Here our aim is to prove that the Holder norm \begin{equation} \|f\|_{\Lambda_\gamma}:=\|f\|_{L^{\infty}}+\sup _{x \neq y \in \mathbb{R}^n} \frac{|f(x)-f(y)|}{|x-y|^\gamma} \end{equation} is equivalent to \begin{equation} \|f\|_{\Lambda_\gamma} \approx\left\|P_{\leq 0} f\right\|_{L^{\infty}}+\sup_{k>0} 2^{k \gamma}\left\|P_k f\right\|_{L^{\infty}}, \end{equation} where $P_k$ is the Littlewood-Paley projection. I am now stuck in the proof of \begin{equation} \|f\|_{L^{\infty}}+\sup _{x \neq y \in \mathbb{R}^n} \frac{|f(x)-f(y)|}{|x-y|^\gamma} \lesssim \left\|P_{\leq 0} f\right\|_{L^{\infty}}+\sup_{k>0} 2^{k \gamma}\left\|P_k f\right\|_{L^{\infty}}. \end{equation} In fact, we let $f(x) = P_{\le 0} f+ \sum_{k>0} P_k f$. The previous one $P_{\le 0} f$ is easy to be controlled. As for $\sum_{k>0} P_k f$, we found that \begin{align*} \frac{|P_kf(x)-P_k f(y)|}{|x-y|^\gamma} & \simeq \frac{|P_k^2f(x)-P_k^2 f(y)|}{|x-y|^\gamma} \\ & =\Big| \int \frac{m_k (x-z) -m_k(y-z)}{|x-y|^\gamma} P_kf(z) dz \Big| \\ & = \Big| \int \frac{|m_k (x-z) -m_k(y-z)|}{|x-y|^\gamma} dz \Big| \cdot \| P_k f \|_\infty, \end{align*} where $m_k(x)= 2^{kn} \phi (2^k x)$ and $\hat{\phi}$ is support on an annulus $1 \le |x| \le 4$. Then we see that \begin{align*} \int \frac{|m_k (x-z) -m_k(y-z)|}{|x-y|^\gamma} dz & = \int_{\mathbb{R}^N} \frac{|m_k (z+h) -m_k(z)|}{|h|^\gamma} dz \\ & \lesssim |h|^{1-\gamma}\int_{\mathbb{R}^N} \int_0^1 |\nabla m_k(z+th) dtdz \\ & \lesssim |h|^{1-\gamma} 2^{k}, \quad \text{where} h=y-x, \end{align*} moreover, we also have \begin{align*} \int \frac{|m_k (x-z) -m_k(y-z)|}{|x-y|^\gamma} dz & \lesssim |h|^{-\gamma}, \end{align*} so we see that \begin{align*} \int \frac{|m_k (x-z) -m_k(y-z)|}{|x-y|^\gamma} dz & \lesssim \min\{|h|^{1-\gamma} 2^k, |h|^{-\gamma} \} \simeq 2^{k \gamma}, \end{align*} hence we see that we can only get that \begin{align*} \frac{|P_kf(x)-P_k f(y)|}{|x-y|^\gamma} & = \Big| \int \frac{|m_k (x-z) -m_k(y-z)|}{|x-y|^\gamma} dz \Big| \cdot \| P_k f \|_\infty \lesssim 2^{k \gamma} \| P_k f \|_\infty, \end{align*} but then we can only get that \begin{align*} \sup_{x\not = y}\frac{|\sum_{k>0} P_k f(x)- \sum_{k>0} P_k f(y)|}{|x-y|^\gamma} \lesssim \sum_{k >0} 2^{k \gamma} \| P_k f \|_\infty, \end{align*} and we cannot further bound it by $\sup_{k > 0 }2^{k \gamma} \| P_k f \|_\infty$.

So I wonder is there any improvement about the proof? A simple idea is that I want to improve the estimate to \begin{align*} \frac{|P_kf(x)-P_k f(y)|}{|x-y|^\gamma} & \lesssim 2^{k \gamma'} \| P_k f \|_\infty, \end{align*} for some $\gamma' < \gamma$. But roughly speaking, the above quotation term can be regarded as $D^\gamma P_k f(x)$, so it is almost equal to $2^{k \gamma} P_k(x)$, which I am a little confused.

  • $\begingroup$ From memory, there’s a proof of this in PDEs by Taylor which you may find helpful $\endgroup$
    – JackT
    Mar 17, 2023 at 0:24

1 Answer 1


Follow your proof, $$\sup_{h\ne 0}\frac{|\sum_{k>0}P_kf(x+h)-\sum_{k>0}P_kf(x)|}{|h|^\gamma} \le \sup_{h}\sum_{k>0} \min(|h|^{1-\gamma}2^k, |h|^{-\gamma})\Vert P_kf\Vert_\infty,$$ where one pick an $h$ uniformly for all coefficients $\min(|h|^{1-\gamma}2^k, |h|^{-\gamma})$. Then for a large $k$, this coefficient is far less than $2^{k\gamma}$. One can in fact show that $\sum_{k>0} \min(|h|^{1-\gamma}2^k2^{-k\gamma}, |h|^{-\gamma}2^{-k\gamma})$ is bounded.

  • $\begingroup$ That means using the first term in the minimum for the summation of the first few terms and using the second one for the summation of the rest. Nice proof! $\endgroup$
    – Chushamm
    Mar 18, 2023 at 13:00
  • $\begingroup$ Très bien! Thanks to Prof. Great Chushannn, who inspired me on this question! Follow this idea, it can be bounded by $(C_1 2^N h)^{1-\gamma} + ( C_2 2^N h)^{-\gamma}$ for some constants $C_1$ and $C_2$ and any $N$. We can easily see that it is bounded by picking N such that these two terms almost equal. $\endgroup$
    – Tianqqqi
    Mar 18, 2023 at 13:11

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