let \(k\in\mathbb{N}\), \(k\ge 2\), and let \(f\in\mathcal{P}_d(\mathbb{R}^n)\). Then \[\begin{align} &\int_{[-1,1]^n}|\nabla f|_0^{2k-2} \sum_{i,j=1}^n (1-x_i^2)(1-x_j^2) (\partial_{x_ix_j}^2f)^2\,dx + (k-1)^2 \int_{[-1,1]^n} |\nabla f|_0^{2k-4} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2\,dx \\ &\le 512k^4d^2 \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1, 1]^n)}^{2k}. \end{align}\]
Proof. For \((j_1,\ldots,j_k)\in\{1,\ldots,n\}^k\), set \[Q_{j_1,\ldots,j_k}:= \partial_{x_{j_1}}f\cdot\ldots\cdot\partial_{x_{j_k}}f \in\mathcal{P}_{k(d-1)}(\mathbb{R}^n).\] For \(j\in\{1,\ldots,n\}\), let \(m_j=m_j((j_1,\ldots,j_k))\) be the number of occurrences of \(j\) in \((j_1,\ldots,j_k)\). Then \[\sum_{j=1}^n m_j=k \quad\text{and}\quad \prod_{\ell=1}^k(1-x_{j_\ell}^2) = \prod_{j=1}^n(1-x_j^2)^{m_j}.\] By Lemma 7, applied to the polynomial \(Q_{j_1,\ldots,j_k}\) and the measure \(\mu_{\boldsymbol{m}}\), \({\boldsymbol{m}}=(m_1, \ldots, m_n)\), we obtain \[\begin{align} &\int_{[-1,1]^n} \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{i=1}^n(1-x_i^2)(\partial_{x_i}Q_{j_1,\ldots,j_k})^2\,dx \\ &\le k(d-1)\bigl(k(d-1) + 2\max_{1\le j\le n}m_j+1\bigr) \int_{[-1,1]^n} \prod_{\ell=1}^k(1-x_{j_\ell}^2)Q_{j_1,\ldots,j_k}^2\,dx \\ &\le 2k^2d^2 \int_{[-1,1]^n} \prod_{\ell=1}^k(1-x_{j_\ell}^2)Q_{j_1,\ldots,j_k}^2\,dx. \end{align}\] Summing over all \((j_1,\ldots,j_k)\), we obtain \[\label{eq-MB-weighted} \sum_{j_1,\ldots,j_k=1}^n \int_{[-1,1]^n} \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{i=1}^n(1-x_i^2) \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2\,dx \le 2k^2d^2 \int_{[-1,1]^n}|\nabla f|_0^{2k}\,dx.\tag{14}\]
Now we compute the left hand side. For fixed \(i\), we have \[\partial_{x_i}Q_{j_1,\ldots,j_k} = \sum_{s=1}^k \partial_{x_i x_{j_s}}^2f \prod_{\substack{\ell=1\\ \ell\ne s}}^k \partial_{x_{j_\ell}}f.\] Hence \[\bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2 = \sum_{s=1}^k \bigl(\partial_{x_i x_{j_s}}^2f\bigr)^2 \prod_{\substack{\ell=1\\ \ell\ne s}}^k \bigl(\partial_{x_{j_\ell}}f\bigr)^2 + \sum_{\substack{s,t=1\\ s\ne t}}^k \partial_{x_i x_{j_s}}^2f\, \partial_{x_i x_{j_t}}^2f \partial_{x_{j_s}}f\,\partial_{x_{j_t}}f \prod_{\substack{\ell=1\\ \ell\ne s,t}}^k \bigl(\partial_{x_{j_\ell}}f\bigr)^2.\] First, the diagonal part gives \[\begin{align} &\sum_{j_1,\ldots,j_k=1}^n \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{s=1}^k \bigl(\partial_{x_i x_{j_s}}^2f\bigr)^2 \prod_{\substack{\ell=1\\ \ell\ne s}}^k \bigl(\partial_{x_{j_\ell}}f\bigr)^2 \\ &= k \biggl(\sum_{j=1}^n (1-x_j^2) \bigl(\partial_{x_i x_j}^2f\bigr)^2\biggr) \biggl( \sum_{m=1}^n (1-x_m^2)(\partial_{x_m}f)^2 \biggr)^{k-1} \\ &= k|\nabla f|_0^{2k-2} \sum_{j=1}^n (1-x_j^2) \bigl(\partial_{x_i x_j}^2f\bigr)^2 . \end{align}\] Second, the off-diagonal part gives \[\begin{align} &\sum_{j_1,\ldots,j_k=1}^n \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{\substack{s,t=1\\ s\ne t}}^k \partial_{x_i x_{j_s}}^2f\, \partial_{x_i x_{j_t}}^2f \partial_{x_{j_s}}f\,\partial_{x_{j_t}}f \prod_{\substack{\ell=1\\ \ell\ne s,t}}^k \bigl(\partial_{x_{j_\ell}}f\bigr)^2 \\ &= k(k-1) \biggl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_i x_j}^2f \biggr)^2 \biggl( \sum_{m=1}^n (1-x_m^2)(\partial_{x_m}f)^2 \biggr)^{k-2} \\ &= k(k-1)|\nabla f|_0^{2k-4} \biggl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_i x_j}^2f \biggr)^2. \end{align}\] Therefore, for every fixed \(i\), \[\begin{align} &\sum_{j_1,\ldots,j_k=1}^n \prod_{\ell=1}^k(1-x_{j_\ell}^2) \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2 \\ &= k|\nabla f|_0^{2k-2} \sum_{j=1}^n (1-x_j^2) \bigl(\partial_{x_i x_j}^2f\bigr)^2 + k(k-1)|\nabla f|_0^{2k-4} \biggl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_i x_j}^2f \biggr)^2. \end{align}\] Multiplying by \(1-x_i^2\) and summing over \(i=1,\ldots,n\), we obtain \[\begin{align} &\sum_{j_1,\ldots,j_k=1}^n \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{i=1}^n(1-x_i^2) \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2 \\ &= k|\nabla f|_0^{2k-2} \sum_{i,j=1}^n (1-x_i^2)(1-x_j^2) (\partial_{x_i x_j}^2f)^2 \\ &+ k(k-1)|\nabla f|_0^{2k-4} \sum_{i=1}^n(1-x_i^2) \biggl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_i x_j}^2f \biggr)^2. \end{align}\]
We now estimate the term with \(\nabla|\nabla f|_0^2\). Since \[|\nabla f|_0^2 = \sum_{j=1}^n(1-x_j^2)(\partial_{x_j}f)^2,\] we have \[\partial_{x_i}|\nabla f|_0^2 = -2x_i(\partial_{x_i}f)^2 + 2\sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_ix_j}^2f .\] Hence, on the cube \([-1, 1]^n\), \[\begin{align} \bigl|\nabla|\nabla f|_0^2\bigr|_0^2 &= \sum_{i=1}^n(1-x_i^2) \bigl(\partial_{x_i}|\nabla f|_0^2\bigr)^2 \\ &\le 8|\nabla f|_0^2|\nabla f|^2 + 8\sum_{i=1}^n(1-x_i^2) \Bigl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_ix_j}^2f \Bigr)^2. \end{align}\] Multiplying by \(|\nabla f|_0^{2k-4}\) and integrating, we get \[\begin{align} &\int_{[-1,1]^n} |\nabla f|_0^{2k-4} \bigl|\nabla|\nabla f|_0^2\bigr|_0^2\,dx \\ &\le 8\int_{[-1,1]^n} |\nabla f|_0^{2k-2}|\nabla f|^2\,dx + 8\int_{[-1,1]^n} |\nabla f|_0^{2k-4} \sum_{i=1}^n(1-x_i^2) \Bigl( \sum_{j=1}^n (1-x_j^2)\partial_{x_j}f\,\partial_{x_ix_j}^2f \Bigr)^2 dx. \end{align}\] Therefore, applying 14 , we obtain \[\begin{align} &\int_{[-1,1]^n}|\nabla f|_0^{2k-2} \sum_{i,j=1}^n (1-x_i^2)(1-x_j^2) (\partial_{x_ix_j}^2f)^2\,dx + (k-1)^2 \int_{[-1,1]^n} |\nabla f|_0^{2k-4} \bigl|\nabla|\nabla f|_0^2\bigr|_0^2\,dx \\ &\le 8\int_{[-1,1]^n} \sum_{j_1,\ldots,j_k=1}^n \prod_{\ell=1}^k(1-x_{j_\ell}^2) \sum_{i=1}^n(1-x_i^2) \bigl(\partial_{x_i}Q_{j_1,\ldots,j_k}\bigr)^2\, dx \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+ 8(k-1)^2\int_{[-1,1]^n} |\nabla f|_0^{2k-2}|\nabla f|^2\,dx \\ &\le 16k^2d^2\bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1, 1]^n)}^{2k} + 8(k-1)^2 \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1, 1]^n)}^{2k-2}\|\nabla f\|_{L^{2k}([-1, 1]^n)}^{2}. \end{align}\] Finally, by Lemma 5 \[\|\nabla f\|_{L^{2k}([-1,1]^n)} \le 8kd\, \||\nabla f|_0\|_{L^{2k}([-1,1]^n)},\] which implies the announced estimate since \[16k^2d^2 + 512k^2d^2(k-1)^2\le 512k^4d^2.\] The lemma is proved. ◻
The case \(k=1\) follows from Lemma 7 with \(\alpha_1=\ldots=\alpha_n=0\): \[\bigl\||\nabla f|_0\bigr\|_{L^2([-1,1]^n)} \le \sqrt{d(d+1)}\,\|f\|_{L^2([-1,1]^n)} \le \sqrt{2}d\,\|f\|_{L^2([-1,1]^n)}.\] By Lemma 5, we also have \[\|\nabla f\|_{L^2([-1, 1]^n)}\le 8\sqrt{2}d^2\,\|f\|_{L^2([-1,1]^n)}\le 12d^2\,\|f\|_{L^2([-1,1]^n)}.\]
Let now \(k\ge2\). We first prove the estimate for the weighted gradient.
