From: Joubert on
Suppose f is continuous and vanishes at infinity. Suppose you know that
|(g_k)f - f| --> 0 in the SUP norm for k going to infinity (g_k are
infinitely differentiable compactly supported functions with support on
the ball of radius K centered in the origin and such that g_k(x)=1 if x
belongs to the the previously mentioned ball, 0 otherwise.) What can we
say about the convergence of
|x^a||(g_k)f - f)| for a > 1 (k going to infinity) ??

The book claims that from the convergence in SUP norm of |(g_k)f - f|
one can deduce uniform convergence on compacta and then POINTWISE
convergence of |x^a||(g_k)f - f)|. My question is: why can't I just say
that |x^a||(g_k)f - f)| converges uniformly to zero as k goes to
infinity? What is it that makes uniform convergence fail for
|x^a||(g_k)f - f)|?
From: José Carlos Santos on
On 13-08-2010 14:59, Joubert wrote:

> Suppose f is continuous and vanishes at infinity. Suppose you know that
> |(g_k)f - f| --> 0 in the SUP norm for k going to infinity (g_k are
> infinitely differentiable compactly supported functions with support on
> the ball of radius K

Did you mean _k_ here?

> centered in the origin and such that g_k(x)=1 if x
> belongs to the the previously mentioned ball, 0 otherwise.)

This makes no sense. Such a function cannot be even continuous.

> What can we say about the convergence of
> |x^a||(g_k)f - f)| for a > 1 (k going to infinity) ??
>
> The book

Which book?

> claims that from the convergence in SUP norm of |(g_k)f - f|
> one can deduce uniform convergence on compacta and then POINTWISE
> convergence of |x^a||(g_k)f - f)|. My question is: why can't I just say
> that |x^a||(g_k)f - f)| converges uniformly to zero as k goes to
> infinity?

Because it is not true. Take, for instance, f(x) = x/(1 + x^2).

Best regards,

Jose Carlos Santos