BV functions in Hilbert spaces

Abstract

We study the basic theory of BV functions in a Hilbert space X endowed with a (not necessarily Gaussian) probability measure $\nu$. We present necessary and sufficient conditions in order that a function $u\in L^p(X, \nu )$ is of bounded variation. We also discuss the De Giorgi approach to BV functions through the behavior as $t\rightarrow 0$ of $\int _X \Vert \nabla T(t)u\Vert \,d\nu$, for a smoothing semigroup T(t). Particular attention is devoted to the case where u is the indicator function of a sublevel set $\{x:\; g(x)<r\}$ of a real Borel function g. We give several examples, for different measures $\nu$ such as weighted Gaussian measures, infinite products of non Gaussian measures, and invariant measures of some stochastic PDEs such as reaction-diffusion equations and Burgers equation.

Appendices

Appendix A: An alternative proof of Theorem 3.3

We give an alternative proof of the implication $V_z(u)<+\infty \Longrightarrow u\in BV_z(X, \nu )$, the other implication being a simple consequence of the respective definitions. It consists of two steps. In the first step we consider the case where u has bounded support, and in the second step we consider the general case.

Proof

First step: the case $u\equiv 0$ in $X{\setminus } B(0,r)$.

We recall the notation used in Sect. 3: for $\varphi \in C^1_b(X)$ we set $T_{u,z}\varphi = \int _X u\,\partial _z^* \varphi \, d\nu$, where $\partial _z^* \varphi = \langle R\nabla \varphi , z\rangle - v_z\varphi$. We claim that

$$\begin{aligned} |T_{u,z}\varphi | \le V_z(u)\Vert \varphi \Vert _{L^{\infty }(B(0,r))}, \quad \varphi \in C^1_b(X). \end{aligned}$$

(A.1)

Indeed, for any $\varepsilon >0$ let $\theta _{\varepsilon } \in C^1(\mathbb R)$ be such that

$$\begin{aligned} \Vert \theta _{\varepsilon }\Vert _{\infty } \le 1, \quad \theta _{\varepsilon } \equiv 1\; \mathrm{in} \;[0, r^2], \quad \theta _{\varepsilon } \equiv 0\; \mathrm{in} \;[(r+\varepsilon )^2, +\infty ), \end{aligned}$$

and we set

$$\begin{aligned} \eta _{\varepsilon } (x) := \theta _{\varepsilon } (\Vert x\Vert ^2), \quad x\in X. \end{aligned}$$

For any $\varphi \in C^1_b(X)$, since $u\equiv 0$ in $X{\setminus } B(0, r)$ and $\varphi \equiv \varphi \eta _{\varepsilon }$ in B(0, r), we have

$$\begin{aligned} T_{u,z}\varphi = \int _X u\, \partial _z^* \varphi \,d\nu = \int _{B(0,r)} u\, \partial _z^*( \varphi \eta _{\varepsilon } )\,d\nu = \int _{X} u\, \partial _z^*( \varphi \eta _{\varepsilon } )\,d\nu = T_{u,z}(\varphi \eta _{\varepsilon } ). \end{aligned}$$

Therefore,

$$\begin{aligned} |T_{u,z}\varphi | = |T_{u,z}(\varphi \eta _{\varepsilon } )| \le V_z(u) \Vert \varphi \eta _{\varepsilon } \Vert _{\infty } \le V_z(u) \Vert \varphi \Vert _{L^{\infty }(B(0,r+ \varepsilon ))}, \quad \varepsilon >0. \end{aligned}$$

Since $\varphi$ is uniformly continuous, $\lim _{\varepsilon \rightarrow 0} \Vert \varphi \Vert _{L^{\infty }(B(0,r+ \varepsilon ))} = \Vert \varphi \Vert _{L^{\infty }(B(0,r))}$, and (A.1) follows.

Now we endow B(0, r) with the weak topology. Since B(0, r) is closed, convex and bounded, it is weakly compact. We denote by $C_w( B(0,r))$ the space of the weakly continuous functions (namely, continuous with respect to the weak topology) from B(0, r) to $\mathbb R$.

The elements of $\mathcal {FC}^1_b(X)$ are weakly continuous functions. Their restrictions to B(0, r) constitute an algebra $\mathcal A$ that separates points and contains the constants. By the Stone-Weierstrass Theorem, $\mathcal A$ is dense in $C_w( B(0,r))$.

Every $f\in \mathcal A$ is the restriction to B(0, r) of a function $\widetilde{f} \in \mathcal {FC}^1_b(X)$. We set

$$\begin{aligned}\widetilde{T}_{u,z}f := T_{u,z}\widetilde{f} . \end{aligned}$$

$\widetilde{T}_{u,z}$ is well defined: if f is the restriction to B(0, r) of both $\widetilde{f}_1$ and $\widetilde{f}_2$, we have $T_{u,z}\widetilde{f}_1= T_{u,z}\widetilde{f}_2$ by estimate (A.1). Still (A.1) implies $|\widetilde{T}_{u,z}f | \le V_z(u)\Vert f\Vert _{L^{\infty }(B(0,r))}$ for every $f\in \mathcal A$. So, $\widetilde{T}_{u,z}$ is bounded on $\mathcal A$, and therefore it has a linear bounded extension (still called $\widetilde{T}_{u,z}$) to the whole $C_w( B(0,r))$. By the Riesz Theorem, such extension has an integral representation: there exists a (unique) real valued measure $\mu$ on the Borel sets of B(0, r) with respect to the weak topology, such that

$$\begin{aligned} |\mu |(B(0,r)) = \Vert \widetilde{T}_{u,z}\Vert = V_z(u), \end{aligned}$$

and

$$\begin{aligned} \widetilde{T}_{u,z}\varphi = - \int _{B(0,r)} \varphi \,d\mu , \quad \varphi \in C_w( B(0,r)). \end{aligned}$$