We apply Lemma 8 with \[r=2k-2, \quad q=k-1.\] Then \(r\ge q\ge1\), and we obtain \[\begin{align} &\bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k} \le 2\bigl\|f|\nabla f|_0^{k-1}\bigr\|_{L^2([-1, 1]^n)} \biggl( \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+ \int_{[-1,1]^n}|\nabla f|_0^{2k-2} \sum_{j,k=1}^n (1-x_j^2)(1-x_k^2) (\partial_{x_jx_k}^2f)^2\,dx \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+ (k-1)^2\int_{[-1,1]^n} |\nabla f|_0^{2k-4} \bigl|\nabla |\nabla f|_0^2\bigr|_0^2 \,dx \biggr)^{1/2}. \end{align}\] By Lemma 9, \[\begin{align} &\int_{[-1,1]^n}|\nabla f|_0^{2k-2} \sum_{i,j=1}^n (1-x_i^2)(1-x_j^2) (\partial_{x_ix_j}^2f)^2\,dx + (k-1)^2 \int_{[-1,1]^n} |\nabla f|_0^{2k-4} \bigl|\nabla|\nabla f|_0^2\bigr|_0^2 \,dx \\ &\le 512k^4d^2 \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k}. \end{align}\] Therefore, \[\int_{[-1,1]^n}|\nabla f|_0^{2k}\,dx \le 2\bigl\|f|\nabla f|_0^{k-1}\bigr\|_{L^2([-1,1]^n)} \Bigl( (1+512k^4d^2)\bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k} \Bigr)^{1/2}.\] Since \(k\ge2\) and \(d\ge1\), \[1+512k^4d^2\le 23^2k^4d^2.\] Hence \[\int_{[-1,1]^n}|\nabla f|_0^{2k}\,dx \le 46k^2d\, \bigl\|f|\nabla f|_0^{k-1}\bigr\|_{L^2([-1,1]^n)} \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^k.\] By Hölder’s inequality, \[\bigl\|f|\nabla f|_0^{k-1}\bigr\|_{L^2([-1,1]^n)} \le \|f\|_{L^{2k}([-1,1]^n)} \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{k-1}.\] Thus, \[\bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k} \le 46k^2d\, \|f\|_{L^{2k}([-1,1]^n)} \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)}^{2k-1},\] which implies \[\bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)} \le 46k^2d\,\|f\|_{L^{2k}([-1,1]^n)}.\] Finally, Lemma 5 gives \[\|\nabla f\|_{L^{2k}([-1,1]^n)} \le 8kd\, \bigl\||\nabla f|_0\bigr\|_{L^{2k}([-1,1]^n)} \le 368k^3d^2\,\|f\|_{L^{2k}([-1,1]^n)}.\] This concludes the proof. 0◻
Lemma 10. Let \(n, d\in \mathbb{N}\) and \(p\ge 1\). Assume that \[\label{eq-cube-MB} \|\nabla g\|_{L^p([-1,1]^n)} \le C(p)d^2\|g\|_{L^p([-1,1]^n)} \quad \forall g\in\mathcal{P}_d(\mathbb{R}^n).\tag{15}\] Then, for every \(a_j<b_j\), \(j=1,\ldots,n\), and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), \[\int_{[a_1,b_1]\times\ldots\times[a_n,b_n]} \Bigl( \sum_{j=1}^n (b_j-a_j)^2 \bigl(\partial_{x_j} f(x)\bigr)^2 \Bigr)^{p/2} \,dx \le \bigl(2C(p)d^2\bigr)^p \int_{[a_1,b_1]\times\ldots\times[a_n,b_n]} |f(x)|^p\,dx.\]
Proof. Let \[L(y) = \Bigl( \frac{b_1-a_1}{2}y_1+\frac{a_1+b_1}{2}, \ldots, \frac{b_n-a_n}{2}y_n+\frac{a_n+b_n}{2} \Bigr)\] and set \[g(y):=f(L(y)).\] Then \(g\in\mathcal{P}_d(\mathbb{R}^n)\) and \[\partial_{y_j}g(y) = \frac{b_j-a_j}{2}\, \partial_{x_j}f(L(y)).\] Hence \[|\nabla g(y)|^2 = \frac{1}{4} \sum_{j=1}^n (b_j-a_j)^2 \bigl(\partial_{x_j}f(L(y))\bigr)^2.\] Applying 15 to \(g\), we get \[\int_{[-1,1]^n} \Bigl( \frac{1}{4} \sum_{j=1}^n (b_j-a_j)^2 \bigl(\partial_{x_j}f(L(y))\bigr)^2 \Bigr)^{p/2} \,dy \le \bigl(C(p) d^2\bigr)^p \int_{[-1,1]^n}|f(L(y))|^p\,dy.\] Changing variables \(x=L(y)\) on both sides gives the claimed estimate. ◻
Lemma 11. Let \(n\in\mathbb{N}\), let \(\varrho_1,\ldots,\varrho_n\) be bounded probability densities on \(\mathbb{R}\), and set \[M:=\max_{1\le j\le n}\|\varrho_j\|_\infty.\] For \(t_j\ge0\), set \[E_j(t_j):=\{s\in\mathbb{R}\colon \varrho_j(s)\ge t_j\}, \quad L_j(t_j):=\lambda(E_j(t_j)),\] where \(\lambda\) is the Lebesgue measure on \(\mathbb{R}\). Then, for every \(p\ge2\) and every Borel vector field \(\eta=(\eta_1,\ldots,\eta_n)\colon\mathbb{R}^n\to[0,+\infty)^n\), one has \[\label{eq-cake} \int_{[0,\infty)^n} \int_{E_1(t_1)\times\cdots\times E_n(t_n)} \Bigl(\sum_{j=1}^n L_j(t_j)^2 \eta_j(x)\Bigr)^{p/2} \,dx\,dt \ge \frac{1}{M^p} \int_{\mathbb{R}^n} \Bigl(\sum_{j=1}^n\eta_j(x)\Bigr)^{p/2} \prod_{i=1}^n \varrho_i(x_i)\,dx.\tag{16}\]
