Counterexample for $L^p$ Inclusion The general question is : disprove that $L^p(\mathbb R)\subset L^q(\mathbb R)$ and $L^q(\mathbb R)\subset L^p(\mathbb R)$ for $q<p\leq\infty$
I managed to find counterexamples for the finite cases and the case in which $L^1\subset L^\infty$ is proven wrong with the function $f(x)=1/\sqrt{x}$ defined on $[0,1]$ 
But I have difficulties to disprove $L^\infty\subset L^1$. Could anyone give me hints on how to find such a function? 
 A: Here you have a counterexample at once. Consider the function $f_{\alpha}(x) = \min \left\{1, \frac{1}{x^\alpha}\right\}$, $x\in \mathbb{R}$ and an arbitrary $\alpha \in \mathbb{R}$.
Assume $\alpha$ is such that $\frac{1}{p}<\alpha<\frac{1}{q}$. The $L^p$- norm is
$$\|f_{\alpha}\|_{L^p(\mathbb{R})} = 2^{1/p}\left(1- \frac{1}{1-p\alpha}\right)^{1/p}<\infty.$$
Nevertheless, the $L^q$-norm for this $\alpha$ is infinity since $\alpha \ngtr \frac{1}{q}$.
For the extreme case ($p=\infty$, $q=1$), you can also use the function $f_{\alpha}$ since it is always bounded, and hence belongs to $L^\infty (\mathbb{R})$ but if $\alpha \in (0,1)$, it is never in $L^1(\mathbb{R})$ since the tails are "too fat".
Of course, any constant function works as well, since they are indeed bounded and of course, not globally integrable.
For the opposite case, you can do the same, mutatis mutandis, using the function $g_{\alpha}(x) = \frac{1}{x^\alpha}$, $x\in (0,1)$ and $0$ otherwise.
A: Take a non-zero constant function on $\mathbb{R}$.