In particular,

$$\begin{aligned} T_{u,z}\varphi = - \int _{B(0,r)} \varphi \,d\mu , \quad \varphi \in \mathcal {FC}^1_b(X). \end{aligned}$$

We recall that the Borel subsets of B(0, r) with respect to the weak topology coincide with the Borel subsets of B(0, r) with respect to the strong topology. So, $\mu$ is a Borel measure on B(0, r), that we extend in an obvious way to all Borel sets of X, setting

$$\begin{aligned} m_z(B) := \mu (B\cap B(0,r)), \quad B\in \mathcal B(X). \end{aligned}$$

Therefore we have

$$\begin{aligned} T_{u,z}\varphi = - \int _{X} \varphi \,dm_z , \quad \varphi \in \mathcal {FC}^1_b(X). \end{aligned}$$

(A.2)

To finish Step 1 we have to extend the validity of formula (A.2) to all $\varphi \in C^1_b(X)$. For $\varphi \in C^1_b(X)$ we consider the cylindrical approximations $\varphi _n(x) := \varphi (P_nx)$. For every $n\in \mathbb N$, (A.2) yields

$$\begin{aligned} T_{u,z}\varphi _n = - \int _{X} \varphi _n \,dm_z . \end{aligned}$$

(A.3)

We have $\lim _{n\rightarrow \infty } \varphi _n(x) = \varphi (x)$ for every $x\in X$, and $|\varphi _n(x) |\le \Vert \varphi \Vert _{\infty }$. Therefore, the right-hand side of (A.3) goes to $- \int _{X} \varphi \,dm_z$ as $n\rightarrow \infty$. Moreover, by Lemma 2.1(i) we have $\lim _{n\rightarrow \infty } \partial \varphi _n/\partial (R^*z)(x)$ $=$ $\partial \varphi / \partial (R^*z)(x)$ for every $x\in X$ and $|\partial \varphi _n/\partial (R^*z)(x)| \le \Vert \nabla \varphi \Vert _{\infty } /\Vert R^*z\Vert$. It follows $|\partial ^*_z \varphi _n(x) |\le \Vert \nabla \varphi \Vert _{\infty } /\Vert R^*z\Vert + |v_z(x)| \,\Vert \varphi \Vert _{\infty }$. By the Dominated Convergence Theorem, the left-hand side of (A.3) goes to $T_{u,z}\varphi$ as $n\rightarrow \infty$. So, letting $n\rightarrow \infty$ in (A.3) yields $T_{u,z}\varphi = - \int _{X} \varphi \,d\mu _z$, and recalling (3.9) the statement is proved.

Before going to Step 2, we prove an intuitive property of the measure $m_z$ of Step 1. We show that if $u\equiv 0$ on $B(0,r_0)$ for some $r_0<r$, then , where is the open ball centered at 0 with radius r. To this aim we recall that

Fixed any $\varepsilon >0$ let $\varphi \in C_b(X)$ be such that $\Vert \varphi \Vert _{\infty } \le 1$, $\varphi (x)= 0$ for $\Vert x\Vert \ge r_0$ and

By Lemma 2.1(iii) there exists a double-indexes sequence of $C^1_b(X)$ functions $(\widetilde{\varphi }_{k,n})$ such that $|\widetilde{\varphi }_{k,n}(x)\Vert \le \Vert \varphi \Vert _{\infty }$ for every $x\in X$, $\lim _{k\rightarrow \infty } \lim _{n\rightarrow \infty } \widetilde{\varphi }_{k,n}(x) = \varphi (x)$ for every x, and $\widetilde{\varphi }_{k,n}(x) = 0$ if $\Vert x\Vert \ge r_0$. By the Dominated Convergence Theorem we get

$$\begin{aligned} \lim _{k\rightarrow \infty } \lim _{n\rightarrow \infty } \int _X \widetilde{\varphi }_{k,n}\,dm_z = \int _X \varphi \,dm_z \end{aligned}$$

so that there are $k(\varepsilon )$, $n(\varepsilon )\in \mathbb N$ such that

Since $\widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )}$ belongs to $C^1_b(X)$ we have

$$\begin{aligned} \int _X \widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )} \,dm_z = T_{u,z}( \widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )}) = \int _X u\, \partial _z^*( \widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )} )d\nu \end{aligned}$$

and the last integral is zero, because u vanishes $\nu$-a.e in $B(0,r_0)$ and $\widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )}$ vanishes in $X{\setminus } B(0,r_0)$, so that $\partial _z^*( \widetilde{\varphi }_{k(\varepsilon ),n(\varepsilon )} )$ vanishes $\nu$-a.e. in $X{\setminus } B(0,r_0)$. Therefore for every $\varepsilon >0$, which implies .

Second step: the general case.

Let $\theta \in C^1(\mathbb R)$ be an even function supported in $(-1,1)$, such that $\theta \equiv 1$ in a neighborhood of 0, and such that $\theta (\xi ) = 1-\theta (1-\xi )$ for every $\xi \in (0,1)$. For instance, one can take the even extension of the function that is equal to 1 in [0, 1/4], to $[\cos (2\pi (\xi -1/4)) +1]/2$ for $\xi \in [1/4, 3/4]$, to 0 for $\xi \ge 3/4$.

