Relation between $L^1(\mathbb R)$ and $L^2(\mathbb R)$

Question

I know that if $1\le p<q<\infty$ then $L^p\supset L^q $ and $l^p \supset l^q$.

But what is the relation between $L^1(\mathbb R)$ and $L^2(\mathbb R)$? I guess there is no relation, i.e. $L^1(\mathbb R) \subsetneq L^2(\mathbb R)$ and $L^2(\mathbb R) \subsetneq L^1(\mathbb R)$.

Can you give an example to show this?

Answer 1

Let $f, g \colon \mathbb{R} \to \mathbb{R}$ with $$ f(x) = \begin{cases} \frac{1}{x} & \text{if $x > 1$}, \\ 0 & \text{otherwise}, \end{cases} \quad\text{and}\quad g(x) = \begin{cases} \frac{1}{\sqrt{x}} & \text{if $0 < x < 1$}, \\ 0 & \text{otherwise}. \end{cases} $$ Then $f \in L^2(\mathbb{R})$ and $g \in L^1(\mathbb{R})$ but $f \notin L^1(\mathbb{R})$ and $g \notin L^2(\mathbb{R})$.