Proof. For each \(j\), let \[M_j:=\|\varrho_j\|_\infty.\] By Fubini’s theorem, \[\int_0^{M_j} L_j(t_j)\,dt_j = \int_{\mathbb{R}}\varrho_j(s)\,ds = 1.\] Since \(L_j\) is nonincreasing on \([0,M]\), for every \(0<\tau<M_j\) we have \[\frac{1}{\tau}\int_0^\tau L_j(t_j)\,dt_j \ge L(\tau) \ge \frac{1}{M_j-\tau}\int_\tau^{M_j} L(t_j)\,dt_j.\] Therefore, for \(0<\tau<M_j\), \[\begin{align} \frac{1}{\tau}\int_0^\tau L(t_j)\,dt_j &\ge \frac{\tau}{M_j}\cdot\frac{1}{\tau}\int_0^\tau L(t_j)\,dt_j + \frac{M_j-\tau}{M_j}\cdot\frac{1}{M_j-\tau}\int_\tau^{M_j} L(t_j)\,dt_j \\ &= \frac{1}{M_j}\int_0^{M_j} L(t_j)\,dt_j = \frac{1}{M_j}. \end{align}\] Hence, by the Cauchy–Schwarz inequality, for every \(0<\tau\le M_j\), we have \[\int_0^\tau L_j(t_j)^2\,dt_j \ge \tau \biggl(\frac{1}{\tau}\int_0^\tau L_j(t_j)\,dt_j\biggr)^2 \ge \frac{\tau}{M_j^2}.\]
By Tonelli’s theorem, the left-hand side of 16 equals \[\int_{\mathbb{R}^n} \int_0^{\varrho_1(x_1)}\ldots\int_0^{\varrho_n(x_n)} \Bigl(\sum_{j=1}^n L_j(t_j)^2\eta_j(x)\Bigr)^{p/2} \,dt\,dx.\] Fix \(x\in\mathbb{R}^n\) and assume first that \(\varrho_j(x_j)>0\) for all \(j\in\{1,\ldots,n\}\). Since \(p/2\ge1\), Jensen’s inequality gives \[\begin{align} &\int_0^{\varrho_1(x_1)}\ldots\int_0^{\varrho_n(x_n)} \Bigl(\sum_{j=1}^n L_j(t_j)^2\eta_j(x)\Bigr)^{p/2} \,dt_1\ldots dt_n \\ &\ge \prod_{i=1}^n\varrho_i(x_i) \biggl( \frac{1}{\prod_{i=1}^n\varrho_i(x_i)} \int_0^{\varrho_1(x_1)}\ldots\int_0^{\varrho_n(x_n)} \sum_{j=1}^n L_j(t_j)^2\eta_j(x) \,dt_1\ldots dt_n \biggr)^{p/2} \\ &= \prod_{i=1}^n\varrho_i(x_i) \biggl( \sum_{j=1}^n \eta_j(x) \frac{1}{\varrho_j(x_j)} \int_0^{\varrho_j(x_j)}L_j(t_j)^2\,dt_j \biggr)^{p/2} \\ &\ge \prod_{i=1}^n\varrho_i(x_i) \Bigl(\sum_{j=1}^n M_j^{-2}\eta_j(x)\Bigr)^{p/2} \ge \frac{1}{M^p} \prod_{i=1}^n\varrho_i(x_i) \Bigl(\sum_{j=1}^n\eta_j(x)\Bigr)^{p/2}. \end{align}\] If \(\varrho_j(x_j)=0\) for at least one \(j\), then both sides of the pointwise estimate above vanish, and the estimate is trivial. Integrating over \(x\) gives 16 . ◻
Theorem 7. Let \(n,d\in \mathbb{N}\) and \(p\ge2\). Assume that \[\|\nabla g\|_{L^p([-1,1]^n)} \le C(p)d^2\|g\|_{L^p([-1,1]^n)} \quad \forall g\in\mathcal{P}_d(\mathbb{R}^n).\] Let \(\mu=\mu_1\otimes\cdots\otimes\mu_n\), where each \(\mu_j\) has a bounded unimodal density \(\varrho_j\) on \(\mathbb{R}\). Set \[M:=\max_{1\le j\le n}\|\varrho_j\|_\infty.\] Then, for every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^p(\mu)} \le 2C(p)Md^2 \|f\|_{L^p(\mu)}.\]
Proof. For \(t_j\ge0\), set \[E_j(t_j):=\{x\in\mathbb{R}\colon \varrho_j(x)\ge t_j\}, \quad L_j(t_j):=\lambda(E_j(t_j)).\] Since \(\varrho_j\) is unimodal, each nonempty set \(E_j(t_j)\) is an interval. Moreover, for \(t_j>0\) it has finite length, and its endpoints do not affect the integrals below. Therefore, applying Lemma 10 to the rectangle \[E_1(t_1)\times\ldots\times E_n(t_n),\] for \(t_1,\ldots,t_n>0\), we obtain \[\int_{E_1(t_1)\times\cdots\times E_n(t_n)} \Bigl( \sum_{j=1}^n L_j(t_j)^2 \bigl(\partial_{x_j}f(x)\bigr)^2 \Bigr)^{p/2} \,dx \le (2C(p)d^2)^p \int_{E_1(t_1)\times\cdots\times E_n(t_n)} |f(x)|^p\,dx.\] Integrating this inequality over \(t=(t_1,\ldots,t_n)\in[0,\infty)^n\) and using Tonelli’s theorem, the right-hand side becomes \[(2C(p)d^2)^p \int_{\mathbb{R}^n}|f(x)|^p \prod_{j=1}^n\varrho_j(x_j)\,dx = (2C(p)d^2)^p \|f\|_{L^p(\mu)}^p.\] By Lemma 11, applied to \[\eta_j(x):=\bigl(\partial_{x_j}f(x)\bigr)^2,\] the left-hand side is bounded from below by \[\frac{1}{M^p} \int_{\mathbb{R}^n} |\nabla f(x)|^p\,d\mu(x) = \frac{1}{M^p}\|\nabla f\|_{L^p(\mu)}^p.\] Combining the two bounds gives \[\frac{1}{M^p}\|\nabla f\|_{L^p(\mu)}^p \le (2C(p)d^2)^p\|f\|_{L^p(\mu)}^p.\] Taking the power \(1/p\) gives the claimed estimate. ◻
Combining Theorem 7 and Theorem 4 proves Theorem 5.