The set of functions $\{\theta _k:\;k\in \mathbb Z\}$ defined by $\theta _k(\xi ) := \theta (\xi -k)$, $k\in \mathbb Z$, is a $C^1$ partition of 1 in $\mathbb R$. Each $\theta _k$ is supported in $(k-1, k+1)$, and therefore every $\xi \in \mathbb R$ belongs to the support of at most two of them. Define

$$\begin{aligned} \eta _k(x) := \theta _k(\Vert x\Vert ), \quad k\in \mathbb N\cup \{0\}. \end{aligned}$$

Every $\eta _k$ belongs to $C^1_b(X)$, since every $\theta _k$ belongs to $C^1_b(\mathbb R)$ and it is constant in a neighborhood of 0. Moreover

$$\begin{aligned} \sum _{k=0}^{\infty } \eta _k(x) =1, \quad x\in X, \end{aligned}$$

where for every $x\in X$ the series converges because it has at most two nonzero addenda; more precisely, if $k\le \Vert x\Vert \le k+1$ we have $\eta _h(x)=0$ for $h\le k-1$ and for $h\ge k+2$. So, we have

$$\begin{aligned} u(x) = \sum _{k=0}^{\infty } \eta _k (x) u(x) =: \sum _{k=0}^{\infty } u_k(x), \quad x\in X, \end{aligned}$$

where $u_k := u\eta _k$ vanishes outside $\{x: \; k-1< x< k+1\}$. In particular, its support is contained in $B(0, k+1)$ and for every $\varphi \in C^1_b(X)$ we have

$$\begin{aligned} \displaystyle \bigg | \int _X u\,\eta _k \partial ^*_z \varphi \,d\nu \bigg |= & {} \bigg | \int _X u (\partial ^*_z (\varphi \eta _k) -\varphi \partial _{R^*z}\eta _k)\,d\nu \bigg | \\ \\ \displaystyle\le & {} V_z(u)\Vert \varphi \eta _k\Vert _{\infty } + \bigg | \int _X u \,\varphi \,\partial _{R^*z}\eta _k\,d\nu \bigg | \le V_z(u) \Vert \varphi \Vert _{\infty } + C_k\Vert \varphi \Vert _{\infty } \end{aligned}$$

where

$$\begin{aligned} C_k = \int _{\{x: \; k-1< \Vert x\Vert<k+1\}} |u\, \partial _{R^*z}\eta _k| d\nu \le \int _{\{x: \; k-1< \Vert x\Vert < k+1\}}|u | d\nu \Vert \theta '\Vert _{\infty }. \end{aligned}$$

By Step 1 there exists a Borel measure $m_k$, supported in $\{x: \; k-1\le x\le k+1\}$, whose total variation does not exceed $V_{z}(u) + C_k$, such that

$$\begin{aligned} T_{u_k,z}\varphi = \int _X u\,\eta _k \partial ^*_z \varphi \,d\nu =- \int _{\{x: \; k-1< \Vert x\Vert < k+1\}}\varphi \,dm_k, \quad \varphi \in C^1_b(X). \end{aligned}$$

(A.4)

The consequent estimate $|m_k|(X) \le V_z(u) + C_k$ is not enough for our aim. To improve it we recall (3.9), that yields

$$\begin{aligned} |m_k|(X) = V_z(\eta _ku) = \sup \bigg \{ \int _X u\,\eta _k\, \partial ^*_z \varphi \,d\nu : \; \; \varphi \in C^1_b(X), \; \Vert \varphi \Vert _{\infty } \le 1 \bigg \}. \end{aligned}$$

For every $k\in \mathbb N\cap \{0\}$ let $\varphi _k \in C^1_b(X)$ be such that $\Vert \varphi _k\Vert _{\infty } \le 1$ and such that

$$\begin{aligned} |m_k| (X) \le \int _X u\,\eta _k\, \partial _z^* \varphi _k\,d\nu + \frac{1}{2^k}. \end{aligned}$$

(A.5)

Set

$$\begin{aligned} \varphi (x) := \sum _{k=1}^{\infty } \eta _k(x) \varphi _k(x), \quad x\in X. \end{aligned}$$

The function $\varphi$ is well defined and it belongs to $C^1_b(X)$, since every x has a neighborhood where all the summands vanish, except for at most two of them. Moreover,

$$\begin{aligned} |\varphi (x) | \le \sum _{k=1}^{\infty } \eta _k(x) |\varphi _k(x)| \le \sum _{k=1}^{\infty } \eta _k(x) = 1, \quad x\in X, \end{aligned}$$

so that $\Vert \varphi \Vert _{\infty } \le 1$. Then we have

$$\begin{aligned} \int _X u\,\partial ^*_z\varphi \,d\nu \le V_z (u). \end{aligned}$$

(A.6)

On the other hand, the left hand side is equal to $\int _X u\, \partial ^*_z (\sum _{k=1}^{\infty } \eta _k \varphi _k)\,d\nu$, and recalling that $\partial ^*_z( \eta _k \varphi _k) = \eta _k \partial ^*_z \varphi _k + \varphi _k\partial _{R^*z}\eta _k$ we get

$$\begin{aligned} \int _X u\,\partial ^*_z\varphi \,d\nu = \sum _{k=1}^{\infty }\int _X u\, \eta _k \,\partial ^*_z\varphi _k \, d\nu + \sum _{k=1}^{\infty }\int _Xu \, \varphi _k \, \partial _{R^*z} \eta _k \, d\nu , \end{aligned}$$

where for every k we have

$$\begin{aligned} \bigg | \int _Xu\, \varphi _k \,\partial _{R^*z} \eta _k \, d\nu \bigg | \le \int _{\{x:\, k-1<\Vert x\Vert<k\} } |u \, \partial _{R^*z} \eta _k| d\nu \le \int _{\{x:\, k-1<\Vert x\Vert <k\} } |u| d\nu \Vert \theta ' \Vert _{\infty } , \end{aligned}$$

so that, recalling (A.5) and summing up,

$$\begin{aligned} \displaystyle \int _X u\,\partial ^*_z\varphi \,d\nu\ge & {} \displaystyle \sum _{k=1}^{\infty }\left( |m_k|(X) - \frac{1}{2^k} \right) - \sum _{k=1}^{\infty } \Vert u\Vert _{L^1 (B(0, k+1){\setminus } B(0,k-1))} \Vert \theta ' \Vert _{\infty } \\ \\\ge & {} \displaystyle \sum _{k=1}^{\infty } |m_k|(X) - 1 -2\Vert u\Vert _{L^1 (X, \nu )} \Vert \theta ' \Vert _{\infty } \end{aligned}$$

that implies, through (A.6),

$$\begin{aligned} \sum _{k=1}^{\infty } |m_k| (X)\le V_z(u) +1 + 2\Vert u\Vert _{L^1 (X, \nu )} \Vert \theta ' \Vert _{\infty } . \end{aligned}$$

