RUS  ENG
Полная версия
ВИДЕОТЕКА
Международная конференция по функциональным пространствам и теории приближения функций, посвященная 110-летию со дня рождения академика С. М. Никольского
25 мая 2015 г. 17:05, г. Москва, МИАН

Norm convolution inequalities in $L_p$

E. D. Nursultanova, S. Yu. Tikhonovb, N. T. Tleukhanovac

a Kazakhstan Branch of Lomonosov Moscow State University
b Centre de Recerca Matemàtica
c L. N. Gumilev Eurasian National University

Аннотация: Let $1\leq p\leq\infty$, $L_p\equiv L_p(\mathbb {R})$ and let the convolution operator be given by
$$ (Af)(x)=(K*f)(x)=\int_{{\mathbb R}} K(x-y) f(y) dy, \qquad K\in L_{\text{loc}}. $$

Let $d>0$ and let
– $M_1$ be the set of all intervals of length $\leq d$;
– $M_2$ be the set of all measurable sets $e\subset[-d,d]$ such that $\operatorname{diam}(e)=\sup_{x,y\in e}|x-y|\leq d$;
– $W_1$ be the set of all finite arithmetic progressions of integer numbers;
– $W_2$ be the set of all finite sets $w\subset{\mathbb Z}$ such that $\min_{i,j\in w}|i-j|\geq 2$.
Now we define the sets $\mathfrak{L}_{d}, \mathfrak{U}_{d}, \mathfrak{V}_{d}$ as follows:
\begin{align*} \mathfrak{L}_d&=\biggl\{E=\bigcup_{k\in w}(e+kd): e\in M_1, \, w\in W_1\biggr\}, \\ \mathfrak{U}_d&= \biggl\{E= \bigcup_{k\in w}(e_k+kd): e_k\in M_2, \, w\in W_2, \, |e_k|=|e_j|, \, k,j\in w \biggr\}, \\ \mathfrak{V}_{d}&= \biggl\{E=\bigcup_{x\in e}(x+w(x)d): e\in M_2, \,w(x)\in W_2, \, |w(x)|=|w(y)|, \, x,y\in e\biggr\}, \end{align*}
where $|e|$ is the measure of a set $e\in M_i$ and $|w|$ is the number of elements of $w \in W_i$. Note that $\mathfrak{L}_d\subset\mathfrak{U}_d\cap\mathfrak{V}_d$. If $E\in\mathfrak{L}_d$, then $|E|=|e||w|$, where $e$, $w$ are the sets from the representation of $E$. Similarly, this property holds for $E\in \mathfrak{U}_d$ and $E\in\mathfrak{V}_d$.
Theorem. Let $1<p<q<\infty$. If for some $d>0$ we have either
$$ \sup\limits_{E\in \mathfrak{U}_{d}}\frac{1}{|E|^{1/p-1/q}}\int_{E}|K(x)|\,dx\leq D $$
or
$$ \sup\limits_{E\in \mathfrak{V}_{d}}\frac{1}{|E|^{1/p-1/q}}\int_{E}|K(x)|\,dx\leq D, $$
then the operator $Af=K*f$ is bounded from $L_p(\mathbb R)$ to $L_q(\mathbb R)$ and
$$ \|A\|_{L_p\rightarrow L_q}\leq C(p,q) D, $$
where $C(p,q)$ depends on $p$ and $q$.
Theorem. Let $1<p<q<\infty$, $d>0$, and the operator $Af=K*f$ be bounded from $L_p({\mathbb R})$ to $L_q({\mathbb R})$. If for any $B>0$ we have
$$ \sup_{\substack{E\in \mathfrak{L}_d\\ |E|\leq B}}\frac{1}{|E|^{1/p-1/q}}\biggl|\int_{E}K(x)\,dx\biggr|\leq C(B)<\infty, $$
then
$$ \sup_{E\in \mathfrak{L}_d}\frac{1}{|E|^{1/p-1/q}}\biggl|\int_{E}K(x)\,dx\biggr|\leq C(p,q)\|A\|_{L_p\rightarrow L_q}. $$

Corollary. Let $1<p\leq q<\infty$ and $\lambda = 1-(\frac1p-\frac1q)$. Let also
$$ \mathcal{K}(x)= \frac{e^{i|x|^a}}{|x|^b}\mspace{2mu}, $$
where $a\neq 0,$ $a \neq 1$, and $b\neq \lambda$. If
$$ \max(q, p')>\frac{a}{\lambda-b}>0, $$
then the operator $Af=\mathcal{K}*f$ is not bounded from $L_p$ to $L_q$.

Язык доклада: английский

Список литературы
  1. R. O'Neil, “Convolution operators and $L(p, q)$ spaces”, Duke Math. J., 30 (1963), 129–142  crossref  mathscinet  zmath
  2. V. D. Stepanov, Some topics in the theory of integral convolution operators, Dalnauka, Vladivostok, 2000
  3. P. Sjölin, “Regularity of solutions to the Schrödinger equation”, Duke Math. J., 55:3 (1987), 699–715  crossref  mathscinet  zmath  isi

© МИАН, 2026