Let $\Omega$ be a bounded set in $\mathbb{R}^n$ and let $1\leq q<r\leq p < \infty$. Prove that in the linear vector space $L^p(\Omega)$ the norms of $L^q(\Omega)$ and $L^r(\Omega)$ are not equivalent.
Two norms $||.||_1$ and $||.||_2$ are equivalent in there exists positive numbers $\alpha$ an $\beta$ such that $\alpha||x||_1 \leq ||x||_2 \leq \beta ||x||_1$ for all $x\in X$
I have no idea what funcion in $L^p$ would satisfy this. Any help is appreciated! Thanks
Hint: Consider functions of the form \begin{align} f(x) = \frac{1}{|x-a|^\alpha} \end{align} where $a \in \Omega$ and $\alpha>0$.
For instance $\mathbb{R}^n$ where $n\geq 3$, let $q = 1$ and $p=2$. If $\alpha = n/2$, then we see that \begin{align} \int_{\Omega} \frac{dx}{|x-a|^{n/2}} \leq \int_{B(a, r)} \frac{dx}{|x-a|^{n/2}} = \int^r_0 \int_{|x-a|=\tau} \frac{1}{|x-a|^{n/2}} dS d\tau = \int^r_0 \frac{dx}{\tau^{n/2-1}}<\infty \end{align} where $B(a, r) \supset \Omega$ but \begin{align} \int_{\Omega} \frac{dx}{|x-a|^n} \geq \int_{B(a,r_2)} \frac{dx}{|x-a|^n} =\int^{r_2}_0\frac{1}{\tau}\ d\tau = \infty \end{align} where $\Omega \supset B(a, r_2)$, which means $f \in L^1(\Omega)$ but not in $L^2(\Omega)$, i.e. there doesn't exist constant $C>0$ such that \begin{align} || f||_{L^2(\Omega)} \leq C|| f||_{L^1(\Omega)}. \end{align}