$L^p$ integrable but not $L^q$ integrable

Question

Does there exist a continuous function on $[0, \infty)$ such that it is in $L^p(0,\infty)$ for some $p\in [1,2]$ but is not in $L^q(0,\infty)$ for any $q\in (2, 2/(2-p))$?

Thanks!

Answer 1

For any $n\in\mathbb N$, put $\delta_n=\frac{e^{-n}}{n^2}$. Now, let $f:[0,\infty)\to\mathbb R$ be the continuous function defined as follows: $f(t)\equiv e^n$ on $\left[n-\frac{\delta_n}2,n+\frac{\delta_n}2\right]$, $f(t)\equiv 0$ outside $\bigcup_{n\in\mathbb N} [n-\delta_n, n+\delta_n]$, and $f$ is linear on the remaining intervals. Then $f$ is in $L^1(0,\infty)$ because $$\int_0^\infty \vert f(t)\vert dt= \sum_{n=1}^\infty \frac32\delta_n\times e^n=\frac32\sum_{n=1}^\infty\frac{1}{n^2}<\infty\, .$$ On the other hand, $f$ is not in any $L^q$, $q>1$ since $$\int_0^\infty \vert f(t)\vert^q dt\geq \sum_{n=1}^\infty \delta_n\times e^{qn}=\sum_{n=1}^\infty \frac{e^{(q-1)n}}{n^2}=\infty\, . $$