Understanding Int(0,infinite)[f(x)]^2dx Convergence

[eljose79]

A question ...

Let suppose we have a function f(x) so

Int(0.infinite)f(x)dx=N N finite number then my question is ..would be this f(x) a L^2(0,infinite) function so

Int(0,infinite)[f(x)]^2dx is convergent?..thanks.

 

[matt grime]

Yes, I think so.

If the infinite integral exists, then except on a set of measure zero it must be that f tends to zero, so pick T such that |f| is less than 1 for almost all x greater than T and the answer becomes clear.

 

[HallsofIvy]

Am I misreading this? The question asserts that the integral of f from 0 to infinity exists and asks if it follows that the integral of f² exists. I don't see how that implies that, except on a measure 0, f "tends" to 0. (In fact, I don't see what "tends" to means. At each point f is either 0 or it is not!)

What about f(x)= x^-1/2 for 0< x< 1 and f(x)= 0 for x>=1.

The integral from 0 to infinity is 2, a finite number but
f²= 1/x for 0< x< 1, 0 for x>=1 which cannot be integrated from 0 to 1.

 

[matt grime]

OK, am wrong - i was thinking of the problems of the wrong kind in the integral (an infinitely long bit such as the tail of 1/x rather than singular values). I wasn't particularly happy with it there either. I was thinking of bounded functions on R.

Tends to (in this strange sense) means that for all epsilon, there exists and R such that |f|<epsilon for all x>R except for a subset of zero measure. Ie tends to in the usual sense except that there may be a measure zero set on which there are arbitrary values, such as sin(x)/x for x not in Z and 1 for x in Z.