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From If $f : [a, \infty) → \Bbb R$ is monotonically decreasing and the integral $\int_0^\infty f(x) \,dx$ is convergent, then $\lim_{x→\infty} f(x) = 0$.

, I know if $f$ $[a,\infty) \rightarrow \mathbb{R}$ is monotonically decreasing and the integral $\int_0^{\infty} f(x) dx<\infty$, the $\lim_{x \rightarrow \infty}f(x)=0$.

How about relaxing monotonically decreasing condition? For example, suppose

$f: [0,\infty) \rightarrow (0,\infty)$ be a continuously differentiable function with $\sup_{t>0} |f'(t)|<\infty$ and $\int_0^{\infty} f(t)dt <\infty$. Then I want to show $\lim_{t \rightarrow \infty}f(t) =0$.

First I know since $\sup_{t>0} |f'(t)|<\infty$, $f'(t)$ is bounded hence $f$ is uniformly continuous, but it seems it does not related to $\lim_{x \rightarrow \infty}f(x)=0$.

I tried to do the similar things in If $f : [a, \infty) → \Bbb R$ is monotonically decreasing and the integral $\int_0^\infty f(x) \,dx$ is convergent, then $\lim_{x→\infty} f(x) = 0$. but cannot finish the proof.

phy_math
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  • https://math.stackexchange.com/questions/92105/f-uniformly-continuous-and-int-a-infty-fx-dx-converges-imply-lim-x
    might give some insight.     – F. Conrad May 13 '21 at 14:25

1 Answers1

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Hint: If $\lim_{t\to\infty} f(t)=0$ fails, then there exists $\epsilon>0$ and a sequence $t_n \to \infty$ such that $f(t_n)\ge \epsilon.$ If you space the $t_n$ far enough apart, uniform continuity implies there is $\delta>0$ such that $f>\epsilon/2$ on each interval $[t_n,t_n+\delta].$

zhw.
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  • I got it, since $f>\frac{\epsilon}{2}$, the integral on the interval $|\int f(x) dx| \geq \frac{\epsilon}{2} \delta$, and taking $t_n \rightarrow \infty$ it reduces the diverge of $\int^{\infty} f(x) dx$, so contradicition! Thanks – phy_math May 13 '21 at 16:48