Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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?

share|cite|improve this question
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

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 $\| u \|_{L^\infty} \le C_1 + C_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.