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Set-up.

For a bounded continuous function $u \colon \mathbb{R}^n \to \mathbb{R}$, the $\gamma$-Hölder semi-norm of $u$ is $$ \begin{eqnarray} [u]_{C^\gamma} &=& \sup \left\{\frac{|u(x) - u(y)|}{|x-y|^\gamma} : x,y \in U, x \neq y \right\} \\ &=& \inf \left\{ C \geq 0 : |u(x) - u(y)| \leq C |x-y|^{\gamma} \text{ for all } x,y \in U \right\}. \end{eqnarray} $$

The Problem

Fix $1 \leq p < \infty$, $0 < \gamma \leq 1$, and $0 < \lambda < 1$ such that $$ 0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}. $$ I am trying to prove that there is a constant $C$ such that $$ \|u\|_{L^\infty} \leq C \|u\|_{L^p}^{\lambda} [u]_{C^\gamma}^{1-\lambda} $$ for every compactly supported $C^{1}(\mathbb{R}^n)$ function $u$.

My Strategy

My plan is to use the interpolation result for Lebesgue spaces: For every $q,r$ satisfying $p < q < r \leq \infty$ and $$ \frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r}, $$ we have $$ \|u\|_{L^q} \leq \|u\|_{L^{p}}^{\lambda} \|u\|_{L^r}^{1-\lambda} $$ for every $u \in L^p \cap L^r$.

Solving for $1/r$ in terms of $q$ and using the relationship between $p$, $\gamma$, and $n$, we find $$ \frac{1}{r} = \frac{1/q - \lambda/p}{1-\lambda} = \frac{1/q}{1-\lambda} - \frac{\gamma}{n} $$ So, by letting $q \to \infty$, we have $r \to -n / \gamma$, and $$ \frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r}, $$ goes to $$ 0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}. $$ Meanwhile, for $u \in L^{\infty}$ (which certainly holds when $u$ is compactly supported and $C^1$), we have that $\lim_{q \to \infty} \|u\|_{L^q} =\|u\|_{L^\infty}$.

So if I could prove that $\|u\|_r \to [u]_{C^{\gamma}}$ as $q \to \infty$ (i.e., as $r \to -n / \gamma$), I'd be done. The problem is that I don't know how to prove this. Moreover, I'm not sure if this is even the right approach to prove the desired inequality.

Can someone please help me out?

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    There is no reason that $\|u\|_r$ should tell you anything about $[u]_{C^\gamma}$ on its own (they measure completely different things). Also, your Lebesgue interpolation inequality only holds for $p < q < r$, but you are taking $q \to \infty$ and $r \to -n/\gamma$ (does $r < 1$ even make sense?).2011-11-17
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    You may have to assume that $u \in C^\gamma$ (instead of $u$ continuous) and use some kind of Sobolev embedding.2011-11-17
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    @Jeff: It was a mistake to not assume $u \in C^{\gamma}$. Thank you for pointing it out.2011-11-17
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    @Jeff: You are right that $q \rightarrow \infty$ and $r \rightarrow -n/\gamma$ is not compatible with the requirement $p in the Lebesgue interpolation inequality. This is one reason why I am not so sure my strategy will really work.2011-11-17
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    @Jeff: Could you please elaborate a bit more on your comment about using a Sobolev embedding?2011-11-17
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    The only way that I know how to relate $L^p$ norms and Holder norms is the Sobolev embeddings (http://en.wikipedia.org/wiki/Sobolev_embedding), but I don't see exactly how you would apply them here.2011-11-17
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    Is this a homework problem? My first thoughts were Sobolev embeddings, too. Are you sure, this should be an upper bound for the $ L^\infty $ norm of $ u $ ? With Sobolev embedding you run in trouble for $ p = \infty $. I think there's a true inequality like this: $ \|u\|_{L^r} \le C \|u\|_{L^q}^\lambda\|u\|_{C^{0,\gamma}}^{(1-\lambda)}$ on a compactly supported open set $ \Omega \subset \mathbb{R^n} $ and where $ \|u\|_{C^{0,\gamma}}=\|u\|_{C^0} + [u]_{C^{0,\gamma}}$. I hope there's no mistake.2011-11-18

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