For \(m\in\mathbb{N}\) and \(f\in C^\infty_P(\mathbb{R}^n)\), set \[L_m f := \Delta f-2m\sum_{j=1}^n x_j^{2m-1}\partial_{x_j}f.\] For \(j\in\{1,\ldots,n\}\), define \[\delta_{m,j}f := 2m x_j^{2m-1}f-\partial_{x_j}f.\] For a vector field \(u\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\), define \[\delta_m u := \sum_{j=1}^n\delta_{m,j}u_j = 2m\sum_{j=1}^n x_j^{2m-1}u_j-\operatorname{div}u.\] Then \[\delta_m\nabla f=-L_m f.\]
For \(f,g\in C^\infty_P(\mathbb{R}^n)\), integration by parts gives \[\int_{\mathbb{R}^n}\partial_{x_j}f\,g\,d\nu_m^n = \int_{\mathbb{R}^n} f\,\delta_{m,j}g\,d\nu_m^n.\] Consequently, for \(f\in C^\infty_P(\mathbb{R}^n)\) and \(u\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\), \[\label{eq-freud-ibp-1} \int_{\mathbb{R}^n}\langle \nabla f,u\rangle\,d\nu_m^n = \int_{\mathbb{R}^n} f\,\delta_m u\,d\nu_m^n.\tag{17}\]
We will also use the following analogue of the Gaussian identity 5 . Since \[\partial_{x_j}(\delta_m u) = \sum_{k=1}^n\delta_{m,k}\partial_{x_j}u_k + 2m(2m-1)x_j^{2m-2}u_j,\] we obtain \[\begin{align} \label{eq-freud-ibp-2} \int_{\mathbb{R}^n}(\delta_m u)^2\,d\nu_m^n &= \int_{\mathbb{R}^n} \sum_{j=1}^n u_j\partial_{x_j}(\delta_m u)\,d\nu_m^n \\ &= 2m(2m-1) \int_{\mathbb{R}^n} \sum_{j=1}^n x_j^{2m-2}u_j^2\,d\nu_m^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n \partial_{x_k}u_j\,\partial_{x_j}u_k\,d\nu_m^n \notag \\ &\le 2m(2m-1) \int_{\mathbb{R}^n} \sum_{j=1}^n x_j^{2m-2}u_j^2\,d\nu_m^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n (\partial_{x_k}u_j)^2\,d\nu_m^n. \notag \end{align}\tag{18}\]
Theorem 8. Let \(m\in\mathbb{N}\). There exists a constant \(C(m)>0\) such that, for every \(n,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\|\nabla f\|_{L^2(\nu_m^n)} \le C(m)d^{1-\frac{1}{2m}} \|f\|_{L^2(\nu_m^n)}.\]
Proof. We first prove a one-dimensional coefficient estimate. Let \(\{u_j\}_{j\ge0}\) be the orthonormal polynomial system in \(L^2(\nu_m)\), where \(u_j\) has degree \(j\). For \(0\le j<r\), set \[b_{j,r}:= \int_{\mathbb{R}}u_r'u_j\,d\nu_m.\] Since \(u_r'\) has degree at most \(r-1\), we have \[u_r'=\sum_{j=0}^{r-1}b_{j,r}u_j.\] Integrating by parts gives \[\int_{\mathbb{R}}u_r'u_j\,d\nu_m = -\int_{\mathbb{R}}u_ru_j'\,d\nu_m + 2m\int_{\mathbb{R}}u_r(t)u_j(t)t^{2m-1}\,\nu_m(dt).\] The first integral on the right-hand side is zero, since \(u_j'\) has degree at most \(j-1<r\) and \(u_r\) is orthogonal to all polynomials of degree smaller than \(r\). Hence \[b_{j,r} = 2m\int_{\mathbb{R}}u_r(t)u_j(t)t^{2m-1}\,\nu_m(dt).\] Now \(t^{2m-1}u_j(t)\) has degree at most \(j+2m-1\). Therefore it is orthogonal to \(u_r\) whenever \(r>j+2m-1\). Thus \[b_{j,r}=0 \quad \text{unless} \quad j<r<j+2m.\]
Let \(f\in\mathcal{P}_d(\mathbb{R})\) and write \[f=\sum_{r=0}^d a_ru_r.\] Then \[f' = \sum_{r=1}^d a_ru_r' = \sum_{j=0}^{d-1} \Bigl( \sum_{\substack{r\in\{1,\ldots,d\}\\ j<r<j+2m}} a_rb_{j,r} \Bigr)u_j.\] By orthonormality and the Cauchy–Schwarz inequality, \[\begin{align} \|f'\|_{L^2(\nu_m)}^2 &= \sum_{j=0}^{d-1} \Bigl( \sum_{\substack{r\in\{1,\ldots,d\}\\ j<r<j+2m}} a_rb_{j,r} \Bigr)^2 \le (2m-1) \sum_{j=0}^{d-1} \sum_{\substack{r\in\{1,\ldots,d\}\\ j<r<j+2m}} a_r^2 b_{j,r}^2 \\ &\le (2m-1) \sum_{r=1}^d a_r^2\sum_{j=0}^{r-1} b_{j,r}^2 = (2m-1) \sum_{r=1}^d a_r^2\|u_r'\|_{L^2(\nu_m)}^2. \end{align}\] By the one-dimensional Markov–Bernstein inequality for Freud weights 3 , applied to the polynomial \(u_r\), we have \[\|u_r'\|_{L^2(\nu_m)} \le C_1(m)r^{1-\frac{1}{2m}}\|u_r\|_{L^2(\nu_m)} = C_1(m)r^{1-\frac{1}{2m}}.\] Consequently, \[\label{eq-1-dim-MB-Freud-coeff} \|f'\|_{L^2(\nu_m)}^2 \le C_2(m) \sum_{r=1}^d r^{2-\frac{1}{m}}a_r^2.\tag{19}\]
