1 Introduction

Let us recall dimension-free \(L^p\)-gradient estimates involving Ornstein–Uhlenbeck operators (cf. [4,5,6, 21, 25]). Let \((\lambda _k)\) be a strictly increasing sequence of positive numbers such that

$$\begin{aligned} \sum _{k=1}^{\infty } \frac{1}{\lambda _k} < \infty . \end{aligned}$$
(1)

For any \(m \ge 1\) we denote by \(A_m\) the \(m \times m\) diagonal matrix with negative eigenvalues \(-\lambda _k\), \(k=1, \ldots , m\). Let \(f : {{\mathbb {R}}}^m \rightarrow {{\mathbb {R}}}\) be a bounded \(C^2\)-function with all first and second bounded derivatives, i.e., \(f \in C^{ 2}_b({{\mathbb {R}}}^m)\), and denote by \(u\in C^{ 2}_b({{\mathbb {R}}}^m)\) the unique bounded classical solution to

$$\begin{aligned} u(x) - \Big ( \frac{1}{2} \triangle _m u(x) + \langle A_mx , Du(x) \rangle \Big ) = u(x) - \frac{1}{2}\sum _{k=1}^m D^2_{kk} u(x) + \sum _{k =1}^m \lambda _k x_k D_k u(x) = f(x), \end{aligned}$$
(2)

where \(x= (x_1, \ldots , x_m) \in {{\mathbb {R}}}^m\) and \(\langle \cdot , \cdot \rangle \) denotes the standard scalar product in \({{\mathbb {R}}}^m\). Here, \(D_k\) and \(D^2_{kk}\) are first and second partial derivatives with respect to the canonical basis \((e_k)\) in \({{\mathbb {R}}}^m\). The operator we consider is an m-dimensional Ornstein–Uhlenbeck operator, namely \(L_m = \frac{1}{2} \triangle _m + \langle A_m x, D\rangle \).

Then, introduce the Gaussian measure \(\mu _m= {\mathcal {N}}(0, (-2 A_m)^{-1})\) with mean 0 and covariance matrix \((-2A_m)^{-1}\), with density

$$\begin{aligned} \varphi _m(x) := \left( \frac{1}{2\pi }\right) ^{m/2} \left( \prod _{i=1}^m \frac{1}{2\lambda _i}\right) ^{-1/2} \exp \left\{ - \sum _{i=1}^m \lambda _i x_i^2\right\} , \qquad x= (x_1, \ldots , x_m) \in {{\mathbb {R}}}^m\ . \end{aligned}$$

Note that \(L_m\) is a self-adjoint operator on \(L^2({{\mathbb {R}}}^m, \mu _m)\), the usual \(L^2\)-space with respect to \(\mu _m\). See, for instance, [4, 6, 7, 13]. It is known that, if \(1< p< \infty \), there exists a constant \(c_p\), independent of f and the dimension m, such that the following sharp gradient estimate holds:

$$\begin{aligned} \int _{{{\mathbb {R}}}^m} \Big (\sum _{k=1}^m \lambda _k \, (D_k u (y))^2 \Big )^{p/2} \mu _m (\textrm{d}y) \le (c_p)^p \, \int _{{{\mathbb {R}}}^m} |f( y)|^p \mu _m (\textrm{d}y). \end{aligned}$$
(3)

The result follows from the general estimates (11) given in Theorem 5.3 of [4], which extends Proposition 3.5 in [25] (see also the references therein). Note that (3) can be rewritten as

$$\begin{aligned} \Vert (-A_m)^{1/2} Du \Vert _{L^p({{\mathbb {R}}}^m, \mu _m)} \le c_p \Vert f \Vert _{L^p({{\mathbb {R}}}^m, \mu _m)}, \end{aligned}$$
(4)

where

$$\begin{aligned} (-A_m)^{1/2} Du (x) = \sum _{k=1}^m \sqrt{\lambda _k} \, D_k u(x) e_k. \end{aligned}$$

Our main result (cf. Theorem 6 below) shows that, when \(p=\infty \), the dimension-free estimate (4) in general fails to hold. Indeed, we prove the following stronger assertion. Writing \(u =u^{f}\) to stress the dependence of the solution u on f, we show that if \(\lambda _k \sim k^2\) as \(k \rightarrow \infty \), then, choosing \(x=0\), we have

$$\begin{aligned} \sup _{\begin{array}{c} f \in C^{ 2}_b({{\mathbb {R}}}^m) \\ \Vert f\Vert _{\infty } \le 1 \end{array}} |(-A_m)^{1/2} Du^f (0)|_{{{\mathbb {R}}}^m}^2 = \sup _{\begin{array}{c} f \in C^{ 2}_b({{\mathbb {R}}}^m) \\ \Vert f\Vert _{\infty } \le 1 \end{array}} \Big \{ \sum _{k=1}^m \lambda _k \, (D_k u^f (0))^2 \Big \} \rightarrow \infty \;\; \text{ as } \; m \rightarrow \infty . \end{aligned}$$
(5)

In contrast to (5), we point out that, when \(A_m = - \lambda I_m\) with \(\lambda >0\) and \(I_m\) the \(m \times m\) identity matrix, then the following dimension-free \(L^{\infty }\)-gradient estimates

$$\begin{aligned} \Vert (\lambda )^{1/2} D u^f \Vert _{\infty } = \sup _{x \in {{\mathbb {R}}}^m}\, |(\lambda )^{1/2} D u^f (x)|_{{{\mathbb {R}}}^m} \, \le \frac{\pi }{\sqrt{2}}\, \sup _{x \in {{\mathbb {R}}}^m}| f(x)|,\;\; f \in C^{ 2}_b ({{\mathbb {R}}}^m) \end{aligned}$$
(6)

hold true; see Proposition 5.

1.1 Infinite dimensional Ornstein–Uhlenbeck semigroups

Let us comment on the previous dimension-free \(L^p_{\mu }\)-estimate (4). Such kind of inequalities can be deduced from known results for infinite dimensional Ornstein–Uhlenbeck operators. This point of view is of interest in probability because of its connection with SPDEs (see also [8] and the references therein).

