proof: xf(x) -> 0

Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as x->infinity. It should be trivial but I don't see it.

karl

2007-10-07 14:03:00 UTC

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as x->infinity. It should be trivial but I don't see it.

Try to work it out.

mathemaat

2007-10-07 14:16:51 UTC

I have been trying for a very long time, but I don't get to the solution. Obviously f(x) should go faster to zero than x goes to infinity else it wouldn't be integrable. Still, I don't get there. L'hospital also doesn't work since f(x) is arbitrary

Ronald Bruck

2007-10-07 14:08:05 UTC

In article

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as x->infinity. It
should be trivial but I don't see it.

Probably because it's false. You don't even have f(x) --> 0.

--
Ron Bruck

mathemaat

2007-10-07 15:02:58 UTC

I also know that f(x) is non-increasing and f(x)>=0

Jules

2007-10-07 16:49:07 UTC

Post by mathemaat
I also know that f(x) is non-increasing and f(x)>=0

If f(x) * x > e for some x, then what can you say about the integral
of f from x/2 to x? Using this, what happens if there is some e for
which f(x) * x > e for arbitrarily large values of x?

James Burns

2007-10-08 23:21:55 UTC

Ronald Bruck wrote:
:> In article
:> <***@nitrogen.mathforum.org>,
:> mathemaat <***@hotmail.com> wrote:
:>
:>>Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as
:>>x->infinity. It should be trivial but I don't see it.
:>
:> Probably because it's false. You don't even have f(x) --> 0.

Post by Jules

Post by mathemaat
I also know that f(x) is non-increasing and f(x)>=0

If f(x) * x > e for some x, then what can you say about the integral
of f from x/2 to x? Using this, what happens if there is some e for
which f(x) * x > e for arbitrarily large values of x?

This question turned out to be more difficult than I expected.
(1) Did I get it right?
(2) Is there an easier way?

-----------
To prove: Given that INT_0^infinity (f(x)dx) is finite,
f(x) is non-increasing and f(x) >= 0, show that
x*f(x) -> 0 as x -> infinity.

------------
(Let INT_0^infinity (f(x)dx) = M.)

If x*f(x) -> 0 as x -> infinity, then (by definition)
for all eps > 0, exists x' such that for all x > x',
abs( x*f(x) ) < eps.

Assume that that is false:
There exist eps > 0 such that for all x' there exists
x > x', such that x*f(x) > eps.

Select such an eps.
It follows that there exists a sequence ( x_j ),
such that x_j*f(x_j) > eps for all j and
x_j -> infinity as j->infinity. (This last condition
will be contradicted later.)

[By the way, if I could use eps/x as a lower bound for f(x),
this proof would be much shorter and clearer. But what
we really have is f(x) > eps/x_j for all x < x_j,
where x_j is any element of the series above.]

We will use the series (x_j) to define g(x) as a lower
bound for f(x):
g(x) = eps/x_j, for x_{j-1} < x < x_j

Let us calculate:
INT_0^infy (f(x)dx) = M

Post by Jules
= INT_0^infy (g(x)dx)

= SUM_{j=1}^infy ( eps/x_j*( x_j - x_{j-1} ) )
= eps* SUM_{j=1}^infy ( a_j )
where we define a_j = 1/x_j*( x_j - x_{j-1} )

Note that
x_j = 1/(1-a_j)*x_{j-1}
and thus
x_j = x_0*PRODUCT_{k=1}^j ( 1/(1-a_k))
= x_0* exp( SUM_{k=1}^j (-log( 1-a_k)) )
=< x_0* exp( SUM_{k=1}^j ( a_k ) )
since -log(1-x) =< x. But we have seen above that
SUM_{k=1}^j ( a_k )
=< SUM_{k=1}^infy ( a_k )
=< (1/eps)*INT_0^infy (f(x)dx)
= M/eps
Therefore
x_j =< x_0* exp( SUM_{k=1}^j ( a_k ) )
=< x_0* exp( M/eps )
and x_j does /not/ -> infy as j -> infy.

