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Been stuck on it for quite a while now, any help would be appreciated:

Prove or provide counterexample:

Let $f$:[1,$\infty$)$ \rightarrow\Bbb R$ be a continuous function. if $f>0$ and integrable on [1,$\infty$), then there exists some sequence ${x_n}$ $\rightarrow$ $\infty$ such that $\lim_{n\to \infty}f(x_n)=0$


My attempted solution:

This statement seems true as I could not find a counterexample, so here is my solution (I got stuck in the process):

Define $g(x)=\int_{1}^x f(t)dt$,

then: $\lim_{x\to \infty}g(x)=\lim_{x\to \infty}\int_{1}^x f(t)dt=L>0$

By the fundamental theorem of calculus we get:

$g'(x)=f(x)>0$

Hence $g$ is strictly increasing, is bounded above by $L$ and positive.

If only I were able to show that $\lim_{x\to \infty}g'(x)=0$ then my proof would be done but I could not figure out how.

Thank you for your input!

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2 Answers 2

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For any $n\in \mathbb N,$ we must have $\inf_{[n,\infty)} f=0.$ Otherwise for some $n_0,$ $\inf_{[n_0,\infty)} f =c>0.$ We would then have

$$\int_{n_0}^\infty f(x)\,dx \ge \int_{n_0}^\infty c\,dx =\infty,$$

contradiction.

It follows that for each $n,$ there exists $x_n\in [n,\infty)$ such that $f(x_n)<1/n.$ We then have $x_n\to \infty$ and $f(x_n)\to 0$ as desired.

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The thing is that it need not be the case that $g' = f \to 0$. Consider, for instance, the case where in each interval $[n, n+1)$, the graph of $f$ looks like a triangle of height $n$ but width $1/n^3$, outside of which $f$ is $0$. Such an $f$ of course satisfies the claim, but it scuppers your stategy.

On the other hand, we know that $\liminf f = 0$. Otherwise there is some $\varepsilon > 0$ such that $f > \varepsilon$ for large $x$, which makes the integral of $f$ blow up. Since $\liminf$s are approached arbitrarily closely infinitely many times, this tells us that a sequence of $x_n$ such that $f(x_n) \to 0$ must exist.

One can be quite explicit in identifying such a sequence. For naturals $n$, let $m_n := \min_{x \in [n,n+1]} f(x),$ let $y_n$ be the corresponding minimiser (which exists because continuous function on a compact set), and consider the step function $h := \sum m_n \mathbf{1}\{x \in [n, n+1)\}.$

Then we know that $h > 0$ and $h \le f$ is integrable. So we know that $m_n \to 0$. This gives a sequence of values $y_n$ which diverge (since $y_n \ge n$), and for which $f(y_n) \to 0$.

The only problem with the $y_n$ is that some of the consecutive $y_n$ may coincide - for instance if the minima are at the even naturals. However, it is trivially true that $y_n \neq y_{n+2}$ for any $n$, since they live in disjoint sets. The simple fix then is to go in steps of two. So $x_n = y_{2n}$ serves as a sequence that satisfies the claim.

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