To introduce this setting, we replace \({{\mathbb {R}}}^m\) by a real separable Hilbert space H with orthonormal basis \((e_k)_{k \ge 1} \) and inner product \(\langle \cdot , \cdot \rangle \). Then, we consider the unbounded self-adjoint operator \(A : D(A)\subset H\rightarrow H\) such that

$$\begin{aligned} D(A) = \Big \{ x \in H \, :\, \sum _{k \ge 1} (\langle x,e_k \rangle )^2 \, \lambda _k^2 < \infty \Big \},\;\;\;\; A e_k = - \lambda _k e_k, \;\; k \ge 1 \end{aligned}$$
(7)

(cf. [1, 8, 9, 22]). Our condition (1) is equivalent to require that the inverse operator \(A^{-1} : H \rightarrow H\) is a trace class operator. The operator A generates a strongly continuous semigroup \((e^{tA})\) on H, given by \(e^{tA} e_k = e^{-t\lambda _k} e_k\) for any \(t \ge 0\) and \(k \ge 1\). We can define the corresponding Ornstein–Uhlenbeck semigroup \((P_t)\) by

$$\begin{aligned} P_t f (x) = \int _{H} f(e^{tA} x+ \sqrt{I - e^{2tA}}\, y ) \; {\mathcal {N}} \big (0 , - (2 A)^{-1} \big ) \, (\textrm{d}y),\;\; f \in { B}_b (H),\; x \in H, \; t \ge 0 \end{aligned}$$
(8)

where \(f: H \rightarrow {{\mathbb {R}}}\) is a Borel, bounded function (i.e., \(f \in { B}_b (H)\)); \({\mathcal {N}} \big (0 , - (2 A)^{-1} \big )\) stands for the centered Gaussian measure defined on the Borel \(\sigma \)-algebra of H (see Chapter 1 in [7, 9] and Section 2.2); \(I : H \rightarrow H\) is the identity.

Formula (8) is an extension of a well-known formula used in finite dimension. From the probabilistic point of view \((P_t)\) is the transition Markov semigroup of the OU stochastic process \((X_t^x)\) which solves \(dX_t = AX_t \textrm{d}t + dW_t,\) \( X_0 =x\) where W is a cylindrical Wiener process on H; cf. [7, 8, 14]. When \(f\in C^2_b(H)\), i.e. f is bounded, twice Fréchet-differentiable with first and second bounded and continuous derivatives, we consider \(u: H \rightarrow {{\mathbb {R}}}\) given by

$$\begin{aligned} u(x) = R(1, L) f(x) = \int _0^{\infty } e^{-t} (P_t f)(x)\textrm{d}t,\;\; x \in H. \end{aligned}$$
(9)

Following Chapter 6 in [7], u is the generalized bounded solution to \(u - L u =f\), where L is formally given by \( \frac{1}{2} \text {Tr} (D^2) + \langle x , AD\rangle \). Here, we only note that if f is also cylindrical, i.e., there exist \(m \ge 1\) and \({\tilde{f}} \in C^2_b ({{\mathbb {R}}}^m)\) such that

$$\begin{aligned} f(x) = {\tilde{f}} (\langle x, e_{1}\rangle , \ldots , \langle x, e_{m} \rangle ),\;\;\; x \in H, \end{aligned}$$
(10)

then u given in (9) depends only on a finite number of variables, i.e.,

$$\begin{aligned} u(x) = {\tilde{u}} (\langle x, e_{1}\rangle , \ldots , \langle x, e_{m} \rangle ), \ x \in H \end{aligned}$$

(cf. Sect. 2.2). Moreover, \({\tilde{u}}\) solves (2) with f replaced by \({\tilde{f}}\). In addition, if \(f \in C_b^2(H)\), we have that \(u = R(1, L) f \in C^2_b(H)\), and \(Du(x) \in D((-A)^{1/2})\), \(x \in H\) (cf. [9] for stronger results). By Theorem 5.3 of [4] (see also Corollary 5.4 in [4] and Remark 1), there exists a constant \(c_p\) (independent of f) such that

$$\begin{aligned} \Vert (-A)^{1/2} Du \Vert _{L^p(H, \mu )} \le c_p \Vert f \Vert _{L^p(H, \mu )}, \;\;\; 1< p< \infty , \end{aligned}$$
(11)

where \(\mu = {\mathcal {N}} \big (0 , - (2 A)^{-1} \big )\). Moreover, we have \( \Vert D^2 u \Vert _{L^p(H, \mu )} \le c_p \Vert f \Vert _{L^p(H, \mu )}\), i.e.,

$$\begin{aligned} \int _{H} \Big (\sum _{k=1}^{\infty } \, (D_{kk} u (y))^2 \Big )^{p/2} \mu (\textrm{d}y) \le (c_p)^p \, \int _{H} |f( y)|^p \mu (\textrm{d}y). \end{aligned}$$
(12)

It is not difficult to show that (11) implies (4) using cylindrical functions f as in (10); see Remark 8. Estimates (11) and (12) are part of the generalized Meyer’s inequalities proved in [4] using the elliptic Littlewood-Paley-Stein inequalities associated with the OU semigroup \((P_t)\). For applications of the classical Meyer’s inequalities to the Malliavin Calculus we refer to [15, 20, 21, 26] (see also Remark 2). The results given in [4] give a characterization of the domain of the generator of \((P_t)\) in \(L^p(H, \mu )\); see also [5] (the case \(p=2\) was obtained earlier in [6]). We also mention the characterization of the domain of non self-adjoint Ornstein–Uhlenbeck generators given in [17, 19]. Estimates (12) have been used to prove strong uniqueness for a class of SPDEs in [10]. For related results on Ornstein–Uhlenbeck operators in Gaussian harmonic analysis we refer to [2, 3, 13, 18] and the references therein.

Our main result implies that (11) fails to hold for \(p= \infty \), i.e. it is not true that there exists \(C >0\), independent of f, such that

$$\begin{aligned} \sup _{x \in H}\, | (-A)^{1/2} D R(1, L) f (x)|_H \, \le C\, \sup _{x \in H}| f(x)|,\;\; f \in C_b^2(H), \end{aligned}$$
(13)

where we have used \(u= R(1, L) f\) as in (9) (see in particular Corollary 7). This estimate is stated in [22, Theorem 7] which is based on [22, Lemma 6]. However, there is a mistake in the proof of that lemma. In particular, we show that [22, Theorem 7] cannot hold.

Remark 1

Let us recall the notation used in [4] to study general symmetric Ornstein–Uhlenbeck semigroups in Hilbert spaces. For the sake of notational clarity, the operator C used in [4] corresponds to our \(-(2 A)^{-1}\), while our semigroup \((e^{tA})\) corresponds to \((e^{-tA})\) in [4]. They use the Malliavin gradient \(D_I = C^{1/2} D\) (where D is the Fréchet derivative) and \(D_A = \frac{1}{\sqrt{2}} D\). Moreover, the symbol \(D_{A^2} = A D_I\), which is used in the definition of the Sobolev space \(W^{1,p}_{A^2}\) (see Corollary 5.4 in [4]) corresponds to our operator \(\frac{{1}}{\sqrt{2}} (-A)^{1/2} D\).

Remark 2

Let us recall the classical Ornstein–Uhlenbeck semigroup \((S_t)\)

$$\begin{aligned} S_t f(x)= \int _{H} f(e^{-t} x+ \sqrt{1 - e^{-2t}}y) \; \nu \, (\textrm{d}y),\;\; f \in { B}_b (H),\; x \in H, \end{aligned}$$
(14)

where \(\nu \) is a centered Gaussian measure on H (see Sect. 2.2). The classical Meyer’s inequalities give a complete characterization of the domains of \((I-N_p)^{m/2}\) in \(L_p(H,\nu )\) for all \(p \in (1, \infty )\) and \(m=1,2, \ldots \) in terms of Gaussian Sobolev spaces related to \(\nu \). Here, \(N_p\) denotes the generator of \((S_t)\) in \(L_p(H,\nu )\) (see [15, 20, 21]). For a discussion of Meyer’s inequalities in the Malliavin Calculus we refer to [15] and [26, Chapter 4].

