Is there anyway to bound the $L^\infty$ norm by other $L^p$ norm? If $f\in L^\infty(\mathbb R^2)$ (in my particular exercise, $f\in H^2(\mathbb R^2)$, the sobolev space), I want to bound $|f|_{L^\infty}= $ esssup $|f|\leq c|f|_{L^p}$  for some p, what kind of number can p be?
 A: Typically questions of this type can only possibly have one answer and you can find that answer based on a scaling argument.  Unfortunately here, the scaling shows that it just can't happen.  Here's how you can see that:
Take a generic function $f \in L^\infty(\mathbb{R}^2)$ and consider the function $f_\lambda$ where $f_\lambda (x) = f(\frac{x}{\lambda})$.  Let's assume that $\| f\|_{L^p} < \infty$ for whatever $p$ this inequality might hold for.  Notice that $\| f_\lambda \|_{L^\infty} = \| f \|_{L^\infty}$ by construction.  But if $0 < p < \infty$ then $ \| f_\lambda \|_{L^p} = \lambda^{\frac{1}{p}} \|f\|_{L^p}$.  If this inequality holds for some $p$ we must have that for every $f \in L^\infty$ we would have that for every $\lambda$
$$ \| f \|_{L^\infty} = \| f_\lambda \|_{L^\infty} \leq  c \|f_\lambda \|_{L^p} = \lambda^{\frac{1}{p}} c \|f\|_{L^p}$$
Taking $\lambda \to 0$ gives us a contradiction since this inequality would imply that $L^\infty \cap L^p = \{0\}$ which is clearly not true by considering the indicator function of the set $[0,1]$.
