Re: f and f' square integrable

On Thu, 04 Jan 2007 18:19:06 -0800, The World Wide Wade
<waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:

In article <2qRmh.292564$FQ1.248893@attbi_s71>,
Stephen Montgomery-Smith <stephen@xxxxxxxxxxxxxxxxx> wrote:

The World Wide Wade wrote:
In article <Ub%lh.198574$aJ.146919@attbi_s21>,
Stephen Montgomery-Smith <stephen@xxxxxxxxxxxxxxxxx> wrote:


The World Wide Wade 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.

You smoothly chop the tops off the peaks, but not "uniformly". E.g.
sum a_n f*delta_n(x-(3n,0))
where delta_n is a smooth, compactly supported, approximate identity
converging very quickly.


Seems like a little bit of a mess to verify the derivatives work
out as desired.

I don't see why. If delta_n is compactly supported enough, then the
f*delta_n(x-(3n,0)) have disjoint supports. Furthermore, their
derivatives are f'*delta_n(x-(3n,0)). Since convolution by an L_1
function is a bounded operator on L_2, it seems rather easy to me.

Let g be a test function, D a partial derivative, and f your
function above. How do you know D(f*g) = (Df)*g? (I can prove
it, but it's an ad hoc argument. Perhaps I'm missing something
easy.)

Well, we need to know that Df is in fact the distribution
derivative. If we believe that then this seems clear:

By definition, if f is a distribution and phi is a test
function then

<Df, phi> = -<f, D phi>.

That leads immediately to the identity

(Df)*phi = f*(D phi).

Now say f is as above, g is a test function, and h is yet
another test function. Then

(D(f*g))*h = (f*g)*Dh = f*(g*Dh) = f*D(g*h)
= (Df)*(g*h) = ((Df)*g)*h.

Hence D(f*g) = (Df)*g.

That seems pretty straighforward, also perfectly standard.


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

David C. Ullrich
.

Relevant Pages

  • Re: All Distributions (Schwarz) are Integrable.
    ... We consider the case where we know of a Phi with Phi'=phi, and let f be a distribution. ... For what test functions phi does there exist a test function ... Let psi be a fixed test function with int_R psi = 1. ...
    (sci.math)
  • Re: Distribution of an infinite order
    ... exists a compact K s.t. | ... Since the only test function with ... And in case, there is no such an N, we say that u is of infinite order. ... Think about functions all of whose first N derivatives are small ...
    (sci.math)
  • Re: -- Question about distributions
    ... test function phi (i.e. a smooth real or complex-valued function on | ... Since F has compact support there exists N and a compact set K ... where that last norm is the sup over K of the partials of phi of order ...
    (sci.math)
  • Re: Distribution of an infinite order
    ... infinite order defined on some subset of reals. ... for every test function f with compact support in K. ... distribution u defined by ... I would be grateful for you suggestion of a rigorous proof. ...
    (sci.math)
  • Re: Distribution of an infinite order
    ... infinite order defined on some subset of reals. ... for every test function f with compact support in K. ... Since the only test function with support in K is 0, ... distribution would have order N. Instead, I think you want your condition ...
    (sci.math)
Quantcast