uniform convergence - on compacta
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From: Joubert on 13 Aug 2010 09:59 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 13 Aug 2010 10:23
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 |