Remark 3

Estimates like (11) and (12) hold true also in Hölder spaces (see [9, 23] for more details). In particular, for any \(\theta \in (0,1)\), there exists an absolute constant \(c_{\theta }\) only depending on \(\theta \) such that

$$\begin{aligned} \Vert (-A)^{1/2} D R(1, L) f \Vert _{C^{\theta }_b(H, H)} \le c_{\theta } \Vert f \Vert _{ C^{\theta }_b(H)}. \end{aligned}$$
(15)

Remark 4

We do not know if for \(p \not = 2\) the constant \(c_p\) appearing in (3) is an absolute constant (independent of the positive eigenvalues \((\lambda _k))\). Indeed, as we have mentioned before, the dimension-free estimate (3) follows from infinite dimensional estimates like (11) and (12) which are proved in Theorem 5.3 of [4] (extending Proposition 3.5 in [25]). However, Theorem 5.3 uses Lemma 5.1 in [4], whose proof invokes results on sums of operators with bounded imaginary powers (see Theorem 4 and Corollary 2 in [24]). The approach of [11] and [24], which has been also used in [19], does not provide sharp constants in the estimates and so we do not know if \(c_p\) also depends on A. We point out that the estimates given in Proposition 3.5 of [25] provide absolute constants.

2 Notations and main results

Let Q be a symmetric and positive definite \(m \times m\) matrix. We denote by \({\mathcal {N}}(0,Q)\) the Gaussian measure with mean 0 and covariance matrix Q, which has density

$$\begin{aligned} \left( \frac{1}{2\pi }\right) ^{m/2} \frac{1}{\sqrt{\det {Q}}} \exp \left\{ - \frac{1}{2}\ |Q^{-1/2} x|^2\right\} \end{aligned}$$
(16)

with respect to the m-dimensional Lebesgue measure. We first consider for \(\lambda >0\) the equation

$$\begin{aligned} v(x) - \Big ( \frac{1}{2} \triangle _m v(x) - \lambda \langle x , Dv(x) \rangle \Big ) = v(x) - M_m v(x) = f(x), \;\;\; x\in {{\mathbb {R}}}^m, \end{aligned}$$
(17)

with \(M_m = \frac{1}{2} \triangle _m - \lambda \langle x, D \rangle \). We assume that \(f \in C^2_b({{\mathbb {R}}}^m)\). Equation (17) is similar to (2) with \(A_m\) replaced by \(-\lambda I_m\). Using the following Ornstein–Uhlenbeck semigroup \((S_t^m)\)

$$\begin{aligned} S_t^m f(x)= \int _{{{\mathbb {R}}}^m} f(e^{-\lambda t} x+ \sqrt{1 - e^{-2 \lambda t}} \, y ) \; {\mathcal {N}}\big (0, \frac{1}{2 \lambda }I_m \big ) \, (\textrm{d}y),\;\;\; x \in {{\mathbb {R}}}^m, \; t \ge 0, \end{aligned}$$
(18)

we find (cf. (9), and [7])

$$\begin{aligned} v(x) = R(1, M_m) f(x) = \int _0^{\infty } e^{-t} (S_t^m f)(x)\textrm{d}t,\;\; x \in {{\mathbb {R}}}^m. \end{aligned}$$

Then, we have the following

Proposition 5

For any \(\lambda >0\) it holds:

$$\begin{aligned} \sup _{x \in {{\mathbb {R}}}^m}\, \sqrt{\lambda } |D R(1, M_m) f (x)|_{{{\mathbb {R}}}^m} \, \le \frac{\pi }{\sqrt{2}}\, \sup _{x \in {{\mathbb {R}}}^m}| f(x)|,\;\; f \in C^{ 2}_b ({{\mathbb {R}}}^m). \end{aligned}$$
(19)

Proof

Let \(v(x) = R(1, M_m) f \in C^2_b({{\mathbb {R}}}^m)\). We set \(v(x) = u(\sqrt{\lambda } \, x)\) and so, for \(y \in {{\mathbb {R}}}^m\), we get

$$\begin{aligned} u(y) - \frac{\lambda }{2}\triangle u( y) + \lambda \langle y , Du(y) \rangle = f(y / \sqrt{\lambda }) \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\lambda } u(y) - \frac{1}{2}\triangle u( y) + \langle y , Du(y) \rangle = \frac{1}{\lambda } f(y / \sqrt{\lambda }) = {\tilde{f}}(y). \end{aligned}$$

We have

$$\begin{aligned} u(x) = \int _0^{\infty } e^{-\frac{1}{\lambda } \, t} \textrm{d}t \int _{{{\mathbb {R}}}^m} {\tilde{f}}(e^{-t } x + y ) \, {\mathcal {N}} \Big (0, \frac{ 1- e^{- 2t }}{2} \, I_m \Big ) (\textrm{d}y) \end{aligned}$$

and, considering the directional derivative \( \langle Du(x), h\rangle = D_h u(x)\), \(h \in {{\mathbb {R}}}^m\), \(|h|=1\), we get, differentiating under the integral sign,

$$\begin{aligned} D_h u(x) = 2 \int _0^{\infty } e^{-\frac{1}{\lambda } \, t} \int _{{{\mathbb {R}}}^m} {\tilde{f}}(e^{-t } x + y )\frac{e^{- t} }{1 - e^{- 2 t } } \, \langle h, y \rangle \, {\mathcal {N}} \Big (0, \frac{ 1- e^{- 2t }}{2} \, I_m \Big ) (\textrm{d}y) \end{aligned}$$

(cf. Theorem 6.2.2 in [7, 9]). Then, changing variable in the integral over \({{\mathbb {R}}}^m\), we obtain

$$\begin{aligned} \Vert D_h u \Vert _{\infty }&\le 2 \Vert {\tilde{f}}\Vert _{\infty } \int _0^{\infty } \frac{e^{-\frac{1}{\lambda }\, t} \, e^{- t} }{1 - e^{- 2 t } } \textrm{d}t \int _{{{\mathbb {R}}}^m} \Big |\langle h, \big (\frac{ 1- e^{- 2t }}{2} \big )^{1/2}y \rangle \Big | \, {\mathcal {N}} \big (0, \, I_m \big ) (\textrm{d}y) \\&\le \frac{\sqrt{2}}{\lambda } \Vert f\Vert _{\infty } \int _0^{\infty } \frac{e^{- t}}{(1 - e^{- 2 t })^{1/2} } \textrm{d}t \int _{{{\mathbb {R}}}^m} \big |\langle h, y \rangle \big | \, {\mathcal {N}} \big (0,\, I_m \big ) \textrm{d}y \le \frac{\pi }{\lambda \sqrt{2}} \Vert f\Vert _{\infty }. \end{aligned}$$

