Re: f and f' square integrable

In article <s2uip2h6uf3c3705hrv56uakasrqgh8lod@xxxxxxx>,
David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> 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.


Didn't quite get your notation, but with h a suitable function on
[0,oo), we can define g_a(x) on R^2 by g_a(x) = h(|x|^a) for a >
0 and then consider a sum of translates of these g_a's with the
a's small enough, right? Nice.
.

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