| Popis: |
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 $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), $ $ x \in \R^m$. It is known that the following dimension-free estimate holds: $$ \displaystyle \int_{\R^m} \Big (\sum_{k=1}^m \lambda_k \, (D_k u (y))^2 \Big)^{p/2} \mu_m (dy) \le (c_p)^p \, \int_{\R^m} |f( y)|^p \mu_m (dy),\;\;\; 1 < p < \infty; $$ here $\mu_m$ 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 [Chojnowska-Michalik, Goldys, J. Funct. Anal. 182 (2001)]. We show that, if $\lambda_k \sim k^2$, then such estimate does not hold when $p= \infty$. Indeed we prove $$ \sup_{\substack{f \in C^{ 2}_b(\R^m),\;\; \|f\|_{\infty} \leq 1}} \Big \{ \sum_{k=1}^m \lambda_k \, (D_k u^f (0))^2 \Big \} \to \infty \;\; \text {as} \; m \to \infty. $$ This is in contrast to the case of $\lambda_k = \lambda >0$, $k \ge 1$, where a dimension-free bound holds for $p =\infty$. |
