Example for $f \in \mathcal{L}^q(\mathbb{R}, \lambda) \setminus \mathcal{L}^p(\mathbb{R}, \lambda)$ for $q < p$

Question

We consider $q, p \in [1, \infty)$. I'm required to show that $\mathcal{L}^p(\mathbb{R}, \lambda) \subsetneq \mathcal{L}^q(\mathbb{R}, \lambda)$ and $\mathcal{L}^q(\mathbb{R}, \lambda) \subsetneq \mathcal{L}^p(\mathbb{R}, \lambda)$. Let's assume without loss of generality: $q < p$.

I was already able to show the second statement by defining:

$$f: \mathbb{R} \rightarrow \mathbb{R}\hspace{15pt}x \mapsto \mathbb{1}_{[1, \infty)}(x) \cdot x^{-\frac{1}{q}}$$

We get $f \in \mathcal{L}^p(\mathbb{R}, \lambda)\,\setminus\, \mathcal{L}^q(\mathbb{R}, \lambda))$, but I fail to come up with $g \in \mathcal{L}^q(\mathbb{R}, \lambda) \,\setminus\, \mathcal{L}^p(\mathbb{R}, \lambda)$.

Answer 1

How about $x \mapsto 1_{(0,1)}(x) \cdot x^{-\frac 1p}$.

Answer 2

Let $r:=\frac{p+q}2$. Consider the function $$ f(x):=\mathbf 1_{[0,1]} x^{-1/r}, $$ we have $$ \int_0^1 x^{-q/r} dx <\infty $$ but $$ \int_0^1 x^{-p/r} dx =\infty. $$