We now pass to the product measure. For a multi-index \(\mathbf{r}=(r_1,\ldots,r_n)\in\mathbb{N}_0^n\), set \[|\mathbf{r}|:=r_1+\ldots+r_n\] and \[U_{\mathbf{r}}(x):= u_{r_1}(x_1)\ldots u_{r_n}(x_n).\] The system \(\{U_{\mathbf{r}}\colon |\mathbf{r}|\le d\}\) is an orthonormal basis of \(\mathcal{P}_d(\mathbb{R}^n)\) in \(L^2(\nu_m^n)\). Hence every polynomial \(f\in\mathcal{P}_d(\mathbb{R}^n)\) can be written as \[f=\sum_{|\mathbf{r}|\le d}a_{\mathbf{r}}U_{\mathbf{r}}.\]
Fix \(j\in\{1,\ldots,n\}\). For \[\widehat{\mathbf{r}} = (r_1,\ldots,r_{j-1},r_{j+1},\ldots,r_n) \quad \text{and} \quad \widehat x = (x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n),\] write \[U_{\widehat{\mathbf{r}}}(\widehat x) = \prod_{\ell\ne j}u_{r_\ell}(x_\ell).\] Then \[f(x) = \sum_{|\widehat{\mathbf{r}}|\le d} U_{\widehat{\mathbf{r}}}(\widehat x) f_{\widehat{\mathbf{r}}}(x_j),\] where \[f_{\widehat{\mathbf{r}}}(x_j) := \sum_{r=0}^{d-|\widehat{\mathbf{r}}|} a_{(\widehat{\mathbf{r}},r)}u_r(x_j).\] Here \(a_{(\widehat{\mathbf{r}},r)}\) denotes the coefficient corresponding to the multi-index whose \(j\)-th coordinate is \(r\) and whose remaining coordinates are given by \(\widehat{\mathbf{r}}\). Therefore, \[\partial_{x_j}f(x) = \sum_{|\widehat{\mathbf{r}}|\le d} U_{\widehat{\mathbf{r}}}(\widehat x) f_{\widehat{\mathbf{r}}}'(x_j).\] Using the orthonormality of the functions \(U_{\widehat{\mathbf{r}}}\) in \(L^2(\nu_m^{n-1})\), we obtain \[\int_{\mathbb{R}^n} |\partial_{x_j}f|^2\,d\nu_m^n = \sum_{|\widehat{\mathbf{r}}|\le d} \int_{\mathbb{R}}|f_{\widehat{\mathbf{r}}}'|^2\,d\nu_m.\] Applying 19 to each \(f_{\widehat{\mathbf{r}}}\), we get \[\int_{\mathbb{R}^n} |\partial_{x_j}f|^2\,d\nu_m^n \le C_2(m) \sum_{|\widehat{\mathbf{r}}|\le d} \sum_{r=0}^{d-|\widehat{\mathbf{r}}|} r^{2-\frac{1}{m}}|a_{(\widehat{\mathbf{r}},r)}|^2 = C_2(m) \sum_{|\mathbf{r}|\le d} r_j^{2-\frac{1}{m}}|a_{\mathbf{r}}|^2.\] Summing over \(j\in\{1,\ldots,n\}\), we obtain \[\int_{\mathbb{R}^n}|\nabla f|^2\,d\nu_m^n = \sum_{j=1}^n \int_{\mathbb{R}^n}|\partial_{x_j}f|^2\,d\nu_m^n \le C_2(m) \sum_{|\mathbf{r}|\le d} \Bigl(\sum_{j=1}^n r_j^{2-\frac{1}{m}}\Bigr)|a_{\mathbf{r}}|^2.\] Since \(2-1/m\ge1\), we have \[\sum_{j=1}^n r_j^{2-\frac{1}{m}} \le \Bigl(\sum_{j=1}^n r_j\Bigr)^{2-\frac{1}{m}} = |\mathbf{r}|^{2-\frac{1}{m}}.\] Therefore, \[\int_{\mathbb{R}^n}|\nabla f|^2\,d\nu_m^n \le C_2(m) \sum_{|\mathbf{r}|\le d} |\mathbf{r}|^{2-\frac{1}{m}}|a_{\mathbf{r}}|^2 \le C_2(m)d^{2-\frac{1}{m}} \sum_{|\mathbf{r}|\le d}|a_{\mathbf{r}}|^2 = C_2(m)d^{2-\frac{1}{m}} \int_{\mathbb{R}^n}|f|^2\,d\nu_m^n.\] Taking square roots proves the theorem. ◻
Lemma 12. Let \(m\in\mathbb{N}\) and let \(r\ge q>0\). There exists a constant \(C:=C(m,r,q)>0\) such that, for every \(n,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\begin{align} &\|\nabla f\|_{L^{r+2}(\nu_m^n)}^{r+2} \\ &\le C \|f|\nabla f|^{r-q}\|_{L^2(\nu_m^n)} \biggl( \int_{\mathbb{R}^n} |\nabla f|^{2q} \sum_{j=1}^n x_j^{2m-2}(\partial_{x_j}f)^2\,d\nu_m^n + \int_{\mathbb{R}^n} |\nabla f|^{2q}\|D^2f\|_{\rm HS}^2\,d\nu_m^n \biggr)^{1/2}. \end{align}\]
Proof. The proof is the same as in the Gaussian case, with 18 replacing 5 . Let \(\eta\in C^\infty(\mathbb{R})\) satisfy \[\eta(t)=0 \quad \text{for } |t|\le 1, \quad \eta(t)=1 \quad \text{for } |t|\ge 2, \quad 0\le \eta(t)\le 1 \quad \text{for all } t\in\mathbb{R}.\] For \(\varepsilon>0\), set \(\eta_\varepsilon(t):=\eta(t/\varepsilon)\). For \(a>0\), define \[u_{a,\varepsilon} := |\nabla f|^a\eta_\varepsilon(|\nabla f|)\nabla f.\] The expression is understood as zero on the set \(\{|\nabla f|\le \varepsilon\}\). With this convention, we have \(u_{a,\varepsilon}\in C^\infty_P(\mathbb{R}^n,\mathbb{R}^n)\).