The signed measure

$$\begin{aligned} m_z:= \sum _{k=1}^{\infty }m_k \end{aligned}$$

is therefore well defined, and for every $\varphi \in C^1_b(X)$ we have, using (A.4),

$$\begin{aligned} \int _X \varphi \, dm_z = \sum _{k=1}^{\infty } \int _X \varphi \, dm_k = \sum _{k=1}^{\infty } \int _X u\,\eta _k \,\partial _z^* \varphi \, d\nu = \int _X u \, \partial _z^* \varphi \,d\nu = T_{u,z}\varphi . \end{aligned}$$

So, $m_z$ is the measure that we were looking for. We already know that $|m_z|(X) = V_z(u)$, and the proof is complete. $\square$

Appendix B: Proof of the commutation formula (5.14)

Proposition B.1

Let Hypothesis 5.1 hold, and assume in addition that $U\in C^1(X)$ and that $\nabla U$ is Lipschitz continuous. Then (5.14) holds.

Proof

The formal derivation of (5.14) is easy: setting $v(t,x):=(T(t)f)(x)$, v satisfies $\partial v/\partial t = Kv$. Taking the derivative along $e_k$ and using (5.7) we should get

$$\begin{aligned} \frac{\partial }{\partial t} \frac{\partial }{\partial e_k} v = \frac{\partial }{\partial e_k}Kv = K\left( \frac{\partial v}{\partial e_k} \right) - \frac{1}{2\lambda _k} \frac{\partial v}{\partial e_k} - \langle D^2Ue_k, \nabla v\rangle \end{aligned}$$

which yields (5.14). However, we do not know whether $\partial T(t)f/\partial e_k$ is time differentiable and belongs to D(K), even for $f\in \mathcal {FC}^{\infty }_b(X)$. To justify (5.14) we follow a tedious approximation procedure, already used in [20, Sect. 3.2.2] to prove regularity results for the elements of D(K), that allows to use regularity results for PDEs in finite dimension.

Let L be such that

$$\begin{aligned} \Vert \nabla U(x)-\nabla U(y)\Vert \le L\Vert x-y\Vert , \quad x, \;y\in X. \end{aligned}$$

We approach U by $U_n(x) := U(P_nx)$, where $P_n$ is defined by (2.2).

The function

$$\begin{aligned} u_n: \mathbb R^n\mapsto \mathbb R, \quad u_n(\xi ): = U_n\left( \sum _{i=1}^n \xi _ie_i\right) \end{aligned}$$

is such that $U_n(x) = u_n( \langle x, e_1\rangle , \ldots \langle x, e_n\rangle )$, it is convex and it has Lipschitz continuous gradient as well as U. This is not enough for our aims, and we approach it again by

$$\begin{aligned} u^{\varepsilon }_{n}(\xi ) = \int _{\mathbb R^n} u_n(\xi -\varepsilon y)\theta _n(y)dy, \quad \xi \in \mathbb R^n, \end{aligned}$$

where $\theta _n: \mathbb R^n\mapsto \mathbb R$ is any smooth nonnegative compactly supported function with $\int _{\mathbb R^n} \theta _n(y)dy =1$. Then $u^{\varepsilon }_{n}$ is smooth and convex, $\nabla u^{\varepsilon }_{n}$ is Lipschitz continuous with Lipschitz constant $\le L$, so that $\Vert D^2 u^{\varepsilon }_{n}(\xi )\Vert _{{\mathcal L}(\mathbb R^n)} \le L$ for every $n\in \mathbb N$, $\varepsilon >0$, $\xi \in \mathbb R^n$. We introduce the differential operator $K_{n, \varepsilon }$ defined by

$$\begin{aligned} K_{n, \varepsilon }\varphi (\xi ) = {\mathcal L}_n \varphi (\xi ) - \langle \nabla u^{\varepsilon }_{n}(\xi ), \nabla \varphi (\xi )\rangle , \quad \xi \in \mathbb R^n, \end{aligned}$$

where ${\mathcal L}_n$ is the Ornstein–Uhlenbeck operator

$$\begin{aligned} {\mathcal L}_n \varphi (\xi ) = \frac{1}{2} \sum _{i=1}^n \left( \frac{\partial ^2 \varphi }{\partial \xi _i^2}(\xi ) - \lambda _i^{-1}\xi _i \frac{\partial \varphi }{\partial \xi _i} (\xi )\right) , \quad \xi \in \mathbb R^n. \end{aligned}$$

and we consider the Cauchy problem in $\mathbb R^n$,

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial }{\partial t}v_{n, \varepsilon }(t, \xi ) =K_{n, \varepsilon }v_{n, \varepsilon }(t, \xi )\rangle , \quad t>0, \\ \\ v_n(0, \xi ) = v_0(\xi ) \end{array}\right. \end{aligned}$$

