On an inclusion condition between $L^p(\mathbb R)$ for different $p$ s Consider the real line $\mathbb R$ with Lebesgue measure . If $1\le p < q < \infty$ are real numbers and $r \in [p,q]$ then we know that $L^p(\mathbb R) \cap L^q(\mathbb R) \subseteq L^r(\mathbb R)$ . My question is , is any kind of converse true , i.e.
let $p,q,r \in [1,\infty )$ be such that $p<q$ and $L^p(\mathbb R) \cap L^q(\mathbb R) \subseteq L^r(\mathbb R)$ , then is it true that $r \in [p,q]$ ? 
 A: Yes, it is true. There is quite easy way to see this. Let 
$f_s(x)=\frac{1}{|x|^{s/p}}\chi_{|x|>1}$ for any $s>1$. Then 
$f_s\in L^p((\mathbb{R})\cap L^q(\mathbb{R})$. So $f_s\in L^r$ (because $r$ has to satisfy $\frac{rs}{p}>1$), and hence  $r>p/s$ for any $s>1$, which means that it can't be less then $p$. On the other hand, let us define $g_s(s)=\frac{1}{|x|^{s/q}}\chi_{0<|x|<1}$ for any $0<s<1$. Similarly, 
$g_s\in  L^p((\mathbb{R})\cap L^q(\mathbb{R})$. So $g_s\in L^r$, and hence $r<q/s$ 
for any $0<s<1$ (because $\frac{sr}{q}<1$). Therefore $r\in[p,q]$. 
A: Yes. Suppose that $r < p$. Then, there is a function $f \in L^p(\mathbb R) \setminus L^r(\mathbb R)$. Now, this property is preserved after truncation, i.e., $g$ defined via
$g(x) = \max(-1, \min(1, f(x)))$ still satisfies $g \in L^p(\mathbb R) \setminus L^r(\mathbb R)$. Moreover, $g \in L^\infty(\mathbb R)$.
Applying your argument to $g$ yields $g \in L^q(\mathbb R)$
and this shows
$L^p(\mathbb R) \cap L^q(\mathbb R) \nsubseteq L^r(\mathbb R)$.
The case $r > q$ can be handled analogously.
