Let (lambda(k)) be a strictly increasing sequence of positive numbers such thatSigma(infinity)(k=1) 1/lambda(k) < infinity. Let f be a bounded smooth function and denote by u = u(f) the bounded classical solution tou(x) - 1/2 Sigma(m)(k=1) D(kk)(2)u(x) + Sigma(m)(k=1) lambda(k)x(k)D(k)u (x) = f (x), x is an element of R-m.It is known that the following dimension-free estimate holds:integral(Rm) [Sigma(m)(k=1) lambda(k) (D(k)u(y))(2)](p/2) mu m(dy) <= (cp)(p) integral(Rm) |f (y)| p(mu m) (dy), 1 < p < infinitywhere mu(m) is the "diagonal" Gaussian measure determined by lambda(1), . . . , 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) similar to k(2), then such estimate does not hold when p = infinity. Indeed we provesup (f is an element of Cb2 (Rm), ||f ||infinity <= 1){Sigma(m)(k=1) lambda(k) (D(k)uf(0))2} -> infinity as m -> infinity.This is in contrast to the case of lambda(k) = lambda > 0, k = 1, where a dimension-free bound holds for p = infinity.
A counterexample to L∞-gradient type estimates for Ornstein-Uhlenbeck operators
Dolera, E
;Priola, E
2024-01-01
Abstract
Let (lambda(k)) be a strictly increasing sequence of positive numbers such thatSigma(infinity)(k=1) 1/lambda(k) < infinity. Let f be a bounded smooth function and denote by u = u(f) the bounded classical solution tou(x) - 1/2 Sigma(m)(k=1) D(kk)(2)u(x) + Sigma(m)(k=1) lambda(k)x(k)D(k)u (x) = f (x), x is an element of R-m.It is known that the following dimension-free estimate holds:integral(Rm) [Sigma(m)(k=1) lambda(k) (D(k)u(y))(2)](p/2) mu m(dy) <= (cp)(p) integral(Rm) |f (y)| p(mu m) (dy), 1 < p < infinitywhere mu(m) is the "diagonal" Gaussian measure determined by lambda(1), . . . , 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) similar to k(2), then such estimate does not hold when p = infinity. Indeed we provesup (f is an element of Cb2 (Rm), ||f ||infinity <= 1){Sigma(m)(k=1) lambda(k) (D(k)uf(0))2} -> infinity as m -> infinity.This is in contrast to the case of lambda(k) = lambda > 0, k = 1, where a dimension-free bound holds for p = infinity.I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.