Since \(D_h u(y) = \frac{1}{\sqrt{\lambda }} D_h v(\frac{y}{\sqrt{\lambda }} )\), we have \(\Vert D_h u \Vert _{\infty } = \frac{1}{\sqrt{\lambda }} \Vert D_h v \Vert _{\infty } \) and (19) follows. \(\square \)

Let us start the proof of the main estimate (5) concerning equation (2) involving the Ornstein–Uhlenbeck operator \(L_m\). Similarly to the proof of Proposition 5 the solution \(u \in C^{ 2}_b({{\mathbb {R}}}^m)\) to (2) is given by

$$\begin{aligned} u(x)= R(1, L_m)f(x)= \int _0^{\infty } e^{-t} (P_t^m f)(x) \textrm{d}t \end{aligned}$$
(20)

with

$$\begin{aligned} P_t^m f(x)&= \int _{{{\mathbb {R}}}^m} f(e^{tA_m} x+ \sqrt{I_m - e^{2tA_m}}\, y)\; {\mathcal {N}} \Big (0 , -\frac{1}{2} A^{-1}_m \Big ) \, (\textrm{d}y) \\&=\int _{{{\mathbb {R}}}^m} f(e^{tA_m} x + y)\; {\mathcal {N}} \big (0 , Q_t^m \big ) \, (\textrm{d}y), \;\; f \in C^2_b({{\mathbb {R}}}^m), \; x \in {{\mathbb {R}}}^m, \end{aligned}$$

where

$$\begin{aligned} Q_t^m = \int _0^t e^{2 sA_m} ds = (-2 A_m)^{-1}(I_m -e^{2tA_m}),\;\;\; t \ge 0 \end{aligned}$$

(\(Q_t^m\) is a diagonal matrix with positive eigenvalues). Let \(\mu _t^m= {\mathcal {N}} \big (0 , Q_t^m \big )\). By differentiating under the integral sign, the following formula holds for the directional derivative of \(P_t^m f\) along \(h \in {{\mathbb {R}}}^m\):

$$\begin{aligned} D_h P_t^m f(x) = \langle D P_t^m f(x),h \rangle = \int _{{{\mathbb {R}}}^m} \langle \Lambda _t^m h,(Q_t^m)^{-\frac{1}{2}} y\rangle \, f (e^{tA_m}x+y ) \mu _t^m(\textrm{d}y), \; x \in {{\mathbb {R}}}^m, \; t>0, \end{aligned}$$
(21)

where \( \Lambda _t^m = (Q_t^m)^{-1/2}e^{tA_m};\) cf. Theorem 6.2.2 in [7]. Hence, the term

$$\begin{aligned} (-A_m)^{1/2} Du^f (0) = (-A_m)^{1/2} D R(1, L_m)f (0) \in {{\mathbb {R}}}^m \end{aligned}$$

that appears in (5) has components

$$\begin{aligned} \langle (-A_m)^{1/2} Du^f (0),e_k\rangle&= \int _0^{\infty } e^{-t} \textrm{d}t \int _{{{\mathbb {R}}}^m} \langle (-A_m)^{1/2} \Lambda _t^m e_k,(Q_t^m)^{-\frac{1}{2}} y\rangle \, f ( y ) \mu _t^m(\textrm{d}y)\\&= \int _0^{\infty } e^{-t} \textrm{d}t \int _{{{\mathbb {R}}}^m} \langle (-A_m)^{1/2} \Lambda _t^m e_k,y\rangle \, f ((Q_t^m)^{\frac{1}{2}} y ) \, {\mathcal {N}}(0,I_m)(\textrm{d}y) \end{aligned}$$

for \(k = 1, \ldots , m\). An easy calculation shows that

$$\begin{aligned}&|(-A_m)^{1/2} D R(1, L_m)f (0)|^{2} \nonumber \\&\qquad = \sum _{k=1}^m \Big (\int _0^{\infty }\dfrac{\lambda _k e^{-t} e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} f (c_1 (t) x_1, \ldots , c_m(t)x_m)\, x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \Big )^2, \end{aligned}$$
(22)

where, for \(t\ge 0\), \(c_k(t) = \left( \dfrac{1 - e^{-2\lambda _k t}}{2\lambda _k}\right) ^{1/2}\) and \((Q_t^m)^{1/2} = \text {diag}[c_1(t), \dots , c_m(t)]\).

We will prove the following result.

Theorem 6

Let \((\lambda _k)\) be a strictly increasing sequence of positive numbers, such that \(\lambda _k \sim k^2\) as \(k \rightarrow +\infty \). Then, assertion (5) is in force, i.e., taking into account (22), there holds

$$\begin{aligned} \sup _{m \in {{\mathbb {N}}}} \ \sup _{\begin{array}{c} f \in C^{ 2}_b ({{\mathbb {R}}}^m) \\ \Vert f\Vert _{\infty } \le 1 \end{array}}&\sum _{k=1}^m \left( \int _0^{\infty }\dfrac{\lambda _k e^{-t} e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \times \right. \\&\times \left. \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} f (c_1(t)x_1, \dots , c_m(t)x_m)\, x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \right) ^2 = \infty . \end{aligned}$$

The proof of this theorem is given in Section 3.

Finally, we show that Theorem 6 implies that the infinite dimensional estimate (13) cannot hold.

Corollary 7

Under the same assumptions of Theorem 6, there holds

$$\begin{aligned} \sup _{\begin{array}{c} f \in C_b^{2}(H),\;\; \Vert f\Vert _{\infty } \le 1 \end{array}} | (-A)^{1/2} D (R(1, L) f)\, (0)|_H = \infty . \end{aligned}$$

Proof of Corollary 7

Recall that we are considering a real separable Hilbert space H with inner product \(\langle \cdot , \cdot \rangle \). According to Chapter 1 in [7], we can rewrite the OU semigroup \((P_t)\) in (8) as follows

$$\begin{aligned} P_t f (x) = \int _{H} f(e^{tA} x+ y)\; {\mathcal {N}} \big (0 , Q_t \big ) \, (\textrm{d}y),\;\; f \in C^2_b(H),\; x \in H, \end{aligned}$$
(23)

where \(Q_t = \int _0^t e^{2 sA} ds := (-2 A)^{-1}(I-e^{2tA}),\; t \ge 0,\) and A is given in (7). Suppose that \(f \in C^2_b(H)\) is also cylindrical, i.e. that (10) holds for some \(m \ge 1\) and \({\tilde{f}} \in C_b^2 ({{\mathbb {R}}}^m)\). Identifying H with \(l^2\), we have that \(f(e^{tA} x + y) = {\tilde{f}} (e^{tA_m}x^{(m)} + y^{(m)})\), where \(A_m\) is the same matrix given in (2) and (20) and \(h^{(m)} := (\langle h, e_{1}\rangle , \ldots , \langle h, e_{m} \rangle ) \in {{\mathbb {R}}}^m\), for any \(h \in H\). Moreover, we put \( \mu := {\mathcal {N}} \big (0 , - (2 A)^{-1} \big ) = {\mathcal {N}} \big (0 , - (2 A_m)^{-1} \big ) \times \nu _m\), where \(\nu _m = \prod _{k= m+1}^{\infty } N(0, (2 \lambda _k)^{-1} )\); see Theorem 1.2.1 in [7]. It follows that, for any \(x \in H\),

