I've the following statement:
Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ integrable function. If $ \int_0^\infty f(t)dt $ converge, does sequence $ (x_n)\in \mathbb{R} $ exist such that:
$ \lim_{n\rightarrow 0} x_n = \infty $ and $ \lim_{n\rightarrow 0} f(x_n) = 0 $
I think it is true, because in order to the integral to converge it must be near zero infinite number of times so it possible to build sequence such that $ f(x_n)$ will converge to zero. Yet I'm not 100% sure about this or how to find these points. So any help will be great.
Thanks.
Best Answer
The answer is NO without further assumptions on $f$ other than it is integrable. For a counter-example, consider the function
$$f(x) = (-1)^{\left\lfloor x^2 \right\rfloor}$$
For any $k \in \mathbb{Z}_{+}$, define $$F_k = \int_{\sqrt{2k-2}}^{\sqrt{2k}} f(x) dx = -\sqrt{2k-2} + 2\sqrt{2k-1} -\sqrt{2k}$$ Notice for large $k$,${}^{\color{blue}{[1]}}$
$$\begin{align}F_k &= (\sqrt{2k-1} - \sqrt{2k-2})-(\sqrt{2k}-\sqrt{2k-1})\\ &= \frac{1}{\sqrt{2k-1}+\sqrt{2k-2}} - \frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\ &= \frac{2}{(\sqrt{2k-1}+\sqrt{2k-2})(\sqrt{2k}+\sqrt{2k-1})(\sqrt{2k}+\sqrt{2k-2})}\\ &\sim \frac{1}{8\sqrt{2k^3}} \end{align} $$ We find as a sequence in $n$, $\displaystyle\;\int_0^{\sqrt{2n}} f(x) dx = \sum_{k=1}^n F_k$ converges to some limit $\Delta$ as $n \to \infty$.${}^{\color{blue}{[2]}}$
For any $\epsilon > 0$, choose a $N \in \mathbb{Z}_{+}$ so large such that
$$\sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2} \quad\text{ and }\quad \left|\sum_{k=1}^m F_k - \Delta \right| < \frac{\epsilon}{2},\forall m \ge N$$
For any $y > \sqrt{2N}$, let $n \ge N$ be the integer such that $y \in \left[\sqrt{2n},\sqrt{2n+2}\right]$, we have
$$\left|\int_0^y f(x) dx - \int_0^{\sqrt{2n}} f(x) dx\right| = \left|\int_{\sqrt{2n}}^y f(x) dx\right| \le \sqrt{2n+1}-\sqrt{2n}\\ \le \sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2}$$
This leads to $$\left| \int_0^y f(x) dx - \Delta \right| \le \left| \int_{\sqrt{2n}}^{y} f(x) dx \right| + \left| \sum_{k=1}^n F_k - \Delta \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
Since $\epsilon$ is arbitrary, $\displaystyle\;\lim\limits_{y\to\infty}\int_0^y f(x)dx\;$ exists and equal to $\displaystyle\;\Delta = \sum_{k=1}^\infty F_k\;$.
By definition, $|f(x)| = 1$ for all $x$. It is impossible to build a sequence $x_n$ such that $f(x_n)$ converges to $0$.
Random Notes
$\color{blue}{[1]}$ - Another way to see the $F_k \sim O\left(\frac{1}{\sqrt{k^3}}\right)$ dependence goes like this. The expression $$F_k = -\sqrt{2k-2} + 2\sqrt{2k-1} - \sqrt{2k}$$ has the form of $2^{nd}$ order finite difference for function $-\sqrt{x}$ at $x = 2k-1$. By a generalization of mean value theorem to higher order finite differences, there is a $\xi \in (2k-2,2k)$ such that $$F_k = \left.\frac{d^2}{dx^2}(-\sqrt{x})\right|_{x=\xi} = \frac{1}{4\sqrt{\xi^3}} \sim \frac{1}{8\sqrt{2k^3}}$$
$\color{blue}{[2]}$ - It can be shown $\Delta = 2(1-\sqrt{8})\zeta\left(-\frac12\right) \approx 0.76020962521937$.