# How to prove the Littlewood-Paley characterization of Holder norm?

Here our aim is to prove that the Holder norm $$$$\|f\|_{\Lambda_\gamma}:=\|f\|_{L^{\infty}}+\sup _{x \neq y \in \mathbb{R}^n} \frac{|f(x)-f(y)|}{|x-y|^\gamma}$$$$ is equivalent to $$$$\|f\|_{\Lambda_\gamma} \approx\left\|P_{\leq 0} f\right\|_{L^{\infty}}+\sup_{k>0} 2^{k \gamma}\left\|P_k f\right\|_{L^{\infty}},$$$$ where $$P_k$$ is the Littlewood-Paley projection. I am now stuck in the proof of $$$$\|f\|_{L^{\infty}}+\sup _{x \neq y \in \mathbb{R}^n} \frac{|f(x)-f(y)|}{|x-y|^\gamma} \lesssim \left\|P_{\leq 0} f\right\|_{L^{\infty}}+\sup_{k>0} 2^{k \gamma}\left\|P_k f\right\|_{L^{\infty}}.$$$$ In fact, we let $$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.

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

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.
• 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. Mar 18, 2023 at 13:11