(B.1)

where $v_0\in B_b(\mathbb R^n)$ (the space of the Borel bounded functions). Since the drift coefficients are globally Lipschitz, and the dissipativity condition

$$\begin{aligned} - \sum _{i,j=1}^n\left( \delta _{ij} \lambda _i^{-1}+ \frac{\partial ^2u^{\varepsilon }_{n}}{\partial \xi _i\partial \xi _j}(\xi ) \right) \eta _i \eta _j \le 0, \quad \xi , \;\eta \in \mathbb R^n \end{aligned}$$

holds, the regularity results of [14, Ch. 1] and of [32] yield existence of a Markov, strong Feller semigroup $e^{tK_{n, \varepsilon }}$ that maps $B_b(\mathbb R^n)$ into $C^{2+\alpha }_b(\mathbb R^n)$ for every $t>0$, $\alpha \in (0,1)$. Moreover, for $v_0\in C_b(\mathbb R^n)$, $v_n(t, \xi ):= (e^{tK_{n, \varepsilon }} v_0) (\xi )$ is the unique bounded classical solution to (B.1), and since all the coefficients of $K_{n, \varepsilon }$ are smooth, $(t,\xi ) \mapsto (e^{tK_{n, \varepsilon }} v_0)(\xi )\in C^{\infty }((0, +\infty )\times \mathbb R^n)$ by the classical local regularity results in parabolic equations. So, for $t>0$ both sides of the equation in (B.1) are differentiable with respect to the space variables, and

$$\begin{aligned} \frac{\partial }{\partial \xi _k}\frac{\partial v_n}{\partial t} = \frac{\partial }{\partial \xi _k}\bigg ( {\mathcal L}_nv(t, \xi ) - \langle \nabla u^{\varepsilon }_{n}(\xi ), \nabla _{\xi } v_n(t, \xi )\rangle \bigg ) \end{aligned}$$

namely

$$\begin{aligned} \frac{\partial }{\partial t} \frac{\partial }{\partial \xi _k} v_n = \left( {\mathcal L}_n - \frac{1}{2\lambda _k}\right) \frac{\partial }{\partial \xi _k} v_n - \langle \nabla u^{\varepsilon }_{n}(\xi ), \nabla _{\xi }\frac{\partial }{\partial \xi _k}v_n(t, \xi )\rangle - \sum _{i=1}^n \left( \frac{\partial ^2 }{\partial \xi _i\partial \xi _k}u^{\varepsilon }_{n}(\xi ) \right) \frac{\partial }{\partial \xi _i} v_n . \end{aligned}$$

Therefore, if $v_0\in C^1_b(\mathbb R^n)$,

$$\begin{aligned}&\frac{\partial }{\partial \xi _k}(e^{tK_{n, \varepsilon }} v_0) - e^{-t/2\lambda _k}e^{tK_{n, \varepsilon }} \frac{\partial v_0}{\partial \xi _k}\nonumber \\&\quad = - \int _0^t e^{-(t-s)/2\lambda _k}e^{(t-s)tK_{n, \varepsilon }} ( \langle D^2u^{\varepsilon }_{n} \cdot \nabla e^{sK_{n, \varepsilon }} v_0, e_k \rangle )\,ds. \end{aligned}$$

(B.2)

Now we go back from $\mathbb R^n$ to X, setting $f_n(\xi ) = f(\sum _{i=1}^n\xi _i e_i)$ for $f\in B_b(X)$, and

$$\begin{aligned} (T_{n, \varepsilon }(t) f )(x) := (e^{tK_{n, \varepsilon }} f_n)(\langle x, e_1\rangle , \ldots , \langle x, e_n\rangle ). \end{aligned}$$

Then, $T_{n, \varepsilon }(t)$ is a Markov strong Feller semigroup, and it maps $B_b(X)$ into $\mathcal {FC}^2_b(X)\cap C^{\infty }(X)$.

We are going to establish estimates on $T_{n, \varepsilon }(t) f$ that yield estimates on $\nabla T(t)f$ and convergence results. First of all, for $v_0\in C^1_b(\mathbb R^n)$ we have ( [20, Sect. 3.2.2])

$$\begin{aligned} \Vert \nabla e^{tK_{n, \varepsilon }} v_0\Vert _{\infty } \le \Vert \nabla v_0\Vert _{\infty }, \quad t>0, \; v_0\in C^1_b(\mathbb R^n). \end{aligned}$$

(B.3)

Taking into account the dissipativity condition, the procedure of [4, Prop. 2.6] with $\Lambda = \mathbb R^n$, $V\equiv 0$ gives (with the choice $a= 1$)

$$\begin{aligned} \Vert \nabla e^{tK_{n, \varepsilon }} v_0\Vert _{\infty } \le \frac{1}{\sqrt{t}} \Vert v_0\Vert _{\infty }, \quad t>0, \; v_0\in C_b(\mathbb R^n), \end{aligned}$$

so that, by the semigroup law and (B.3),

$$\begin{aligned} \Vert \nabla e^{tK_{n, \varepsilon }} v_0\Vert _{\infty } \le \max \left\{ 1, \frac{1}{\sqrt{t}} \right\} \Vert v_0\Vert _{\infty }, \quad t>0, \; v_0\in B_b(\mathbb R^n), \quad t>0. \end{aligned}$$

(B.4)

By (B.3), for $f\in C^1_b(X)$ we have

$$\begin{aligned} \Vert \nabla T_{n, \varepsilon }(t) f\Vert _{\infty } \le \Vert \nabla f\Vert _{\infty }, \quad t>0, \end{aligned}$$

(B.5)

and by (B.4) for $f\in B_b(X)$ we have

$$\begin{aligned} \Vert \nabla T_{n, \varepsilon }(t) f\Vert _{\infty } \le \max \left\{ 1, \frac{1}{\sqrt{t}} \right\} \Vert f\Vert _{\infty }, \quad t>0. \end{aligned}$$

(B.6)

Moreover, for $k>n$ we have $\frac{\partial }{\partial e_k}T_{n, \varepsilon }(t) f =0$, while for $k\le n$ formula (B.2) yields

$$\begin{aligned}&\frac{\partial }{\partial e_k}T_{n, \varepsilon }(t) f - e^{-t/2\lambda _k}T_{n, \varepsilon }(t) \frac{\partial f}{\partial e_k} \nonumber \\&\quad = - \int _0^t e^{-(t-s)/2\lambda _k}T_{n, \varepsilon }(t-s) ( \langle D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f, e_k \rangle )\,ds \end{aligned}$$