$$\begin{aligned} P_t f (x)= & {} P_t^m ({\tilde{f}})( x^{(m)}) = \int _{{{\mathbb {R}}}^m } {\tilde{f}}( e^{tA_m} x^{(m)} + \sqrt{I_m - e^{2tA_m}}\, y ) \; {\mathcal {N}} \big (0 , - (2 A_m)^{-1} \big ) \, (\textrm{d}y), \\ u(x)= & {} R(1,L)f(x) = \int _0^{\infty } e^{-t} (P_t f)(x)\textrm{d}t = {\tilde{u}} (\langle x, e_{1}\rangle , \ldots , \langle x, e_{m} \rangle ) \end{aligned}$$

where \({\tilde{u}}\) is given in (20) with \({\tilde{f}}\) in place of f therein. Setting \(\mu _m := {\mathcal {N}} \big (0 , - (2 A_m)^{-1} \big )\) and using that \(C^2_b(H)\) contains the cylindrical functions displayed in (10), we get that

$$\begin{aligned} \sup _{\begin{array}{c} {\tilde{f}} \in C^{ 2}_b({{\mathbb {R}}}^m) \\ \Vert {\tilde{f}}\Vert _{\infty )} \le 1 \end{array}} \, | (-A_m)^{1/2} D {\tilde{u}} (0)| \le {\sup _{\begin{array}{c} f \in C^{ 2}_b(H) \\ \Vert f\Vert _{\infty } \le 1 \end{array}}} | (-A)^{1/2} Du (0)|_{H}, \end{aligned}$$
(24)

holds for any \(m \ge 1\). Notice that on the left-hand side of (24) we have \(0 \in {{\mathbb {R}}}^m\) while on the right-hand side we have \(0 \in H\). Thus, as a consequence of Theorem 6, we deduce the assertion. \(\square \)

Remark 8

By the same argument as in the previous proof we get easily

$$\begin{aligned} \sup _{\begin{array}{c} {\tilde{f}} \in C^{ 2}_b({{\mathbb {R}}}^m) \\ \Vert {\tilde{f}}\Vert _{L^p({{\mathbb {R}}}^m, \mu _m)} \le 1 \end{array}} \, \Vert (-A_m)^{1/2} D {\tilde{u}} \Vert _{L^p({{\mathbb {R}}}^m, \mu _m)} \le {\sup _{\begin{array}{c} f \in C^{ 2}_b(H) \\ \Vert f\Vert _{L^p(H, \mu )} \le 1 \end{array}}} \Vert (-A)^{1/2} Du \Vert _{L^p(H, \mu )}, \end{aligned}$$
(25)

for any \(m \ge 1\) and \(p \in (1, +\infty )\). This can be used to deduce that (3) or (4) follow from (11) and (12).

3 Proof of Theorem 6

Let \(\delta \in (0,+\infty )\). Then, put \(S_m = S_m (\delta )\) with

$$\begin{aligned} S_m (\delta ) = \!\!\! \ \sup _{\begin{array}{c} f \in C^{ 2}_b ({{\mathbb {R}}}^m) \\ \Vert f\Vert _{\infty } \le 1 \end{array}} \sum _{k=1}^m \Big (\int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \int _{{{\mathbb {R}}}^m} f (c_1 (t) x_1, \ldots , c_m(t)x_m)\, x_k \, \frac{e^{- \frac{|x|^2}{2}}}{{\sqrt{(2 \pi )^m} } } \textrm{d}x \textrm{d}t \Big )^2\ . \end{aligned}$$

If we show that

$$\begin{aligned} \sup _{m \ge 2} S_m = \infty \end{aligned}$$
(26)

holds under the assumption that \(\lambda _k \sim k^2\) as \(k \rightarrow +\infty \), then the validity of Theorem 6 will follow.

3.1 Two useful lemmas

The following identity will be important. Recall that \(x_k = \langle x, e_k \rangle \), \(k =1, \ldots , m\) where \((e_j)\) denotes the canonical basis in \({{\mathbb {R}}}^m.\)

Lemma 9

For any \(m \ge 2\), \(k \in \{1, \dots , m\}\), \( c = (c_1, \dots , c_m) \in {{\mathbb {R}}^m}\setminus \{0\}\) and \(F \in B_b({{\mathbb {R}}})\), it holds

$$\begin{aligned} I_{m,k}(F)&= \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} F(\langle c , x \rangle ) x_k \, e^{- \frac{|x|^2}{2}} dx \nonumber \\&= \dfrac{2\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \frac{c_k}{| c|}\ \int _0^{+\infty }\!\!\!\int _0^{\pi } e^{-\frac{1}{2}\rho ^2} \rho ^m \cos \vartheta (\sin \vartheta )^{m-2} F(| c| \rho \cos \vartheta ) \textrm{d}\rho \textrm{d}\vartheta . \end{aligned}$$
(27)

Proof

We provide additional details for the sake of completeness. Let us first consider \(m=2\). We introduce the unitary vectors \(\gamma _1 = c/|c|\) and \(\gamma _2 \in {{\mathbb {R}}}^2\) such that \((\gamma _1, \gamma _2)\) is an orthonormal basis in \({{\mathbb {R}}}^2\). Using the polar coordinates with respect to such basis we can write

$$\begin{aligned} x = \rho \cos \theta \, \gamma _1 + \rho \sin \theta \, \gamma _2, \end{aligned}$$

which entails that

$$\begin{aligned} I_{2,k}(F)&= \frac{1}{{2 \pi } } \int _0^{2\pi } \int _0^{+\infty } \rho ^2 F(|c| \rho \cos \theta ) \, ( \cos \theta \, \langle \gamma _1, e_k \rangle + \sin \theta \, \langle \gamma _2 , e_k \rangle ) \, e^{-\frac{\rho ^2}{2}} \textrm{d} \rho \textrm{d} \theta \\&= \frac{1}{{ \pi } } \int _0^{\pi } \int _0^{+\infty } \rho ^2 F(|c| \rho \cos \theta ) \, \cos \theta \, \langle \gamma _1 , e_k \rangle \, e^{-\frac{\rho ^2}{2}} \textrm{d} \rho \textrm{d} \theta ,\;\;\; k =1,2 \end{aligned}$$

since \(\int _0^{2\pi } F(|c| \rho \cos \theta )\, \sin \theta \, \textrm{d} \theta =0\). Indeed, to prove this last identity under the sole assumption that \(F \in B_b({{\mathbb {R}}})\), we just notice that

$$\begin{aligned} \int _0^{2\pi } F(|c| \rho \cos \theta )\, \sin \theta \, \textrm{d} \theta&= \int _0^{\pi } F(|c| \rho \cos \theta )\, \sin \theta \, \textrm{d} \theta + \int _{\pi }^{2\pi } F(|c| \rho \cos \theta )\, \sin \theta \, \textrm{d} \theta \\&= \int _{-1}^{1} F(|c| \rho t)\, \textrm{d} t - \int _{-1}^{1} F(|c| \rho t)\, d t = 0 \end{aligned}$$

holds as a consequence of the change of variable \(\cos \theta = t\). Finally, we get easily (27) for \(m=2\) upon recalling that \(\Gamma (1/2) = \sqrt{\pi }\).

