Is there an example of a function that is $L^{p_2}$ but not $L^{p_1}$ and not $L^{p_3}$?

Question

If $\Omega\subset\Bbb R$ is compact we have $L^{\infty}(\Omega)\subset L^p(\Omega)\subset L^q(\Omega)\subset L^1(\Omega)$ if $p>q$ but it is not true if $\Omega$ is not compact.

The function $g(x)=\frac{1}{x}$ is in $L^p(1,\infty)\ \forall p>1$ but not in $L^1(1,\infty)$ with $\Omega=(1,\infty)$.

For some $p_1<p_2<p_3$ is there an example of a function that is $L^{p_2}$ but not $L^{p_1}$ and not $L^{p_3}$?

Answer 1

You could let $\Omega = (0,\infty)$ and define $$f(x) = \left\{ \begin{array}{cl} x^{-1/p_3} & 0 < x < 1 \\ x^{-1/p_1} & 1 \le x < \infty. \end{array} \right.$$