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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?). –  Jeff Nov 17 '11 at 21:18
    
You may have to assume that $u \in C^\gamma$ (instead of $u$ continuous) and use some kind of Sobolev embedding. –  Jeff Nov 17 '11 at 21:21
    
@Jeff: It was a mistake to not assume $u \in C^{\gamma}$. Thank you for pointing it out. –  fferic Nov 17 '11 at 22:03
    
@Jeff: You are right that $q \rightarrow \infty$ and $r \rightarrow -n/\gamma$ is not compatible with the requirement $p<q<r$ in the Lebesgue interpolation inequality. This is one reason why I am not so sure my strategy will really work. –  fferic Nov 17 '11 at 22:06
    
@Jeff: Could you please elaborate a bit more on your comment about using a Sobolev embedding? –  fferic Nov 17 '11 at 22:07

1 Answer 1

This is my first post on MSE and I'm just a grad student so forgive me if screw anything up. [Disclaimer: This proof is likely to be overkill, however I like that it shows what exactly is going on and these techniques generalize to a lot of different problems.] Hopefully someone else can chime in to validate my answer.

I will only consider the case $U = \Bbb R^n$.

The proof will be similar to the proof of Theorem A.3 in Tao's book "Nonlinear dispersive equations". In Theorem A.3, Tao uses basic Littlewood-Paley theory to prove a fairly general version of the Gagliardo-Nirenberg inequality, which is your equation after your swap out the Hölder space with an appropriate Sobolev space.

First, we consider the case when $\| u \|_{L^p} = \| u \|_{C^\gamma} = 1$ and show that the $L^\infty$ norm is bounded above by a constant. For general $u$ in $L^p \cap C^\gamma$ we can reduce to the first case by considering $v(x) = A u(B x)$ for an appropriate choice of constants $A$ and $B$ which make both of the norms $1$.

The main idea is to decompose $u$ into high and low frequencies. We will use the $L^p$ norm to bound the low frequencies and the Hölder regularity to bound the high frequencies.

To begin, the triangle inequality gives

$$ \| u \|_{L^\infty} \le \sum_{k=0}^{-\infty}\| P_k u \|_{L^\infty} + \sum_{k=1}^{\infty} \| P_k u \|_{L^\infty}. $$

The Bernstein inequalities tell us that $$\| P_k u \|_{L^\infty} \le C 2^{k \frac{n}{p}} \| P_k u \|_{L^p} = C 2^{k \frac{n}{p}}$$ which in turns gives a bound on the low frequencies in terms of a convergent geometric sum $$\sum_{k=0}^{-\infty}\| P_k u \|_{L^\infty} \le C \sum_{k=0}^{-\infty} 2^{k \frac{n}{p}} = C_1 < \infty. $$

To deal with the high frequency terms, we use the part of the Littlewood-Paley decomposition of $C^\gamma$ which states that $$ \sup_{k \in \Bbb Z} 2^{k \gamma} \| P_k u \|_{L^\infty} \le C' \| u \|_{C^\gamma} $$

This leads to a bound on the high frequencies in terms of another convergent geometric sum:

$$\sum_{k=1}^{\infty}\| P_k u \|_{L^\infty} \le \sum_{k=1}^{\infty} 2^{-k \gamma} C' \| u \|_{C^\gamma} = C' \sum_{k=1}^{\infty} 2^{-k \gamma} = C_2 < \infty$$

Together these shows that $\| u \|_{L^\infty} \le C_1 + C_2$.

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