If $f\in L^1(\Bbb R,dx)$ Then prove that for almost every $x\in\Bbb R$ $$\lim_{n\to \infty} f(nx) = 0$$
Be aware this statement is different from the following: $$ f\in L^1(\Bbb R,dx)\implies \lim_{|x|\to \infty} f(x) = 0$$ which is a false statement. Indeed, We set the functionn $f$ ([see its construction here}[1]) \begin{equation} f(x)= \begin{cases} 2^{n/2}\cdot \underbrace{-2^{2n+2}(x-n)^2+2^{n+2}(x-n)}_{P_n}& \text{if} ~~ n\le x\le n+\frac{1}{2^n} ~~\text{for some $n\in\mathbb{N}$}\\ 0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{for some $j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x\lt 0. \end{cases} \end{equation}
($h =1$, $ \varepsilon_n =\frac{1}{2^n}$, $a_n = n$ and $b_n =n+\frac{1}{2^n}$)
One can check that $f$ is an even and continuous function. Further, we have $$ \int_\mathbb{R} f(x)\,dx= 2\sum_{n=0}^{\infty}\int_{n}^{n+\frac{1}{2^n}}2^{n/2} P_n(x) dx = 2\sum_{n=0}^{\infty}-4.2^{n/2} \left[ \frac{2^{2n}}{3}(x-n)^3-\frac{2^n}{2}(x-n)^2\right]_{n}^{n+\frac{1}{2^n}}\\ = \frac43\sum_{n=0}^{\infty} \frac{1}{2^{n/2}}<\infty $$
Morerover, for every $n\in\mathbb{N}$ one has $$f(n) = 0~~~~\text{and} ~~~f(n+\frac{1}{2^{n+1}}) = 2^{n/2} $$
Therefore $$\lim_{|x| \to \infty}|f(x)|\not \to 0 $$ further, $\lim_{|x| \to \infty}|f(x)|$ does not exist.
Whereas in this case for all $x\not \in \{n+\frac{1}{2^{n+1}}, n \in \Bbb N\} $
$$\lim_{j\to \infty} f(jx) = 0$$ I also tried to construct a contradiction with this function Does this integral converge or diverge ? $\int_\Bbb R \left (\frac{2+\cos x}{3}\right )^{x^4}dx?$. Sill we have $$\lim_{j\to \infty} f(jx) = 0~~~~x\not\in \{2n\pi:~~ n \in \Bbb N\}$$ Also notice that this question is differ from
Let $f\in L^1(R)$ set $f_n(x)=\dfrac{f(nx)}{n}$ prove that $\lim_{n \to \infty}f_n=0$ for a.e x
Prove that for almost every $x\in$ $\mathbb{R}$ , $\lim_{n\to\infty}n^{-p}f(nx)=0$ .
where the situation is easy to manage. [1]: https://math.stackexchange.com/q/2143818
