Looking for a bounded continuous $o(1)$ function on $]0, +\infty[$ that is not in $L^{p}(]0, +\infty[)$ for all $0 < p < +\infty$

Question

I am looking for a bounded continuous function $f \colon ]0, +\infty[ \to \mathbb{R}$ such that $f(x) = o(1)$ as $x \to \infty$ and $f \notin L^{p}(]0, +\infty[)$ for all $ 0 < p < +\infty$.

My closest candidate was $x \mapsto 1/x$, which satisfies all conditions but the boundedness on the given interval.

Any idea would be appreciated.

Answer 1

You could take $$ f(x) = \begin{cases} \frac{1}{\ln 2} & x < 2 \\ \frac{1}{\ln(x)} & x \ge 2 \end{cases}.$$

$f$ is in no $L^p$-space for $p < \infty$, since $$\int_2^t \frac{1}{\ln(x)^p} \, dx = \int_{\ln(2)}^{\log(t)} y^{-p} e^y \, dy$$ and $y^{-p}e^y \to \infty$ for $y \to \infty$.