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Consider the Littlewood-Paley decomposition of a tempered distribution $f$: \begin{equation} f = \tilde{\theta}(D)f + \sum_{k \geq 1} \theta(2^{-k}D)f. \end{equation} Here $(\tilde{\theta}, \theta)$ are smooth compactly supported functions ($\theta$ having a support contained in an annulus far from zero), satisfying the following dyadic partition of unity relation: \begin{equation} \tilde{\theta}(\xi) + \sum_{k \geq 1} \theta(2^{-k}\xi) = 1, \quad \forall \xi \in \mathbb{R}. \end{equation} In addition, $\tilde{\theta}(D)$ (as well as $\theta(2^{-k}D)$) are Fourier multipliers (i.e., they are defined by $\mathcal{F}(\tilde{\theta}(D)f)(\xi) = \tilde{\theta}(\xi)f(\xi)$ for all $\xi \in \mathbb{R}$). For $0 < \alpha < 1$, consider $C^{0, \alpha}(\mathbb{R})$ the space of continous and bounded functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the Holder condition, i.e. \begin{equation} \exists c > 0 : \forall (x, y) \in \mathbb{R} \times \mathbb{R}, \quad |f(x) - f(y)| \leq C |x-y|^{\alpha}. \end{equation} Assume that $f$ is continous and bounded and \begin{equation} \| f \|_{C^{0, \alpha}}^{LP} := \| \tilde{\theta}(D)f\|_{L^{\infty}} + \sup_{k \in \mathbb{N}} 2^{k\alpha} \|\theta(2^{-k}D)f \|_{L^{\infty}} < +\infty. \end{equation} I am looking to prove the following estimates. Fix $x, y \in \mathbb{R}$ and let $N \in \mathbb{N}$ be arbitrary. First,

\begin{equation} \sum_{k > N} 2 \|\theta(2^{-k}D)f\|_{L^{\infty}} \leq \frac{\| f \|_{C^{0, \alpha}}^{LP}}{2^{\alpha}-1} \frac{1}{2^{\alpha N}}. \end{equation}

Second,

\begin{equation} \Big|\tilde{\theta}(D)f(x) - \tilde{\theta}(D)f(y) + \sum_{k=1}^{N}[\theta(2^{-k}D)f(x) - \theta(2^{-k}D)f(y)] \Big| \leq C|x-y|\sum_{k=1}^{N} 2^{k(1-\alpha)}\| f \|_{C^{0, \alpha}}^{LP} \end{equation}

for some $C > 0$. Concering the first estimate, I "went in brutally" and I manage to obtain a bound like $\| f \|_{C^{0, \alpha}}^{LP} \sum_{k > N} 2^{1-k\alpha}$, so I don't know how to go further. For the second estimate, I sense an impending use of semi-classical Bernstein inequalities but I fail to see how to apply it to obtain such a bound. Any hint is welcomed.

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1 Answer 1

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I have succeeded (sort of, modulo a factor 2) in proving the first estimate. Recall that when $|q| < 1$, the remainder of the associated geometric series writes \begin{equation} \sum_{k=n}^{\infty} q^k = \frac{q^n}{1-q}. \end{equation} Thus, \begin{align} \sum_{k>N}^\infty 2 \| \theta(2^{-k}D)f\|_{L^{\infty}} &= 2\sum_{k = N+1}^\infty \frac{2^{k\alpha}}{2^{k\alpha}} \| \theta(2^{-k}D)f\|_{L^{\infty}} \\ &\leq 2 \sup_{k \in \mathbb{N}} 2^{k\alpha} \| \theta(2^{-k}D)f\|_{L^{\infty}} \sum_{k=N+1}^\infty \frac{1}{2^{k\alpha}} \\ &\leq 2\|f\|_{C^{0, \alpha}}^{LP} \frac{2^{-(N+1)\alpha}}{1-2^{-\alpha}} = \frac{2\|f\|_{C^{0, \alpha}}^{LP}}{2^{\alpha}-1}\frac{1}{2^{\alpha N}}. \end{align} I still don't know how to prove the second inequality.

Edit: I may have found a way to prove the second inequality (using a mean value inequality then applying some of the hypotheses) - I'll write it down as soon as I have some time.

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