I've started learning some functional analysis and I'm struggling to understand the $L^1 \cap L^\infty$ space. The abstract of the paper https://link.springer.com/chapter/10.1007%2F978-3-319-42758-4_3 discusses it and defines the norm on this space as
$$ \Vert \cdot \Vert_{L^1 \cap L^\infty} := \max\left( \Vert \cdot \Vert_{L^1}, \Vert \cdot \Vert_{L^\infty} \right). $$
I don't see how $L^1 \cap L^\infty$ is different from just the $L^1$ space.
My Reasoning:
For $f \in L^1$ we have
$$ \int_\Omega \vert f(x) \vert d\mu(x) < \infty $$
ie. $f$ is absolutely integrable (a.e.). That implies $\vert f(x)\vert <\infty$ for almost all $x$. Therefore $f \in L^\infty$. On a similar line of thought I would have expected $\Vert \cdot \Vert_{L^1} \geq \Vert \cdot \Vert_{L^\infty}$. That would suggest to me that $L^1 \subseteq L^\infty$ and then their intersection is just $L^1$.
Could somebody please explain why this is not so (maybe using a counter-example)?
Thanks for your help!