Re: f and f' square integrable
- From: The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 31 Dec 2006 14:32:45 -0800
In article <NDHlh.1114208$084.1066831@attbi_s22>,
Stephen Montgomery-Smith <stephen@xxxxxxxxxxxxxxxxx> wrote:
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.
If f(x) = [log(1/|x|)]^a for x near 0, then f is not smooth at 0.
Perhaps you can play around with this to arrive at an example,
although the obvious things don't seem to work.
I don't have an example for R^2, but if n > 2 you can do this:
Take any g in C^oo with support near 0 and g(0) nonzero. For b >
0, set g_b(x) = g(bx). Then int |g_b|^2 = (int |g|^2)/b^n and int
|D(g_b)|^2 = (int |D(g)|^2)/b^(n-2). (All integrals are over R^n
and D is any partial derivative of f.) Letting u denote any unit
vector you like, define f(x) = sum (j=1,oo) g_(2^j)(x - ju). Then
f is C^oo, and f and any Df belong to L^2. But f(ju) = g(0) for
all j, so f does not tend to 0 at infinity.
.
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