Examples of functions that are in $L^p(X,\mu)$ but not in $L^q(X,\mu)$, where $1< p < q <\infty$ for each $p$ and $q$?

Question

It is known that if $\mu(X) < \infty$, we have the inclusion $L^q(X,\mu) \subset L^p(X,\mu)$ when $1<p<q$. This implies $L^p$ is a "bigger" space than $L^q$. (Or is it a bad idea to think of them as big and small spaces?)

Can we have some examples of functions that "lie" between the big and small $L^p$ spaces, i.e., $f \in L^p$ but $f \notin L^q $ for each $p$ and $q$, $1<p<q<\infty$? Thanks.

Answer 1

On the finite measure space $(0,1]$, I might consider $x^{-\alpha}$ for $\alpha > 0$. This is in $L^p$ if $\alpha p < 1$ and is in $L^q$ if $\alpha q < 1$. I.e., if $p < \frac{1}{\alpha}$ and $q < \frac{1}{\alpha}$, respectively. You should have no challenge finding an interesting $\alpha$ for each choice of $p < q$. For instance, for $p=2$, $q=3$, try $\alpha = 2/5$.

Answer 2

Let $\mu$ be the Lebesgue measure on $[0,1]$.

Let $f=x^{-1/r}$ for some $r \in (p,q)$. Then $\|f\|_p^p = \int_0^1 x^{-p/r}\mathop{dx} \le \int_0^1 x^{-(1-\epsilon)}\mathop{dx}<\infty$ but $\|f\|_q^q = \int_0^1 x^{-q/r} \mathop{dx} \ge \int_0^1 x^{-(1+\epsilon)} = \infty$.