How can we prove f(x) approaches 0 as x approaches infinity in this scenario?

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Discussion Overview

The discussion revolves around proving that a function f(x) approaches 0 as x approaches infinity, given certain conditions on its derivative and the convergence of its integral. The scope includes mathematical reasoning and exploration of theorems related to calculus and analysis.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant proposes that since f'(x) is bounded, it implies that the function cannot change rapidly near a local maximum, suggesting a connection to the behavior of the integral of |f(x)|.
  • Another participant questions how a lower bound could assist in the proof, indicating a preference for an epsilon-delta approach.
  • A different participant argues that establishing a lower bound for the integral can demonstrate that if the integral of |f(x)| is sufficiently small, then the supremum of |f(x)| must also be small.
  • One participant expresses confusion about utilizing the boundedness of the derivative, considering the implications of continuity and maximum values within closed intervals.
  • Another participant suggests finding a function that decreases at the fastest rate to establish a comparison with f(x), implying that f(x) must be at least as large as this function.
  • One participant articulates a method involving the Mean Value Theorem but struggles to connect it to the overall proof.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to prove the statement, with no consensus reached on the method to be employed. Various strategies are proposed, but uncertainty remains regarding their implementation and effectiveness.

Contextual Notes

Participants note limitations in their reasoning, particularly regarding the application of the Mean Value Theorem and the implications of bounded derivatives. There is also a lack of clarity on how to effectively utilize the boundedness of the derivative in conjunction with the convergence of the integral.

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let f(x) be continuously differentiable in [0,infinity) such that the derivative f'(x) is bounded. suppose that the integral \int_{a}^{\infty}|f(x)|dx converges. prove that f(x)->0 when x->infinity.


now, here's what i did:
f'(x) is bounded then for every x>=0 |f'(x)|<=M for some M>0.
now bacause the integral \int_{a}^{\infty}|f(x)|dx converges for some a>0, then by cauchy criterion we have that for every e>0 there exists B such that for every b1>b2>B>0 \int_{b2}^{b1}|f(x)|dx&lt;e now we can use the inequality: |\int_{b2}^{b1}f(x)dx|&lt;=\int_{b2}^{b1}|f(x)|dx&lt;e.
now, the crucial point here is that i need to show that for every e>0 there exists M'>0 such that for every x>M' |f(x)|<e.
now i think i can use the mean value theorem for integral here, so there exists a point x in (b2,b1) such that \int_{b2}^{b1}f(t)dt=f(x)(b1-b2) and then we have that for every x greater than B
|f(x)|(b1-b2)<e so |f(x)|<e/(b1-b2) but this doesn't work cause b1 and b2 aren't constants, so perhaps here i should use that the derivative is bounded, so by lagrange theorem we have that: |f(b1)-f(b2)|<=M(b1-b2) but still i don't see how to connect both of these theorems to prove this statement.
any hints?
 
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If the derivative is bounded, it means the function can't changes that quickly near a local max. So you can give a lower bound for the integral over a region in terms of the sup of f(x) in that region.
 
how does a lower bound will help me here?
i think i need to prove it by epsilon delta, unless there's another way?
 
Trust me, you want a lower bound. It will help you show that if the integral of |f(x)| is small enough, then sup{|f(x)|} must be small too.
 
ah, ok i understand.
but still i don't see how to use here the boudedness of the derivative, i mean: |f'(x)|<=M for some M>0, so bacause f(x) is continuous it is bounded so if sup(f(x))=max(f(x) in some closed interval, what to do from here?
 
The idea is that the function can only fall from its max so quickly. Find the function that does this the fastest, and you know the actual function is at least as big.
 
sorry but i don't see how to implement it here, i mean if we have a closed interval then sup=max in this interval and the function is not greater than this value for every x in the interval now if i were to use the boudedness, i would get by the mvt that |f(b1)-f(b2)|<=M(b1-b2) but f(x)<=max(f(x))
|f(x)|<=|f(x)-f(b2)|+|f(b2)-f(b1)|+|f(b1)|<=M(x-b2)+M(b1-b2)+max(f(x)) but still stuck.
 

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