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If $f(x) > 0$ is continuous at $[0, +\infty]$ and $\displaystyle \int_0^{+\infty} \frac{1}{f(x)} dx$ is convergent, please prove $\displaystyle \lim_{\lambda \to \infty} \frac{1}{\lambda} \int_0^\lambda f(x) dx = +\infty$.

However, I think if we can prove $\displaystyle \lim_{x \to \infty} f(x) = \infty$, and then we can prove the question. But I don't know how to prove it.

Can somebody help me solve it? Or can somebody give me some hints? thank you!

2012ssohn
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python3
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    Well, you know that the function is eventually smaller than x. Can you use that? – user137500 Mar 25 '14 at 02:40
  • Here's a sketch: I would use l'Hopital's with respect to $\lambda$. You would derivate the numerator (iff $f(x) \rightarrow \infty$) and denominator and get that $\lim \frac{f(\lambda)}{1} = \infty$ by assumption. So, yes it is sufficient to prove this. Suppose that $\exists; \epsilon>0$ such that $\forall N$, $\forall x > N$, where $g(x) \geq \epsilon$. But then the undergraph of $g(x)$ has infinite measure. So, yes $\lim \frac{1}{f(x)} = 0$ and hence $\lim f(x) = \infty$ as $f(x)>0$, $\forall x \geq 0$. – Christopher K Mar 25 '14 at 02:49

2 Answers2

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Hint: Use the Cauchy-Schwarz inequality.

details:

indeed,

$$ \sqrt{T} = \int_0^T \sqrt{\frac{f(x)}T}\frac 1{\sqrt{f(x)}} dx \le \left( \frac 1T\int_0^T f(x) dx\right)^{1/2} \left(\int_0^T \frac 1{f(x)} dx \right)^{1/2}\\ \le M^{1/2}\left(\frac 1T \int_0^T \frac 1{f(x)} dx \right)^{1/2} $$

Phira
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mookid
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Assume f is bounded above by M ==> f(x) < M, then 1/f(x) > 1/M, then Int(0 to n)1/f(x)dx > n/M. Taking limit as n --> infinity we get Int(0 to infinity)1/f(x)dx = infinity- contradiction. So f is unbounded, and we can take it from here.

DeepSea
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