(B.7)

where

$$\begin{aligned} U^{\varepsilon }_{n} (x) := u^{\varepsilon }_{n}(\langle x, e_1\rangle , \ldots , \langle x, e_n\rangle ). \end{aligned}$$

Since $\Vert D^2 u^{\varepsilon }_{n}(\xi )\Vert _{{\mathcal L}(\mathbb R^n)} \le L$ for every $\xi \in \mathbb R^n$,

$$\begin{aligned} \Vert D^2 U^{\varepsilon }_{n}(x)\Vert _{{\mathcal L}(X)} \le L, \quad \varepsilon >0, \; n\in \mathbb N, \; x\in X. \end{aligned}$$

(B.8)

We rewrite (B.7) as

$$\begin{aligned} P_n( \nabla T_{n, \varepsilon }(t) f )(x)= & {} e^{-tA}(\mathbf{T}_{n, \varepsilon }(t)P_n\nabla f (x) ) \nonumber \\&- \int _0^t e^{-(t-s)A} (\mathbf{T}_{n, \varepsilon }(t-s) ( P_n D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f ) )(x)\,ds . \end{aligned}$$

(B.9)

At every $x\in X$, both sides of (B.9) belong to $D(Q^{-1/2})= D(-2A)^{1/2}$. Using the estimate $\Vert (-2A)^{1/2} e^{tA}\Vert _{{\mathcal L}(X)} \le C(1+ t^{-1/2})$ and (B.5), (B.8), for $f\in C^1_b(X)$ we get

$$\begin{aligned}&\Vert Q^{-1/2} e^{-tA}(\mathbf{T}_{n, \varepsilon }(t) P_n \nabla f (x))\Vert _X \le C(1+ t^{-1/2})\Vert \mathbf{T}_{n, \varepsilon }(t)P_n \nabla f (x)\Vert _X \\&\quad \le C(1+ t^{-1/2}) \Vert \nabla f\Vert _{\infty },\\&\quad \quad \displaystyle \left\| Q^{-1/2} \int _0^t e^{-(t-s)A} \mathbf{T}_{n, \varepsilon }(t-s) ( P_n D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f )(x) \,ds \right\| \\&\quad \displaystyle \le \int _0^t C(1+ (t-s)^{-1/2}) \Vert \langle D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f\Vert _{\infty } ds\\&\quad \le \int _0^t C(1+ (t-s)^{-1/2}) L \Vert \nabla f\Vert _{\infty } ds . \end{aligned}$$

Summing up, and recalling that $(I-P_n) ( \nabla T_{n, \varepsilon }(t) f ) \equiv 0$, we obtain

$$\begin{aligned} \Vert Q^{-1/2}\nabla T_{n, \varepsilon }(t) f(x)\Vert \le C_1(t+t^{-1/2}) \Vert \nabla f\Vert _{\infty }, \quad t>0, \;x\in X, \; f\in C^1_b(X), \end{aligned}$$

(B.10)

with $C_1$ independent of n and $\varepsilon$.

Now we let $n\rightarrow \infty$, $\varepsilon \rightarrow 0$. Since, for every $f\in C_b(X)$ and $t>0$, $x\mapsto (T_{n, \varepsilon }(t)f)(x)$ is in $\mathcal {FC}^2_b(X)$ and it depends only on $\langle x, e_1\rangle , \ldots , \langle x, e_n\rangle$, it belongs to D(K) and by (5.7) we have

$$\begin{aligned} \frac{d}{dt} (T(t)f - T_{n, \varepsilon }(t)f) = K(T(t)f - T_{n, \varepsilon }(t)f) - \sum _{i=1}^n \left( \frac{\partial U}{\partial e_i}-\frac{\partial U^{\varepsilon }_{n} }{\partial e_i}\right) \frac{\partial }{\partial e_i} T_{n, \varepsilon }(t)f, \quad t>0 \end{aligned}$$

so that

$$\begin{aligned} T(t)f - T_{n, \varepsilon }(t)f = -\int _0^t T(t-s)\left( \sum _{i=1}^n \left( \frac{\partial U}{\partial e_i}-\frac{\partial U^{\varepsilon }_{n}}{\partial e_i} \right) \frac{\partial }{\partial e_i} T_{n, \varepsilon }(s)f\right) ds, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial }{\partial e_k}( T(t)f - T_{n, \varepsilon }(t)f) = - \frac{\partial }{\partial e_k}\int _0^tT(t-s)\left( \sum _{i=1}^n \left( \frac{\partial U}{\partial e_i}-\frac{\partial U^{\varepsilon }_{n} }{\partial e_i} \right) \frac{\partial }{\partial e_i}T_{n, \varepsilon }(s)f\right) ds, \end{aligned}$$

which we rewrite as

$$\begin{aligned} P_n \nabla ( T(t)f - T_{n, \varepsilon }(t)f) = - P_n \nabla \int _0^t T(t-s)( \langle P_n (\nabla U - \nabla U^{\varepsilon }_{n} ), \nabla T_{n, \varepsilon }(s)f \rangle ) ds. \end{aligned}$$

We recall that T(t) is a contraction semigroup in $L^2(X, \nu )$. Using (B.6) we get

$$\begin{aligned}&\Vert T(t)f - T_{n, \varepsilon }(t)f \Vert _{L^2(X, \nu )} \nonumber \\&\quad \le \displaystyle \int _0^t \Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)}\Vert \nabla T_{n, \varepsilon }(s)f\Vert _{L^2(X, \nu ;X)} ds\nonumber \\&\quad \le \max \{(t, t^{1/2}\} \Vert f\Vert _{\infty } \Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)}. \end{aligned}$$

(B.11)

