# Is $L^p \cap L^q$ dense in $L^r$?

It is known that $L^p \cap L^q \subset L^r$, where $1 \le p \le r \le q \le \infty$. Are all of these inclusions dense? I.e., do we have \begin{equation*} \overline{L^p \cap L^q} = L^r \end{equation*} (closure w.r.t. $\Vert \cdot \Vert_r$)?

To prove this, one would have to show the existence of a sequence of functions $f_n \in L^p \cap L^q$ such that $f_n \stackrel{L^r}{\longrightarrow} f$, given arbitrary $f \in L^r$. I know how this works in the case $p=1$, $q=\infty$. But is it even true in the most general case?

If it is true, as a consequence, any function in $L^p$ could be approximated by functions $f_n \in L^q$ for any $1 \le q \le \infty$. This sounds quite like a utopia to me, as it would make a lot of things I am currently working on easier. I don't see why it would be wrong, however. Many thanks in advance!

## 1 Answer

Let $f\in\mathbb L^r$, and define $f_n:=f\chi{\{n^{-1}\leqslant |f|\leqslant n\}}.$ We have $f_n\in L^p\cap L^q$ for each $n$, and by monotone convergence, $\lVert f-f_n\rVert_r\to 0$ as $n$ goes to infinity.