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)