Re: f and f' square integrable

David C. Ullrich wrote:
On Sat, 06 Jan 2007 18:50:23 GMT, Stephen Montgomery-Smith
<stephen@xxxxxxxxxxxxxxxxx> wrote:


David C. Ullrich wrote:

On Sun, 31 Dec 2006 14:32:45 -0800, The World Wide Wade
<waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:



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.


Huh. WIthout thinking too hard about it, I assumed that this
construction gave a counterexample for R^2. I guess it doesn't.

You can do something similar in R^2. First a one-variable fact:

Suppose that a > 0 and psi(t) = phi(t^a) for t > 0. Then
a change of variables shows that

int_0^infinity t |psi'(t)|^2 dt

= a int_0^infinity t |phi'(t)|^2 dt

(you may want to check that...).

For x in R^2 and a > 0 let x^a be the vector
pointing in the same direction with |x^a|
= |x|^a:

x^a = |x|^(a-1) x.

Now suppose that g is a smooth function in R^2,
g(0) <> 0, and g has support in the unit ball.
Also assume that g is constant in some neighborhood of the origin. And assume that
g is radial. For a > 0 let

g_a(x) = g(x^a).

Then ||g_a||_2 -> 0 as a -> infinity, and
integrating in polar coordinates and using
the one-variable fact shows that the L^2
norm of the gradient of g_a also tends to 0
as a -> 0.

So a sum of translates of suitably chosen
g_a's gives a counterexample.

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

David C. Ullrich

Hi David,

I just looked over your example. It really is rather beautiful.


Just noticed a typo - of course the first "as a -> infinity"
should be "as a -> 0".

I like it because it's "soft" - doesn't rely on the specific
norm of a specific function, works just because something
tends to zero, the rate at which it tends to zero being
irrelevant.

Also it's seemed to me for some years that a sum of translates
of dilates of one function "should" give an example here if
one existed - the last time the question came up I noticed
myself that that didn't work and ended up giving an example
based on the logarithm. Nice to see that the idea _does_
work, with a sort of quasi-dilation in place of an ordinary
dilation. Only remaining problem is that I really don't feel
I understand _why_ it works - that one-variable fact is just
a meaningless calculation, as far as my current understanding
goes.

I think this paper is relevant:

Krbec, M.; Schmeisser, H.-J.
Limiting imbeddings. The case of missing derivatives.
Ricerche Mat. 45 (1996), no. 2, 423--447.
.

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