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Let $f:\mathbb{R}\to\mathbb{R}$ such that $f\ge 0$, monotonically decreasing and $\int_0^\infty f(x) \ dx < \infty$. Prove that $f(x) = O(1/x)$.

So basically, both $f(x)$ and $1/x$ are non-negative and monotonically decreasing. We know that $\int_0^\infty \frac{1}{x} = \infty$ but $\int_0^\infty \lt \infty $ - so there must be some $N$ such that for every $x>N$: $f(x) \le \frac{1}{x}$ which implies $f(x) = O(\frac{1}{x})$.

Is that valid/rigorous enough?

Thanks.

Elimination
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  • The "so there must be" step is not really justified -- it the thing it concludes happens to be true, but you're going to need a more detailed argument why it needs to be true. For example, exactly how do you use the assumption that $f(x)$ is monotonically decreasing? You have to use it somehow, otherwise the conclusion isn't true. – hmakholm left over Monica Jun 08 '15 at 14:11
  • Your conclusion that there is an $N>0$ such that $x>N$ implies $f(x)\le 1/x$ seems to be pulled out of thin air. Suppose that for all $N>0$ there is an $x>N$ so that $f(x)\ge 1/x$, does this imply $\int_0^\infty f(x),dx=\infty$? – Tim Raczkowski Jun 08 '15 at 14:12
  • That's a stronger critiera than $O(g(x))$ anyway, to show $f(x) = O(\frac1{x})$, you only need to bound it at $\frac{c}{x}$ for some $c$, doesn't have to be $1$. – Barry Jun 08 '15 at 14:14
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    @Barry: True -- but the proof of this case I can imagine right away does start by picking a $c$, and $c=1$ is as good a choice as any. (In fact the assumptions are enough to prove that $f(x)=o(\frac1x)$). – hmakholm left over Monica Jun 08 '15 at 14:20

1 Answers1

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For $t > 0$,

$$tf(t) = \int_0^{t}f(t)\, dx \le \int_0^t f(x)\, dx \le \int_0^\infty f(x)\, dx.$$

Thus $0 \le f(t) \le M/t$ for all $t > 0$, where $M = \int_0^\infty f(x)\, dx$. So $f(x) = O(1/x)$.

kobe
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