But we assumed that x_j -> infy, which we could do
if x*f(x) does /not/ -> 0 as x -> infy. Contradiction.

Therefore x*f(x) -> 0 as x -> infy.

----------------

Jim Burns

David C. Ullrich

2007-10-08 12:03:40 UTC

Post by mathemaat
I also know that f(x) is non-increasing and f(x)>=0

Hint:

int_0^x f(t) dt >= x f(x).

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

David C. Ullrich

Vladimir Dergachev

2007-10-07 15:30:02 UTC

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as x->infinity.
It should be trivial but I don't see it.

This should depend strongly on the definition of integral you use - is it
Riemann, Lebesgue or something else ?

Consider the following example:

let f(x)=sum gn(x)

where gn(x) is a step function of the following shape:

gn(x)= n when |x-n|<1/(n*2^n) and 0 otherwise.

Note that f(x) and all gn(x) are non-negative.

It would thus be very reasonable to define

Int f(x) = sum Int(gn(x)) = sum 1/2^n = 1

For this to make sense for a large class of functions one would have to
extend Riemann or Lebesgue definitions somehow, or create a brand new
integral (say for positive functions only).

For example, you can start with Lebesgue measure on R^2 and for a positive
function f(x) define int(f(x)) as the measure of the area enclosed between
0 and f(x). One can easily check that usual properties of the integral are
preserved.

best

Vladimir Dergachev

f***@yahoo.com

2007-10-08 06:56:14 UTC

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as x->infinity.
It should be trivial but I don't see it.

This should depend strongly on the definition of integral you use - is it
Riemann, Lebesgue or something else ?
let f(x)=sum gn(x)
gn(x)= n when |x-n|<1/(n*2^n) and 0 otherwise.
Note that f(x) and all gn(x) are non-negative.
It would thus be very reasonable to define
Int f(x) = sum Int(gn(x)) = sum 1/2^n = 1

Not only would it be reasonable, this *is* the value of the integral
when you are using either the Lebesgue or improper Riemann integral.
For the Lebesgue integral use the monotone convergence theorem; for
the improper integral evaluate Int(f(x),x=0..t) explicitly and let t -

Post by Vladimir Dergachev
infinity.
For this to make sense for a large class of functions one would have to
extend Riemann or Lebesgue definitions somehow, or create a brand new
integral (say for positive functions only).

How so?

Post by Vladimir Dergachev
For example, you can start with Lebesgue measure on R^2 and for a positive
function f(x) define int(f(x)) as the measure of the area enclosed between
0 and f(x). One can easily check that usual properties of the integral are
preserved.

In fact, you recover the Lebesgue integral again, do you not?

Vladimir Dergachev

2007-10-09 01:57:11 UTC

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as
x->infinity. It should be trivial but I don't see it.

Post by Vladimir Dergachev
infinity.

Improper Riemann integral - yes.

The definition of Lebesgue integral that I read required bounded functions
(up to measure 0). Yes, of course, you can define improper integral for
Lebesgue as well.

Post by Vladimir Dergachev
For this to make sense for a large class of functions one would have to
extend Riemann or Lebesgue definitions somehow, or create a brand new
integral (say for positive functions only).

How so?

Improper integral is an extension.

Post by Vladimir Dergachev
For example, you can start with Lebesgue measure on R^2 and for a
positive function f(x) define int(f(x)) as the measure of the area
enclosed between 0 and f(x). One can easily check that usual properties
of the integral are preserved.

In fact, you recover the Lebesgue integral again, do you not?

The values will agree on the set of all Lebesgue integrable functions (as
what would be the point if they did not ?) but, depending on what
definition you use, the integral defined as above might cover more
functions.

best

Vladimir Dergachev

Dave Seaman

2007-10-09 03:23:11 UTC

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as
x->infinity. It should be trivial but I don't see it.

Post by Vladimir Dergachev
infinity.

Improper Riemann integral - yes.
The definition of Lebesgue integral that I read required bounded functions
(up to measure 0). Yes, of course, you can define improper integral for
Lebesgue as well.