In the general case of \(m \ge 3\), we consider an orthonormal basis \((\gamma _k)\) of \({{\mathbb {R}}}^m\) where \(\gamma _1 = c/|c|\). Then, we introduce polar coordinates with respect to \((\gamma _k)\). Let \(\rho = |x|\). Proceeding similarly to [12, Sect. 5.9], we have, for \(x \not = 0\),

$$\begin{aligned} x = \rho \cos \theta _1 \gamma _1 + \rho \sin \theta _1\cos \theta _2 \gamma _2 + \ldots + \rho \sin \theta _1 \cdots \sin \theta _{m-2} \sin \theta _{m-1} \gamma _m, \end{aligned}$$

where \(\rho >0\) (radial distance), \(\theta _1 , \ldots , \theta _{m-2} \in [0, \pi ]\) (latitudes; \(\theta _1\) is the angle between x and \(\gamma _1\)) and \(\theta _{m-1} \in [0, 2 \pi ]\) (longitude). Let \(\theta = (\theta _1, \ldots , \theta _{m-1})\). Denote by

$$\begin{aligned} J(\rho , \theta ) = \rho ^{m-1} (\sin \theta _1)^{m-2} (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) \end{aligned}$$

the Jacobian determinant. Moreover, set \(\gamma _i^{(k)} =\langle \gamma _i, e_k \rangle \), for \(i,k =1, \ldots , m\). Let

$$\begin{aligned} \xi _1 (\theta )= & {} \cos \theta _1 , \;\; \xi _2 (\theta ) = \sin \theta _1\cos \theta _2, \; \ldots ,\\ \xi _{m-1} (\theta )= & {} \sin \theta _1 \cdots \sin \theta _{m-2} \cos \theta _{m-1},\;\;\; \xi _{m} (\theta ) = \sin \theta _1 \cdots \sin \theta _{m-2} \sin \theta _{m-1}. \end{aligned}$$

We infer that

$$\begin{aligned} I_{m,k}(F)&= \frac{1}{\sqrt{(2 \pi )^m} } \int _0^{\infty } \int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} \rho e^{-\frac{\rho ^2}{2}} F(|c|\, \rho \cos \theta _1) \, \left( \sum _{i=1}^{m} \xi _i(\theta ) \gamma _i^{(k)} \right) J(\rho , \theta ) \textrm{d} \rho \textrm{d} \theta \nonumber \\&= \frac{1}{\sqrt{(2 \pi )^m} } \int _0^{\infty } \int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} \rho e^{-\frac{\rho ^2}{2}} F(|c|\, \rho \cos \theta _1) \, \xi _1(\theta ) \gamma _1^{(k)} \, J(\rho , \theta ) \textrm{d} \rho \textrm{d} \theta \end{aligned}$$
(28)

using that

$$\begin{aligned} \frac{1}{\sqrt{(2 \pi )^m} } \int _0^{\infty } \int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} \rho e^{-\frac{\rho ^2}{2}} F(|c|\, \rho \cos \theta _1) \, \left( \sum _{i=2}^{m} \xi _i(\theta ) \gamma _i^{(k)} \right) J(\rho , \theta ) \textrm{d} \rho \textrm{d} \theta =0. \end{aligned}$$
(29)

In order to prove (29) we check that if \(\rho >0\) then

$$\begin{aligned} \int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} F(|c|\, \rho \cos \theta _1) \, \xi _i(\theta ) J(\rho , \theta ) \textrm{d} \theta =0,\;\;\; 2 \le i \le m. \end{aligned}$$
(30)

If \(i = m\), we find that

$$\begin{aligned}&\int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} \, F(|c|\, \rho \cos \theta _1)\, \xi _m(\theta )\, (\sin \theta _1)^{m-2} (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) \textrm{d}\theta \\&\quad = \int _0^{\pi } F(|c|\, \rho \cos \theta _1) (\sin \theta _1)^{m-2} \sin \theta _1 \textrm{d} \theta _1 \times \\&\qquad \times \int _{[0, \pi ]^{m-3} \times [0, 2\pi ]} \, \sin \theta _2 \cdots \sin \theta _{m-1}\, (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) \textrm{d}\theta _2 \cdots \textrm{d} \theta _{m-1} =0 \end{aligned}$$

by the Fubini theorem, since \(\int _0^{2 \pi } \sin \theta _{m-1} \textrm{d} \theta _{m-1} =0.\) Similarly we obtain that (30) holds with \(i = m-1\). Note that up to now we have already proved (30) when \(m=3\). Let \(m \ge 4\). We check (30) when \(2 < i \le m-2\). We have

$$\begin{aligned}&\int _{[0, \pi ]^{m-2} \times [0, 2\pi ]} \, F(|c|\, \rho \cos \theta _1)\, \xi _i(\theta )\, (\sin \theta _1)^{m-2} (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) \textrm{d}\theta \\&\quad = \int _0^{\pi } F(|c|\, \rho \cos \theta _1) (\sin \theta _1)^{m-2} \sin \theta _1 \textrm{d} \theta _1 \times \\&\qquad \times \int _{[0, \pi ]^{m-3} \times [0, 2\pi ]} \, \sin \theta _2 \cdots \cos \theta _{i}\, (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) \textrm{d}\theta _2 \cdots \textrm{d} \theta _{m-1} =0, \end{aligned}$$

because \(\int _0^{ \pi } \cos \theta _{i} \, (\sin \theta _{i})^{m-1 -i} \textrm{d} \theta _{i} =0.\) Similarly, for \(i=2\), we get

$$\begin{aligned}&\int _0^{\pi } F(|c|\, \rho \cos \theta _1) (\sin \theta _1)^{m-2} \sin \theta _1 \textrm{d} \theta _1 \times \\&\qquad \times \int _{[0, \pi ]^{m-3} \times [0, 2\pi ]}\, \cos \theta _{2}\, (\sin \theta _2)^{m-3} \cdots (\sin \theta _{m-2}) d\theta _2 \cdots \textrm{d} \theta _{m-1} =0. \end{aligned}$$