A direct computation gives \[\delta_m u_{a,\varepsilon} = -|\nabla f|^a\eta_\varepsilon(|\nabla f|)L_mf - \bigl( a|\nabla f|^{a-2}\eta_\varepsilon(|\nabla f|) + |\nabla f|^{a-1}\eta_\varepsilon'(|\nabla f|) \bigr) \langle D^2f\nabla f,\nabla f\rangle .\] Therefore, \[\delta_m u_{r,\varepsilon} = |\nabla f|^{r-q} \bigl( \delta_m u_{q,\varepsilon} - (r-q)|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) \langle D^2f\nabla f,\nabla f\rangle \bigr).\]
Using the integration by parts formula 17 , we obtain \[\int_{\mathbb{R}^n} |\nabla f|^{r+2}I_{\{|\nabla f|\ge 2\varepsilon\}}\, d\nu_m^n \le \int_{\mathbb{R}^n} |\nabla f|^{r+2}\eta_\varepsilon(|\nabla f|)\, d\nu_m^n = \int_{\mathbb{R}^n} f\,\delta_m u_{r,\varepsilon}\,d\nu_m^n.\] Hence, using \[\langle D^2f\nabla f,\nabla f\rangle^2 \le |\nabla f|^4\|D^2f\|_{\rm HS}^2,\] and applying the Cauchy–Schwarz inequality, we obtain \[\begin{align} &\int_{\mathbb{R}^n} |\nabla f|^{r+2}I_{\{|\nabla f|\ge 2\varepsilon\}}\, d\nu_m^n \\ &\le C_1(r,q) \|f|\nabla f|^{r-q}\|_{L^2(\nu_m^n)} \biggl( \int_{\mathbb{R}^n} (\delta_m u_{q,\varepsilon})^2\,d\nu_m^n + \int_{\mathbb{R}^n} |\nabla f|^{2q} \|D^2f\|_{\rm HS}^2\, d\nu_m^n \biggr)^{1/2}. \end{align}\]
By 18 , \[\begin{align} \int_{\mathbb{R}^n} (\delta_m u_{q,\varepsilon})^2\,d\nu_m^n &\le C_2(m) \int_{\mathbb{R}^n} |\nabla f|^{2q}\sum_{j=1}^n x_j^{2m-2}(\partial_{x_j} f)^2\, d\nu_m^n + \int_{\mathbb{R}^n} \sum_{j,k=1}^n \bigl(\partial_{x_k}(u_{q,\varepsilon})_j\bigr)^2\, d\nu_m^n. \end{align}\]
It remains to estimate the last term. As in the Gaussian case, \[\partial_{x_k}(u_{q,\varepsilon})_j = |\nabla f|^q\eta_\varepsilon(|\nabla f|) \partial_{x_kx_j}^2f + \bigl(q|\nabla f|^{q-2}\eta_\varepsilon(|\nabla f|) + |\nabla f|^{q-1}\eta_\varepsilon'(|\nabla f|)\bigr) \langle \nabla\partial_{x_k}f,\nabla f\rangle \partial_{x_j}f\] and \[\sum_{j,k=1}^n \bigl(\partial_{x_k}(u_{q,\varepsilon})_j\bigr)^2 \le C(q) \bigl( 1+ \|\eta'\|_\infty^2 I_{\{\varepsilon\le |\nabla f|\le 2\varepsilon\}} \bigr) |\nabla f|^{2q}\|D^2f\|_{\rm HS}^2.\] Combining the preceding estimates and passing to the limit as \(\varepsilon\to0\), with Lebesgue’s dominated convergence theorem applied as in the Gaussian case, gives the desired estimate. ◻
Applying the preceding lemma with \(r=2k-2\) and \(q=k-1\), we obtain the following estimate.
Corollary 3. Let \(m,k\in\mathbb{N}\), \(k\ge2\). There exists a constant \(C=C(k,m)>0\) such that, for every \(n,d\in\mathbb{N}\) and every \(f\in\mathcal{P}_d(\mathbb{R}^n)\), one has \[\begin{align} &\|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k} \\ &\le C\|f|\nabla f|^{k-1}\|_{L^2(\nu_m^n)} \biggl( \int_{\mathbb{R}^n} |\nabla f|^{2k-2} \sum_{j=1}^n x_j^{2m-2}(\partial_{x_j}f)^2\,d\nu_m^n + \int_{\mathbb{R}^n} |\nabla f|^{2k-2}\|D^2f\|_{\rm HS}^2\,d\nu_m^n \biggr)^{1/2}. \end{align}\]
Lemma 13. Let \(m\in\mathbb{N}\). There exists a constant \(C(m)>0\) such that, for every \(n, D\in\mathbb{N}\), every \(j\in\{1,\ldots,n\}\), and every \(f\in\mathcal{P}_D(\mathbb{R}^n)\), one has \[\int_{\mathbb{R}^n}x_j^{2m-2}|f(x)|^2\,d\nu_m^n \le C(m)D^{1-\frac{1}{m}} \|f\|_{L^2(\nu_m^n)}^2.\]
Proof. By Fubini’s theorem, it is sufficient to prove the one-dimensional estimate \[\label{eq-one-dim-mult-needed} \int_{\mathbb{R}}t^{2m-2}|g(t)|^2\,\nu_m(dt) \le C(m)D^{1-\frac{1}{m}} \|g\|_{L^2(\nu_m)}^2\tag{20}\] for every \(g\in\mathcal{P}_D(\mathbb{R})\).
