You can show that for large n we have for some
1. since df(x)/dx =0 for large x, therefore, df(x)/dx=-a/x^p, a>0, p>0
2. integral above so that: f(x)=bx^(1-p)+c<1, b=-a/(1-p), 0<c<1,
3. as 0<f(x)<1, we must have: (1-p)<0, or p>1
4. Therefore: x*f(x)*df(x) < x*1*(-ax^(-p))=-a*x^(1-p)
5. when p>1, or 1-p<0, -a*x^(1-p)=0
I am not comfortable with the proof:
For step 1, how do i know: df(x)/dx=-ax^(-p)? why not df(x)/dx=exp(-ax)? or any other function, and its compund functions (there are infinite of such functions), that equal 0 as x approaches infinite
For step2, can I integrate the df(x) as this is approximate only
Step,3,4,5 are ok to me, but not sure they are really reasonably rigorous
Thank you very much!
Maybe this is a better one...
1. Suppose Lim ABS(x*df)>a, a>0.
2. Since df<0 (which means always -df>0 ), x>0, we have: lim df<-a/x. Integral to have: f(x)<-a*ln(x)+c, or Lim(f(x)=negatiive infinite.
3. However 0<f(x)<1, the contradiction shows: Lim ABS(x*df)=0, in turn, Lim(x*df*f)=0
I see your problem, if ....then seems to fit all requirements... (for a>1)
And (using L'hopital)
But what I think they mean...is that is sufficiently close to all functions for large . (that is all f' that fit the requirements...)
That is, for all we can find such that for .
So that the same conclusion holds for all f'...