A counterexample to \(L^{\infty }\)-gradient type estimates for Ornstein–Uhlenbeck operators

Abstract

Let \((\lambda _k)\) be a strictly increasing sequence of positive numbers such that \({\sum _{k=1}^{\infty } \frac{1}{\lambda _k} < \infty }\). Let f be a bounded smooth function and denote by \(u= u^f\) the bounded classical solution to

$$\begin{aligned} u(x) - \frac{1}{2}\sum _{k=1}^m D^2_{kk} u(x) + \sum _{k =1}^m \lambda _k x_k D_k u(x) = f(x),\quad x \in {{\mathbb {R}}}^m . \end{aligned}$$

It is known that the following dimension-free estimate holds:

$$\begin{aligned} \displaystyle \int _{{{\mathbb {R}}}^m}\! \left[ \sum _{k=1}^m \lambda _k \, (D_k u (y))^2 \right] ^{p/2} \!\! \!\!\!\! \mu _m (\textrm{d}y) \le (c_p)^p \!\! \int _{{{\mathbb {R}}}^m} \!\! |f( y)|^p \mu _m (\textrm{d}y),\;\;\; 1< p < \infty \end{aligned}$$

where \(\mu _m\) is the “diagonal” Gaussian measure determined by \(\lambda _1, \ldots , \lambda _m\) and \(c_p > 0\) is independent of f and m. This is a consequence of generalized Meyer’s inequalities [4]. We show that, if \(\lambda _k \sim k^2\), then such estimate does not hold when \(p= \infty \). Indeed we prove

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

This is in contrast to the case of \(\lambda _k = \lambda >0\), \(k \ge 1\), where a dimension-free bound holds for \(p =\infty \).