# Tag Info

## New answers tagged harmonic-analysis

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With $\sum_{\nu \ge 0}\phi_\nu(\xi)=1$ be a Littlewood-Paley partition of unity we find that $u=\sum_{\nu \ge 0}\phi_\nu(D)u$ and thus since $$\Vert u\Vert_{B^\alpha_{\infty, \infty}}=\sup_{\nu\in \mathbb N} 2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty},$$ we get $$\Vert u\Vert_{L^\infty}\le \sum_{\nu \ge 0}2^{\nu \alpha}\Vert\phi_\nu(D)u\Vert_{L^\infty}... 5 Here are just a few observations to discard some trivial cases. I'll talk directly about the Istratescu formulation and will prefer the probabilistic language, but that shouldn't be a problem after everything that has been posted already. First of all, the problem is equivalent to asking what is the best constant in the inequality E[|X-X'|^p]^{1/p}\le C_{p,... 5 Let me prove that such constant C always exists. It is not hard to find such \alpha, \beta that the inequality$$x^p\leqslant \alpha x^q+\beta$$holds for all positive x and turns into equality if and only if x=2. Then$$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\beta.$$Since all three differences |f-g|, |g-h|, |f-h| ... 3 The solution here mainly follows the lines of the previous answer. For brevity, write p for p(x). The condition that p is bounded from above will not be used. We will only use the condition p\ge1. If f(t):=t^p\ln(e+t) for p=p(x)\ge1 and all real t\ge0, then \begin{equation*} f^{-1}(u)\approx g(u):=\frac{u^{1/p}}{\ln^{1/p}(e+u^{1/p})} \tag{1}... 0 As a supplement and extension of Iosif's answer, let me mention that, more generally, if$$f(t) \sim t^{pq} (\ell(t^q))^p \qquad \text{as } t \to \infty$$for a slowly varying function \ell, then$$f^{-1}(s) \sim s^{1/(pq)} (\ell^\#(s^{1/p}))^{1/q} \qquad \text{as } s \to \infty,$$where \ell^\# is another slowly varying function: the de Bruijn conjugate ... 12 The optimal exponent is k. Such examples are given by sparse power series. This is actually trivial in the case k=0 (which was not included in the OP). Then we can simply take f(z)=\sum j^{-2} z^{N(j)}, say. This is obviously bounded, and the coefficients a_n will not satisfy |a_n|\lesssim n^{-\epsilon} for any \epsilon>0 if N(j) increases ... 1 If f(t):=t^p\ln(e+t) for some real p>0 and all real t\ge0, then \begin{equation*} f^{-1}(u)\approx g(u):=\frac{u^{1/p}}{\ln^{1/p}(e+u^{1/p})} \label{1}\tag{1} \end{equation*} for real u\ge0, where the symbol \approx is used in the sense defined in your post. Indeed, the function f\colon[0,\infty)\to[0,\infty) is continuous and strictly ... 1 Come on, for the first property just use the fact that$$ \int_{-n}^n K(x) dx = \sum_{j = -n+1}^{n-1} \phi(j) + \tfrac12(\phi(-n)+\phi(n)) $$has a finite limit as n \to \infty, together with convergence of$$ \biggl| \int_{-a}^a K(x) dx - \int_{-\lfloor a\rfloor}^{\lfloor a\rfloor} K(x) dx \biggr| \leqslant |\phi(-\lfloor a\rfloor-1)| + |\phi(-\lfloor a\...

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Maybe here there is some usefull information, I'm reaserching these same field but still learning https://arxiv.org/pdf/0805.2531.pdf

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Make a sequence of functions of period $2\pi$ whose limit is of period $\pi$. Your map is discontinuous.

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