## New answers tagged harmonic-analysis

1

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\...

1

$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$The answer is yes to each of your two questions.
Let $a_n:=\phi(n)$. Then
\begin{equation*}
K(x)=\sum_{n\in\Z}a_n R(x-n).
\end{equation*}
Note that for all $j\in\Z$ we have $K(j)=a_j$ and $K$ linear (or, more exactly, affine) on the ...

1

$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$Let $a_n:=\phi(n)$. Then
\begin{equation}
K(x)=\sum_{n\in\Z}a_n 1(n-1/2\le x<n+1/2).
\end{equation}
So, $K(x)=a_0=0$ if $1/2\le x<1/2$. So, for $\ep\in(0,1/2)$,
\begin{equation}
I_\ep:=\int_{1/\ep<|x|<\ep}K(x)\,dx=\int_{|x|<\ep}K(x)\,dx
=\sum_{n\in\Z}a_n J_n,
\end{equation}
...

0

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

1

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

fa.functional-analysis ca.classical-analysis-and-odes probability-distributions harmonic-analysis periodic-functions

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