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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
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