$\overline{L^1(\mathbb{R}) \cap L^p(\mathbb{R})} = L^p$ for $1 < p < 2$?

Question

Is $\displaystyle \overline{L^1(\mathbb{R}) \cap L^p(\mathbb{R})} = L^p(\mathbb{R})$ for $1 < p < 2$ ? I know this holds for $p = 2$. But, I am not able to find any source about the case when $1 < p < 2$.

Answer 1

Simple functions in $L_1$ are dense in any $L_p$ ($1\leq p<\infty$). For any function $f\in L_p$ there is a sequence of simple functions $f_n$ such the $|f_n|\nearrow|f|$. Then $|f_n-f|^p\leq 2^p|f|$ and by monotone convergence $\|f_n-f\|_p\xrightarrow{n\rightarrow\infty}0$. Notice that $f_n\in L_1\cap L_p$, for $f_n$ is simple and $|f_n|\leq |f|$.

Answer 2

For $f \in L^p$ we can define $$ f_n = f1_{B(0,n)}1_{|f| < n} \in L¹ \cap L^p.$$ Then $(f_n)$ converges pointwisely to $f$ with domination $f$ in $L^p$. Thus by the dominated convergence theorem the convergence is in $L^p$.

Answer 3

If you admit the fact that the space $C_{0}$ of continuous functions with compact support is dense in $L^{p}$, then the statement goes through immediately:

$C_{0}\subseteq L^{1}\cap L^{p}\subseteq L^{p}$, then $L^{p}=\overline{C_{0}}\subseteq\overline{L^{1}\cap L^{p}}\subseteq\overline{L^{p}}$.