stupid question from integrate?



FWIW,
  The general fact is that if $f \in L_p, g \in L_q$ with
  $1/p + 1/q = 1$, then $f \times g \in L_1$.  This is the Holder
  inequality.

I doubt that anything better can be said without many technicalities.

-cheers
  -sen

On Tue, 12 Dec 2006, miguel lopez wrote:

> Daniel Lakeland <dlakelan <at> street-artists.org> writes:
>
>>
>> On Tue, Dec 12, 2006 at 08:43:29AM -0800, Daniel Lakeland wrote:
>>>
>>> Too early to do mathematics  However now you've got me going.
>>>
>>> integrate(abs(f(x) * g(x)), 0, inf) can be broken into the following
>>> infinite sum
>>>
>>> sum(integrate(abs(g(x)*f(x)),n*p,(n+1)*p),n,0,inf)
>>>
>>> I think some kind of integration by parts might be helpful here.
>>>
>>> eventually hopefully bounding the integral by k * integrate(f(x),x,0,inf)
>>>
>>> of course if k = 0 then things wouldn't work.
>>>
>>> Don't have time to investigate further at the moment, perhaps at lunch...
>>>
>>> I'm glad that some of you are entertained by this
>>
>> I see that if it's ever to work, it's also going to require that not
>> only k > 0 but also that abs(g(x) * f(x)) > 0 over some measurable
>> interval within each of the sub intervals so that the two functions
>> don't kill each other by careful trickery of the type being sent to me
>> off list...
>>
>
>   This is ultramaxima 60 seconds time:
>
>   Your theorem looks false.
> The idea of a counterexample is f(x) a periodic pulse very narrow (you are
> sampling g(x)) f(x) = 1 in [-e,e], 0 in the rest.  Then integrate(f(x)g(x)) is
> about 1/2e sum(g(x_i,1,inf) so this depends of the peeks g(x_i).   So if your
> sample take small values of g(x) this converges and if your sample takes the
> peeks its diverge.  I think you can fill the missing details.
>
>  Time to work and walk
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