Re: f and f' square integrable

On Sun, 31 Dec 2006 15:06:34 GMT, Stephen Montgomery-Smith
<stephen@xxxxxxxxxxxxxxxxx> wrote:

Fedor wrote:
Stephen Montgomery-Smith a écrit :


Fedor wrote:

Hi all,

suppose that f:R^n -->R is a smooth function such that f^2 and ( f '
)^2 are integrable over R^n. Is it true that f(x) tends towards 0 when
|x| tends towards infinity ? It is easy for n=1 but is it true for the
general case ?

Regards,
fedor


I think not.

First consider a function like f(x)=log(1/|x|)^a for 0<a<1/2. Check
that f and f' are square integrable in R^2.

Next consider sum a_n f(x-(2n,0)) where a_n is square summable.


oh Thank you ! I must had thought to this classical function that shows
that functions in the Sobolev space H^1 are not necessary continuous. I
was asking this question because I was looking for a simple proof of
the trace theorem for Sobolev space. I guess there are no really simple
proof for this theorem .. (that theorem that says that one can define
the trace of a function f \in H^1(Omega) if the boundary of Omega is
regular)


Yes, my motivation was to remember the Sobolev embedding theorems, which
in 2D would only guarantee that f is in VMO (the "continuous" version of
BMO, but I cannot remember what the "V" stands for).

Vanishing.

This is where I
started looking.


************************

David C. Ullrich
.