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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?

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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 $[0,1]$.

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  • $\begingroup$ Thank you. What about $||f||_{L^\infty}\leq sup_{x}|f|$? $\endgroup$ – Zhixia Zhang Dec 15 '12 at 15:39
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    $\begingroup$ That doesn't make sense since $L^\infty$ functions are not defined pointwise. They are equivalence classes of functions up to changing the value on sets of measure zero. $\endgroup$ – Chris Janjigian Dec 15 '12 at 16:10

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