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:


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!


2 Answers 2


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

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


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.