0

Given that f(x) (a positive function) is integrable from 0 to infinity, we were to show that as x tends to infinity, limits of f(x) and its derivative are zero.

Attempt: Let X1 and X2 are very large numbers. Integration of f(x) from X1 to X2 would be less than epsilon, e. And f(x) at these points would be L (limit value). So, L(X1-X2) less than e. But X1 and X2 are large numbers, so, their difference is infinity minus infinity, which could be anything. To ensure its value to be less than e, L must be zero.

I am not able to proceed for the derivative of f(x).

aarbee
  • 8,246
  • 1
    Isn't the very statement you are trying to prove false? (as a function being "with higher and higher, narrower and narrower triangles" shows) Who is asking you to prove it? (see e.g. http://math.stackexchange.com/questions/1231906/integrability-of-f-does-not-necessarily-imply-convergence-of-fx-to-0-as?noredirect=1&lq=1 ) – Clement C. Nov 14 '16 at 15:21
  • It's a class question. And I wonder if the link cited above represents the same question. Because here the integration is from 0 to infinity, and there they are taking some different intervals. – aarbee Nov 14 '16 at 15:59
  • That does not make a difference, really. What is your definition of integrability -- and are you certain the question does not ask you "True or false: ...?" – Clement C. Nov 14 '16 at 16:34
  • Actually, it was an MCQ type question, and I have listed the correct options ( as per the key). And the proof listed above was briefed by the teacher. – aarbee Nov 14 '16 at 17:30
  • Again, I am sorry, but there are positive measurable (and even continuous) functions on $[0,\infty)$ which do not tend to $0$ at $\infty$. (the most standard example being given in the link above, with non-negative instead of positive, which changes nothing). – Clement C. Nov 14 '16 at 21:22
  • Ok, in the next class, I can try to point this out. I wish I knew a simpler example because I might not get enough time to present a big example. – aarbee Nov 15 '16 at 04:26

0 Answers0