We have verified (30) and so (28) holds. We rewrite (28) as follow

$$\begin{aligned} I_{m,k}(F)&= R_m \, \frac{ \gamma _1^{(k)} }{\sqrt{(2 \pi )^m} } \int _0^{\infty } \int _{0}^{\pi } \rho ^m e^{-\frac{\rho ^2}{2}} F(|c|\, \rho \cos \theta _1) \,\cos \theta _1\, (\sin \theta _1)^{m-2} \textrm{d} \rho \textrm{d} \theta _1 \nonumber \\ \gamma _1^{(k)}&= \frac{c_k}{|c|}, \end{aligned}$$
(31)

where \(R_m = 2 \pi \) if \(m=3\) and if \(m>3\)

$$\begin{aligned} R_m&= \int _{[0, \pi ]^{m-3} \times [0, 2\pi ]} \, (\sin \theta _2)^{m-3} (\sin \theta _3)^{m-4} \cdots \sin \theta _{m-2} d\theta _2 \cdots \textrm{d} \theta _{m-1}\\&= 2 \pi \prod _{j=1}^{m-3} \int _0^{\pi } (\sin \phi )^j \textrm{d} \phi = 2 \pi \prod _{j=1}^{m-3} B\left( \frac{j+1}{2}, \frac{1}{2} \right) = 2 \pi \prod _{j=1}^{m-3} \frac{ \Gamma \left( \frac{j+1}{2}\right) \Gamma (\frac{1}{2})}{\Gamma \left( \frac{j+2}{2} \right) }. \end{aligned}$$

We have used the Beta function \(B(\cdot , \cdot )\) (cf. page 103 of [28]). Hence since \(\Gamma (1/2) = \sqrt{\pi }\), we get \(R_m = 2 \pi (\sqrt{\pi })^{m-3} \Big ( {\Gamma \big ( \frac{m-1}{2} \big )} \Big )^{-1}\). Inserting \(R_m\) in (31) we obtain (27), i.e.,

$$\begin{aligned} I_{m,k} (F)= \dfrac{2\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \frac{c_k}{| c|}\ \int _0^{+\infty }\!\!\!\int _0^{\pi } e^{-\frac{1}{2}\rho ^2} \rho ^m \cos \vartheta (\sin \vartheta )^{m-2} F(| c| \rho \cos \vartheta ) \textrm{d}\rho \textrm{d}\vartheta . \end{aligned}$$

\(\square \)

Lemma 10

If \(F \in B_b({{\mathbb {R}}})\) verifies \(F(x) = - F(-x)\) for any \(x \in {{\mathbb {R}}}\), then we have, for any \(m \ge 2\), \(k \in \{1, \dots , m\}\), \( c = (c_1, \dots , c_m) \in {{\mathbb {R}}^m}\setminus \{0\}\),

$$\begin{aligned} I_{m,k} (F) = \dfrac{ 4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \frac{c_k}{| c|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m \textrm{d}\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} F(|c| \rho x) \textrm{d}x \end{aligned}$$
(32)

(cf. (27)). In the special case of \(F= F_0 := \mathbb {1}_{(0, \infty )} - \mathbb {1}_{(- \infty , 0)}\), we obtain

$$\begin{aligned} I_{m,k}(F_0) = \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} F_0(\langle c, x\rangle ) x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x = \frac{\sqrt{2}}{\sqrt{\pi }} \frac{c_k}{| c|}. \end{aligned}$$
(33)

Proof

By changing variable \(x= \cos \theta \) and using that \(F(x) = - F(-x)\), \(x \not =0\), we have

$$\begin{aligned} \int _0^{\pi } \cos \vartheta (\sin \vartheta )^{m-2} F(|c| \rho \cos \vartheta )\textrm{d}\vartheta = 2 \int _{0}^1 x (1-x^2)^{\frac{m-3}{2}} F(|c| \rho x) \textrm{d}x\ . \end{aligned}$$

Whence,

$$\begin{aligned} I_{m,k} (F) = \dfrac{2 \cdot 2\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \frac{c_k}{| c|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m d\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} F(|c| \rho x) \textrm{d}x. \end{aligned}$$

Let us assume that \(F =F_0 =\mathbb {1}_{(0, \infty )} - \mathbb {1}_{(- \infty , 0)} \). We find

$$\begin{aligned} I_{m,k} (F_0) = \dfrac{4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \frac{c_k}{| c|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m d\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} \textrm{d}x. \end{aligned}$$

Using that \(\int _0^1 x (1-x^2)^{\frac{m-3}{2}} \textrm{d}x = \frac{1}{m-1}\) and \(\int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m \textrm{d}\rho = \Gamma \left( \frac{m+1}{2}\right) 2^{\frac{m-1}{2}}\) we find

$$\begin{aligned} I_{m,k}(F_0)&= \dfrac{4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \Gamma \left( \frac{m+1}{2}\right) 2^{\frac{m-1}{2}} \, \frac{1}{m-1} \, \frac{c_k}{| c|} = \frac{\sqrt{2}}{\sqrt{\pi }} \frac{c_k}{| c|}, \end{aligned}$$

since \(x \Gamma (x) = \Gamma (x+1)\), \(x>0\) and this finishes the proof. \(\square \)

3.2 Proof of assertion (26)

Recall that \(c_k(t) := \left( \dfrac{1 - e^{-2\lambda _k t}}{2\lambda _k}\right) ^{1/2}\) for \(k \in \{1, \dots , m\}\) and \(t \ge 0\). Set \(c(t) = (c_1(t), \dots , c_m(t)) \in {{\mathbb {R}}}^m\). Fix \(m \ge 2\) and \(\delta >0\), and put \( S_m := S_m(\delta )\). Then, for \(m \ge 2\), define

$$\begin{aligned} A_m := \frac{{2}}{{\pi }} \sum _{k=1}^m \left( \int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}}\, \frac{c_k(t)}{|c(t)|} \, \textrm{d}t\right) ^2. \end{aligned}$$
(34)

We prove that \(\lim _{m \rightarrow \infty } S_m = \infty \) in two steps.

I step. We prove

$$\begin{aligned} S_m \ge A_m,\;\;\; \forall \ m \ge 2. \end{aligned}$$
(35)

We start by constructing an approximating sequence of smooth functions for \(F_0 := \mathbb {1}_{(0, \infty )} - \mathbb {1}_{(- \infty , 0)}\). For any \(n \ge 1\), consider a non-decreasing \(F_n \in C^{ 2}_b ({{\mathbb {R}}}_+) \) such that \(F_n (y) =0\) if \(0 \le y \le 1/(n+1)\) and \(F_n (y) = 1\) if \(y \ge 1/n\). Then, extend each \(F_n\) to an odd function on \({{\mathbb {R}}}\) by the rule \(F_n(x) = - F_n(-x)\) if \(x<0\), and define

$$\begin{aligned} f_n (x_1, \ldots , x_m) = F_n (x_1 + \ldots + x_m), \;\;\; x_1, \ldots , x_m \in {{\mathbb {R}}}. \end{aligned}$$

It is clear that each \(f_n \in C^{ 2}_b({{\mathbb {R}}}^m)\) and \(\Vert f_n\Vert _{\infty } \le 1\). Whence,

$$\begin{aligned} S_m&\ge \sup _{n \ge 1}\sum _{k=1}^m \Big (\int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} f_n (c_1 (t) x_1, \ldots , c_m(t)x_m)\, x_k \, e^{-\frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \Big )^2 \\&= \sup _{n \ge 1} \sum _{k=1}^m \Big (\int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} F_n(\langle c(t) , x \rangle ) x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \Big )^2. \end{aligned}$$