For \(m=1\), this is immediate. Assume that \(m\ge2\). If \(\|g\|_{L^2(\nu_m)}=0\), there is nothing to prove. Otherwise, since \[\frac{d}{dt}\bigl(tg(t)^2e^{-|t|^{2m}}\bigr) = g(t)^2e^{-|t|^{2m}} + 2tg(t)g'(t)e^{-|t|^{2m}} - 2m t^{2m}g(t)^2e^{-|t|^{2m}},\] integration over \(\mathbb{R}\) gives \[2m\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt) = \|g\|_{L^2(\nu_m)}^2 + 2\int_{\mathbb{R}}tg(t)g'(t)\,\nu_m(dt).\] By Hölder’s inequality with exponents \(2m\), \(\frac{2m}{m-1}\), and \(2\), \[\left|\int_{\mathbb{R}}tg(t)g'(t)\,\nu_m(dt)\right| \le \Bigl(\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt)\Bigr)^{\frac{1}{2m}} \|g\|_{L^2(\nu_m)}^{1-\frac{1}{m}} \|g'\|_{L^2(\nu_m)}.\] By the one-dimensional \(L^2\) Markov–Bernstein inequality for Freud weights 3 , \[\|g'\|_{L^2(\nu_m)} \le C_1(m)D^{1-\frac{1}{2m}}\|g\|_{L^2(\nu_m)}.\] Therefore, \[\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt) \le \|g\|_{L^2(\nu_m)}^2 + C_1(m)D^{1-\frac{1}{2m}} \Bigl(\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt)\Bigr)^{\frac{1}{2m}} \|g\|_{L^2(\nu_m)}^{2-\frac{1}{m}}.\] Set \[a:= \frac{\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt)}{\|g\|_{L^2(\nu_m)}^2}.\] Then \[a \le 1+ C_1(m)D^{1-\frac{1}{2m}}a^{\frac{1}{2m}}.\] If \(a\ge1\), then, since \(D\ge1\), \[a \le 1+ C_1(m)D^{1-\frac{1}{2m}}a^{\frac{1}{2m}} \le (1+C_1(m))D^{1-\frac{1}{2m}}a^{\frac{1}{2m}}.\] Therefore, \[a\le C_2(m)D.\] If \(a<1\), then \[a<1\le (1+C_2(m))D= C_3(m)D.\] Thus, in all cases, \[a\le C_3(m)D.\] Hence \[\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt) \le C_3(m)D\|g\|_{L^2(\nu_m)}^2.\] Finally, by Hölder’s inequality with exponents \(\frac{m}{m-1}\) and \(m\), \[\int_{\mathbb{R}}t^{2m-2}|g(t)|^2\,\nu_m(dt) \le \Bigl(\int_{\mathbb{R}}t^{2m}|g(t)|^2\,\nu_m(dt)\Bigr)^{1-\frac{1}{m}} \|g\|_{L^2(\nu_m)}^{\frac{2}{m}} \le C_4(m)D^{1-\frac{1}{m}}\|g\|_{L^2(\nu_m)}^2.\] This proves 20 , and the product estimate follows by Fubini’s theorem. ◻
The case \(k=1\) is Theorem 8. Assume now that \(k\ge2\). By Corollary 3, we have \[\begin{align} \label{eq-cor-2} &\|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k} \le C_1(k, m)\|f|\nabla f|^{k-1}\|_{L^2(\nu_m^n)} \biggl( \int_{\mathbb{R}^n} |\nabla f|^{2k-2} \sum_{j=1}^n x_j^{2m-2}(\partial_{x_j}f)^2\,d\nu_m^n \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+ \int_{\mathbb{R}^n} |\nabla f|^{2k-2}\|D^2f\|_{\rm HS}^2\,d\nu_m^n \biggr)^{1/2}.\notag \end{align}\tag{21}\] By Hölder’s inequality, \[\|f|\nabla f|^{k-1}\|_{L^2(\nu_m^n)} \le \|f\|_{L^{2k}(\nu_m^n)} \|\nabla f\|_{L^{2k}(\nu_m^n)}^{k-1}.\] For each multi-index \((j_1,\ldots,j_k)\in\{1,\ldots,n\}^k\), let \[Q_{j_1,\ldots,j_k} := \partial_{x_{j_1}}f\ldots\partial_{x_{j_k}}f.\] Then \(Q_{j_1,\ldots,j_k}\in\mathcal{P}_{kd}(\mathbb{R}^n)\). By Lemma 2, \[\sum_{1\le j_1,\ldots,j_k\le n} |\nabla Q_{j_1,\ldots,j_k}|^2 \ge k|\nabla f|^{2k-2}\|D^2 f\|_{\rm HS}^2.\] Applying Theorem 8 to each \(Q_{j_1,\ldots,j_k}\) and summing over all \((j_1,\ldots,j_k)\), we obtain \[\begin{align} \int_{\mathbb{R}^n} |\nabla f|^{2k-2}\|D^2f\|_{\rm HS}^2\,d\nu_m^n &\le \sum_{1\le j_1,\ldots,j_k\le n} \int_{\mathbb{R}^n}|\nabla Q_{j_1,\ldots,j_k}|^2\,d\nu_m^n \\ &\le C_2(k,m)d^{2-\frac{1}{m}} \sum_{1\le j_1,\ldots,j_k\le n} \int_{\mathbb{R}^n}|Q_{j_1,\ldots,j_k}|^2\,d\nu_m^n \\ &= C_2(k,m)d^{2-\frac{1}{m}} \|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k}. \end{align}\] Similarly, \[|\nabla f|^{2k-2} \sum_{j=1}^n x_j^{2m-2}(\partial_{x_j}f)^2 = \sum_{1\le j_1,\ldots,j_k\le n} x_{j_k}^{2m-2}|Q_{j_1,\ldots,j_k}|^2.\] By Lemma 13, applied to each \(Q_{j_1,\ldots,j_k}\), we conclude \[\begin{align} \int_{\mathbb{R}^n} |\nabla f|^{2k-2} \sum_{j=1}^n x_j^{2m-2}(\partial_{x_j}f)^2\,d\nu_m^n &= \sum_{1\le j_1,\ldots, j_k\le n } \int_{\mathbb{R}^n}x_{j_k}^{2m-2}|Q_{j_1, \ldots, j_k}|^2\,d\nu_m^n \\ &\le C_3(k, m)d^{1-\frac{1}{m}} \sum_{1\le j_1,\ldots, j_k\le n } \int_{\mathbb{R}^n}|Q_{j_1, \ldots, j_k}|^2\,d\nu_m^n \\ &= C_3(k, m)d^{1-\frac{1}{m}} \|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k}. \end{align}\]
Substituting these estimates into 21 , we obtain \[\|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k} \le C_4(k,m)d^{1-\frac{1}{2m}} \|f\|_{L^{2k}(\nu_m^n)} \|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k-1}.\] If \(\|\nabla f\|_{L^{2k}(\nu_m^n)}=0\), the result is trivial. Otherwise, dividing by \(\|\nabla f\|_{L^{2k}(\nu_m^n)}^{2k-1}\) gives the claimed estimate. 0◻
ChatGPT was used for language editing, stylistic suggestions, draft wording for selected passages, and help with locating some references. All AI-generated text and suggested references were checked, corrected where necessary, and substantially revised by the author. The author takes full responsibility for the content of the paper.
The author would like to thank Sergey Tikhonov for reading the manuscript and for valuable comments and suggestions.
The research was supported by the AEI grant RYC2023-043616-I and by the Spanish State Research Agency, through the Severo Ochoa and María de Maeztu Program for Centers and Units of Excellence in R&D (CEX2020-001084-M). The author thanks CERCA Programme (Generalitat de Catalunya) for institutional support.