I suspect you didn't read far enough. You are evidently referring to a
simplified definition that served as a preliminary to the real
definition. The general definition does not require the function to be
bounded, or even essentially bounded.

There is more than one way to define the Lebesgue integral, but here is
one that works for positive functions f. The lower Lebesgue integral can
be defined as the supremum of all the integrals of simple functions g
such that 0 <= g <= f. Similarly, the upper Lebesgue integral can be
defined as the infimum of integrals of simple functions g such that f <=
g. If the lower and upper integrals agree, then f is integrable.

The definition matters, because there are important convergence theorems
that depend on the precise definition: the Monotone Convergence Theorem
and the Lebesgue Dominated Convergence Theorem.

Using either theorem, it is easy to come up with examples of unbounded
functions that are Lebesgue integrable. For example, let f(x) =
1/sqrt(x) for x in (0,1], and f(x) = 0 otherwise. Then f is measurable
on R but is unbounded. However, f can be represented as the limit of the
monotone sequence given by

f_n(x) = min(1/sqrt(x), n) for x in (0,1],
f_n(x) = 0 otherwise.

Then we have a monotone sequence 0 <= f_1 <= f_2 <= ..., and f =
lim_{n->oo} f_n. By the Monotone Convergence Theorem, the integral of f
is the limit of the integrals of the f_n, which is 2.

There is no need for improper integrals in the Lebesgue theory.

--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>

Vladimir Dergachev

2007-10-09 05:47:22 UTC

Post by Dave Seaman

Post by mathemaat
Given int 0 to infinity f(x)dx is finite, proof xf(x)->0 as
x->infinity. It should be trivial but I don't see it.

This should depend strongly on the definition of integral you use - is
it Riemann, Lebesgue or something else ?
let f(x)=sum gn(x)
gn(x)= n when |x-n|<1/(n*2^n) and 0 otherwise.
Note that f(x) and all gn(x) are non-negative.
It would thus be very reasonable to define
Int f(x) = sum Int(gn(x)) = sum 1/2^n = 1

Post by Vladimir Dergachev
infinity.

Improper Riemann integral - yes.
The definition of Lebesgue integral that I read required bounded
functions (up to measure 0). Yes, of course, you can define improper
integral for Lebesgue as well.

I suspect you didn't read far enough. You are evidently referring to a
simplified definition that served as a preliminary to the real
definition. The general definition does not require the function to be
bounded, or even essentially bounded.
There is more than one way to define the Lebesgue integral, but here is
one that works for positive functions f. The lower Lebesgue integral can
be defined as the supremum of all the integrals of simple functions g
such that 0 <= g <= f. Similarly, the upper Lebesgue integral can be
defined as the infimum of integrals of simple functions g such that f <=
g. If the lower and upper integrals agree, then f is integrable.
The definition matters, because there are important convergence theorems
that depend on the precise definition: the Monotone Convergence Theorem
and the Lebesgue Dominated Convergence Theorem.
Using either theorem, it is easy to come up with examples of unbounded
functions that are Lebesgue integrable. For example, let f(x) =
1/sqrt(x) for x in (0,1], and f(x) = 0 otherwise. Then f is measurable
on R but is unbounded. However, f can be represented as the limit of the
monotone sequence given by
f_n(x) = min(1/sqrt(x), n) for x in (0,1],
f_n(x) = 0 otherwise.
Then we have a monotone sequence 0 <= f_1 <= f_2 <= ..., and f =
lim_{n->oo} f_n. By the Monotone Convergence Theorem, the integral of f
is the limit of the integrals of the f_n, which is 2.
There is no need for improper integrals in the Lebesgue theory.

Hmm... We did do it in stages, but for some reason I always thought the
definition you gave to be the definition of improper Lebesgue integral.

I am away from (most) of my books at the moment, will lookup
Kolmogorov-Fomin when I get back. Which reference do you use ?

Looking at Wikipedia, they do it your way, but do not define upper integrals
and seem to omit the discussion of convergence.

thank you

Vladimir Dergachev

Dave Seaman

2007-10-09 07:14:43 UTC

Post by Dave Seaman