Moreover, combining the fact that each \(F_n\) is an odd functions with (32), with c replaced by c(t), yields

$$\begin{aligned}&\sup _{n \ge 1} \sum _{k=1}^m \Big (\int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} F_n(\langle c(t) , x \rangle ) x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \Big )^2 \\&\quad = \sup _{n \ge 1} \, \dfrac{ 4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \sum _{k=1}^m \Big ( \frac{c_k(t)}{| c(t)|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m d\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} F_n(|c(t)| \rho x) dx \Big )^2. \end{aligned}$$

Then, using that both \(F_n(x) \le F_{n+1}(x)\) and \(F_n(x) \rightarrow F_0(x)\) hold for any \(x \ge 0\), apply the monotone convergence theorem to get

$$\begin{aligned} S_m&\ge \sup _{n \ge 1} \dfrac{ 4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \sum _{k=1}^m \Big ( \frac{c_k(t)}{| c(t)|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m d\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} F_n(|c(t)| \rho x) \textrm{d}x \Big )^2\\&=\dfrac{ 4\pi (\sqrt{\pi })^{m-3}}{(2\pi )^{m/2} \Gamma \left( \frac{m-1}{2}\right) } \sum _{k=1}^m \Big ( \frac{c_k(t)}{| c(t)|}\ \int _0^{+\infty }\!\!\! e^{-\frac{1}{2}\rho ^2} \rho ^m d\rho \int _0^1 x (1-x^2)^{\frac{m-3}{2}} F_0(|c(t)| \rho x) \textrm{d}x \Big )^2\\&= \sum _{k=1}^m \Big (\int _0^{\delta } \dfrac{\lambda _k e^{-\lambda _k t}}{(1 - e^{-2\lambda _k t})^{1/2}} \frac{1}{\sqrt{(2 \pi )^m} } \int _{{{\mathbb {R}}}^m} F_0(\langle c(t), x \rangle ) x_k \, e^{- \frac{|x|^2}{2}} \textrm{d}x \textrm{d}t \Big )^2 = A_m, \end{aligned}$$

for \(m \ge 2\). In the last line we have used both (32) and (33) with c replaced by c(t). This proves (35).

II step. We prove that

$$\begin{aligned} \lim _{m \rightarrow \infty } A_m = \infty \ , \end{aligned}$$
(36)

thus completing the proof of (26). Recalling the definition of \(c_k(t)\), we have

$$\begin{aligned} A_m = \frac{{2}}{{\pi }} \sum _{k=1}^m \Big ( \int _0^{\delta } \dfrac{\sqrt{\lambda _k} e^{-\lambda _k t}}{\sqrt{2}}\, \frac{1}{|c(t)|}\, \textrm{d}t\Big )^2 \ge \frac{{1}}{{\pi }}\sum _{k=1}^m \lambda _k \left( \int _0^{\delta } \dfrac{ e^{-\lambda _k t}}{|c(t)|}\, \textrm{d}t \right) ^2,\;\; m \ge 2. \end{aligned}$$
(37)

To bound (37) from below, note that

$$\begin{aligned} |c(t)| = \left( \sum _{k=1}^{m} \dfrac{1 - e^{-2\lambda _k t}}{2\lambda _k}\right) ^{1/2} \le \left( \sum _{k=1}^{+\infty } \dfrac{1 - e^{-2\lambda _k t}}{2\lambda _k}\right) ^{1/2} = \left( \int _0^t \left[ \sum _{k=1}^{+\infty } e^{-2\lambda _k s}\right] \textrm{d}s\right) ^{1/2} \end{aligned}$$

holds for any \(t \ge 0\). Now, if there is a positive constant \(c_0\) such that \(\lambda _k \ge c_0 k^2\) for any \(k \ge 1\), then

$$\begin{aligned} \sum _{k=1}^{+\infty } e^{-2\lambda _k s} \le \sum _{k=1}^{+\infty } e^{-2{c_0} k^2 s} \le \int _0^{+\infty } e^{-2{c_0} z^2 s} \textrm{d}z = \sqrt{\frac{\pi }{2{c_0}s}},\;\; s>0, \end{aligned}$$

yielding

$$\begin{aligned} |c(t)|\le \left( \int _0^t \sqrt{\frac{\pi }{2{c_0}s}} \textrm{d}s\right) ^{1/2} = \left( \frac{2\pi t}{{c_0}}\right) ^{1/4}. \end{aligned}$$

Up to now we have found that

$$\begin{aligned} A_m \ge \frac{{1}}{{\pi }} \sum _{k=1}^m \lambda _k \left( \int _0^{\delta } { e^{-\lambda _k t}} \left( \frac{{c_0}}{2\pi t}\right) ^{1/4} \, \textrm{d}t \right) ^2,\;\; m \ge 2. \end{aligned}$$

Now, exploit that

$$\begin{aligned} \int _0^{\delta } t^{-\frac{1}{4}} e^{-\lambda t}\textrm{d}t = \left( \frac{1}{\lambda }\right) ^{\frac{3}{4}} \int _0^{\lambda \delta } s^{-\frac{1}{4}} e^{- s} ds \ge \left( \frac{1}{\lambda }\right) ^{\frac{3}{4}} \int _0^{{c_0}\delta } s^{-\frac{1}{4}} e^{-s}\textrm{d}s \end{aligned}$$

holds for every \(\lambda \ge {c_0}\), to get (after recalling that, in particular, \(\lambda _k \ge c_0\), for any \(k \ge 1\))

$$\begin{aligned} A_m&\ge \frac{1}{\pi } \sqrt{\frac{{c_0}}{2\pi }}\, \sum _{k=1}^m \lambda _k \left( \int _0^{\delta } t^{-\frac{1}{4}} e^{-\lambda _k t}\textrm{d}t\right) ^2\ge \frac{1}{\pi } \sqrt{\frac{{c_0}}{2\pi }}\sum _{k=1}^m \lambda _k \left( \frac{1}{\lambda _k}\right) ^{\frac{3}{2}} \left( \int _0^{{c_0}\delta } s^{-\frac{1}{4}} e^{-s}\textrm{d}s\right) ^2 \\&= \frac{1}{\pi } \sqrt{\frac{{c_0}}{2\pi }} \left( \int _0^{{c_0}\delta } s^{-\frac{1}{4}} e^{-s}\textrm{d}s\right) ^2 \sum _{k=1}^m \frac{1}{\sqrt{\lambda _k}}\ . \end{aligned}$$

Thus, if \(\lambda _k \sim k^2\) as \(k \rightarrow +\infty \), then \(\sum _{k=1}^m \frac{1}{\sqrt{\lambda _k}} \sim \log m\) as \(m \rightarrow +\infty \), and (36) holds. This finishes the proof.