Using (5.8) and (B.5) we get

$$\begin{aligned}&\displaystyle \Vert P_n \nabla ( T(t)f - T_{n, \varepsilon }(t)f) \Vert _{L^2(X, \nu ;X)} \\&\quad \le \displaystyle \int _0^t \frac{C}{\sqrt{t-s}}\Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)}\Vert \nabla T_{n, \varepsilon }(s)f\Vert _{L^2(X, \nu ;X)} ds \\&\quad \le C \sqrt{t} \Vert \nabla f\Vert _{\infty } \Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)}. \end{aligned}$$

So, we estimate

$$\begin{aligned} \Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)}\le & {} \Vert \nabla U -\nabla U_{n} \Vert _{L^2(X, \nu ;X)} + \Vert \nabla U_n -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)} \\ \\\le & {} \displaystyle \left( \int _X \Vert \nabla U(x) - \nabla U(P_nx)\Vert ^2d\nu \right) ^{1/2} + \Vert \nabla U_n -\nabla U^{\varepsilon }_{n} \Vert _{L^{\infty }(X, \nu ;X)} \\ \\\le & {} \displaystyle \left( \int _X \Vert \nabla U(x) - \nabla U(P_nx)\Vert ^2d\nu \right) ^{1/2} + \varepsilon L . \end{aligned}$$

Since $\nabla U$ is continuous and it has at most linear growth at infinity, the first term vanishes as $n\rightarrow \infty$ by the Dominated Convergence Theorem. Therefore, $\lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} \Vert \nabla U -\nabla U^{\varepsilon }_{n} \Vert _{L^2(X, \nu ;X)} =0$, and (B.11) yields

$$\begin{aligned} \lim _{(n, \varepsilon )\rightarrow (\infty , 0)} \Vert T(t) - T_{n, \varepsilon }(t) \Vert _{{\mathcal L}(C_b(X), L^2(X, \nu ))} = 0, \quad t>0. \end{aligned}$$

(B.12)

Moreover, since

$$\begin{aligned}&\Vert \nabla ( T(t)f - T_{n, \varepsilon }(t)f) \Vert _{L^2(X, \nu ;X)} ^2 = \Vert P_n \nabla ( T(t)f - T_{n, \varepsilon }(t)f) \Vert _{L^2(X, \nu ;X)} ^2 \\&\quad + \Vert (I-P_n) \nabla T(t)f\Vert _{L^2(X, \nu ;X)} ^2, \end{aligned}$$

where $\lim _{n\rightarrow \infty } \Vert (I-P_n) \nabla T(t)f\Vert _{L^2(X, \nu ;X)}=0$, we also have

$$\begin{aligned} \lim _{(n, \varepsilon )\rightarrow (\infty , 0)} \Vert \nabla T(t)f - \nabla T_{n, \varepsilon }(t)f \Vert _{L^2(X, \nu ; X)} =0, \quad t>0, \; f\in C_b(X). \end{aligned}$$

(B.13)

Let us use (B.13) to get bounds for $\nabla T(t)f$, when $f\in C^1_b(X)$. By (B.13), there exists a sequence $(\nabla T_{n_k, \varepsilon _k}(t)f(x))$ with $n_k\rightarrow \infty$, $\varepsilon _k\rightarrow 0$, that converges to $\nabla T(t)f(x)$ for $\nu$-a.e.$x\in X$. For such x’s (B.5) yields $\Vert \nabla T(t)f(x)\Vert \le \Vert \nabla f\Vert _{\infty }$. Still for such x’s, the sequence $(\nabla T_{n_k, \varepsilon _k}(t) f(x))$ is bounded in $D(Q^{-1/2})$ by (B.10), and since $D(Q^{-1/2})$ is a Hilbert space, up to a further subsequence $(\nabla T_{n_k, \varepsilon _k}(t) f(x))$ weakly converges to some $h\in D(Q^{-1/2})$; since $X-\lim _{k\rightarrow \infty } \nabla T_{n_k, \varepsilon _k}(t) f(x) = \nabla T (t)f(x)$, we have $h= \nabla T (t)f(x) \in D(Q^{-1/2})$ and

$$\begin{aligned} \Vert \nabla T (t)f(x)\Vert _{D(Q^{-1/2})} \le \liminf _{k\rightarrow \infty } \Vert \nabla T_{n_k, \varepsilon _k}(t) f(x)\Vert _{D(Q^{-1/2})} \le \Vert \nabla f\Vert _{\infty }+ C_2(t+t^{-1/2}) \Vert \nabla f\Vert _{\infty }. \end{aligned}$$

Therefore, for every $t>0$ and $f\in C^1_b(X)$,

$$\begin{aligned} \left\{ \begin{array}{ll} (i) &{} \Vert \nabla T(t)f\Vert _{\infty } \le \Vert \nabla f\Vert _{\infty } , \\ \\ (ii)&{} \Vert Q^{-1/2}\nabla T(t)f\Vert _{\infty } \le C_3(t + t^{-1/2})\Vert \nabla f\Vert _{\infty } , \end{array}\right. \end{aligned}$$

(B.14)

for some $C_3>0$.

Let us go back to (B.7). Using (B.12) and (B.13), the left hand side of (B.7) converges to the left hand side of (5.14) in $L^2(X, \nu )$ as $n\rightarrow \infty$, $\varepsilon \rightarrow 0$. The difference between the respective right hand sides is split as

$$\begin{aligned}&\displaystyle \int _0^t e^{-(t-s)/2\lambda _k}(T_{n, \varepsilon }(t-s)- T (t-s))( \langle D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f, e_k \rangle )\,ds \\ \\&\quad \quad \displaystyle + \int _0^t e^{-(t-s)/2\lambda _k}T (t-s) ( \langle D^2U^{\varepsilon }_{n} \cdot ( \nabla T_{n, \varepsilon }(s)f -\nabla T(s)f), e_k \rangle )\,ds \\ \\&\quad \quad \displaystyle + \int _0^t e^{-(t-s)/2\lambda _k}T (t-s) ( \langle (D^2U^{\varepsilon }_{n} -D^2U) \cdot \nabla T(s)f, e_k \rangle )\,ds \\ \\&\quad := I^{(1)}_{n, \varepsilon }(t) + I^{(2)}_{n, \varepsilon } (t) + I^{(3)}_{n, \varepsilon } (t). \end{aligned}$$

By (B.12), (B.8), and (B.5), for every $s\in (0, t)$ we have

$$\begin{aligned}\lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} \Vert (T_{n, \varepsilon }(t-s)- T (t-s))( \langle D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f, e_k \rangle )\Vert _{L^2(X, \nu )} =0. \end{aligned}$$

Moreover, for every n and $\varepsilon$ we have, by (B.8) and (B.5), and recalling that $T_{n, \varepsilon }(t)$, T(t) are contraction semigroups in $C_b(X)$,

$$\begin{aligned}\Vert (T_{n, \varepsilon }(t-s)- T (t-s))( \langle D^2U^{\varepsilon }_{n} \cdot \nabla T_{n, \varepsilon }(s)f, e_k \rangle )\Vert _{L^{\infty }(X, \nu )} \le L \Vert \nabla f\Vert _{\infty }. \end{aligned}$$

By the Dominated Convergence Theorem,

$$\begin{aligned} \lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} I^{(1)}_{n, \varepsilon }(t) =0. \end{aligned}$$

Similarly, by (B.8) and (B.13), for every $s\in (0, t)$ we have

$$\begin{aligned} \lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} \Vert T (t-s) ( \langle D^2U^{\varepsilon }_{n} \cdot ( \nabla T_{n, \varepsilon }(s)f -\nabla T(s)f), e_k \rangle )\Vert _{L^2(X, \nu )} =0, \end{aligned}$$

while for every n and $\varepsilon$ we have, recalling that T(t) is a contraction semigroup in $L^2(X, \nu )$ and using (B.8), (B.5) and (B.14)(i),

$$\begin{aligned} \Vert T (t-s) ( \langle D^2U^{\varepsilon }_{n} \cdot ( \nabla T_{n, \varepsilon }(s)f -\nabla T(s)f), e_k \rangle )\Vert _{L^2(X, \nu )} \le 2L \Vert \nabla f\Vert _{\infty }, \end{aligned}$$

and again by the Dominated Convergence Theorem,

$$\begin{aligned} \lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} I^{(2)}_{n, \varepsilon } (t)=0. \end{aligned}$$

Concerning $I^{(3)}_{n, \varepsilon }(t)$, we prove that it converges weakly to 0 in $L^2(X, \nu )$. By (B.8), $(U^{\varepsilon }_{n} )$ is bounded in $W^{2,2}(X, \gamma )$, and therefore in $W^{2,2}(X, \nu )$, by a constant independent of n and $\varepsilon$. A sequence $(U^{\varepsilon _k}_{n_k} )$ converges weakly in $W^{2,2}(X, \nu )$, and since $U^{\varepsilon }_{n} \rightarrow U$ as $(n, \varepsilon )\rightarrow (\infty , 0)$, the weak limit is U and we have $U^{\varepsilon }_{n} \rightharpoonup U$ in $W^{2,2}(X, \nu )$, as $(n, \varepsilon )\rightarrow (\infty , 0)$. For every $\psi \in L^2(X, \nu )$, the functional

$$\begin{aligned} \varphi \mapsto F(\varphi ) := \int _X \int _0^t e^{-(t-s)/2\lambda _k}T (t-s) \left( \sum _{j=1}^{\infty } \frac{\partial ^2\varphi }{\partial e_k \partial e_j} \frac{\partial T(t)f}{\partial e_j } \right) \,ds \,\psi \,d\nu \end{aligned}$$

belongs to the dual space of $W^{2,2}(X, \nu )$, since

$$\begin{aligned}\begin{array}{lll} |F(\varphi )| &{} \le &{} \displaystyle \int _0^t \Vert \sum _{j=1}^{\infty } \frac{\partial ^2\varphi }{\partial e_k \partial e_j} \frac{\partial T(s)f}{\partial e_j }\Vert _{L^2(X, \nu )} ds \Vert \psi \Vert _{L^2(X, \nu )} \\ \\ &{} \le &{} \displaystyle \int _0^t \left( \int _X \sum _{j=1}^{\infty } \left( \frac{\partial ^2\varphi }{\partial e_k \partial e_j}\right) ^2 \lambda _j\,d\nu \right) ^{1/2} \Vert Q^{-1/2}\nabla T(s)\Vert _{L^2(X, \nu ;X)} ds \, \Vert \psi \Vert _{L^2(X, \nu )} \\ \\ &{} \le &{} \displaystyle \int _0^t C_3(s+ s^{-1/2})ds\, \Vert \nabla f\Vert _{\infty } \Vert \varphi \Vert _{W^{2,2}(X, \nu )} \Vert \psi \Vert _{L^2(X, \nu )} \end{array} \end{aligned}$$

where we used estimate (B.14)(ii). Therefore, $\lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} F(U^{\varepsilon }_{n}) - F(U) =0$, namely for every $\psi \in L^2(X, \nu )$ we have $\lim _{(n, \varepsilon ) \rightarrow (\infty , 0)} \int _X I^{(3)}_{n, \varepsilon }(t)\, \psi \, d\nu =0$. So, $I^{(3)}_{n, \varepsilon }(t)$ weakly converges to 0 as $(n, \varepsilon ) \rightarrow (\infty , 0)$. Summing up, the right hand side of (B.7) weakly converges to the left hand side of (5.14) in $L^2(X, \nu )$ as $(n, \varepsilon ) \rightarrow (\infty , 0)$, and (5.14) follows. $\square$