On spectral properties of the modified convolution operator

Abstract

We investigated the s-number of the modified convolution operator and obtained the following results

$$c_1\underset{Q?G}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|={?f?}_{M_p^q}=c_3\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(s)ds|,$$

where $1<p<2<q<\mathrm{8}$, $p^{\text{'}}=\frac{p}{p-1}$, G is a set of all segments Q from $[0,1]$, F is a set of all compacts from $[0,1]$, $|Q|$ is the measure of a set Q.

MSC:42A45, 44A35.

1 Introduction

Let $1=p<\mathrm{8}$, $0<q=\mathrm{8}$. We denote by ${\mathfrak{S}}_{p,q}$ the space of all compact operators A, acting in the space $L_2[0,1]$ of all 1-periodic functions square integrable on $[0,1]$ for s-numbers such that the following quasinorm is finite

$${?A?}_{{\mathfrak{S}}_{p,q}}={(\underset{m=1}{\overset{\mathrm{8}}{?}}{s}_m^q(A)m^{q/p-1})}^{1/q},$$

if $q<\mathrm{8}$, and

$${?A?}_{{\mathfrak{S}}_{p,\mathrm{8}}}=\underset{m}{sup}{m}^{\frac{1}{p}}{s}_m\mathtt{\text{if}}q=\mathrm{8}.$$

Recall that the sequence $s_m(A)$ (s-numbers of operator A) are numerated eigenvalues of the operator $\sqrt{A^{*}A}$.

We consider the convolution operator

$$(Af)(y)={?}_0^{1}K(x-y)f(x)dx$$

acting in $L_2[0,1]$. Given a function $f?L_1[0,1]$, we consider also the modified convolution operator

$$(A_{f}f)(y)={?}_0^1(Kf)(x-y)f(x)dx.$$

We say that f belongs to the space $M_{p_0,q_0}^{p_1,q_1}$, if for $A?{\mathfrak{S}}_{p_0,q_0}$, $A_f?{\mathfrak{S}}_{p_1,q_1}$ and

$${?A_f?}_{{\mathfrak{S}}_{p_1,q_1}}=c{?A?}_{{\mathfrak{S}}_{p_0,q_0}},$$

where $c>0$ depends only on $p_0$, $q_0$, $p_1$, $q_1$.

This means that the linear operator $R_f$ defined by the equality $R_f(A)=A_f$ is bounded from ${\mathfrak{S}}_{p_0,q_0}$ to ${\mathfrak{S}}_{p_1,q_1}$. Let

$${?f?}_{M_{p_0,q_0}^{p_1,q_1}}={?R_f?}_{{\mathfrak{S}}_{p_0,q_0}?{\mathfrak{S}}_{p_1,q_1}}.$$

Given that the eigenvalues of the operator $K*f$ coincide with the Fourier coefficients of the kernel K with respect to the trigonometric system, in the case $p_0=p_1=q_0=q_1=p$ this problem reduces to the well-known problem of Fourier series multipliers. Let $K?L_1([0,1])$ and ${\{a_m(K)\}}_{m?Z}$ be the sequence of its Fourier coefficients with respect to the trigonometric system ${\{e^{2pikx}\}}_{k?Z}$. It is assumed that K is such that ${\{a_m(K)\}}_{m?Z}?l_p$, $1=p=\mathrm{8}$. Let $T_f={\{a_m(Kf)\}}_{m?Z}?l_p$. The problem is to determine conditions on the function f ensuring the boundedness of the operator $T_f:l_p?l_p$.

This problem was considered in the works of Stechkin [1], Hirschman [2], Edelstein [3], Birman and Solomyak [4], Karadzhov [5], and others.

We obtain sufficient conditions on a multiplier f ensuring that it belongs to the space $M_{p_0,q_0}^{p_1,q_1}$. These conditions are expressed in terms of Lorentz and Besov spaces. We also construct examples showing the sharpness of the obtained constants for corresponding embedding theorems.

2 Main results

Let f be a µ measurable function which takes finite values almost everywhere and let

$$m(s,f)=\mu \left(\{x:x?[0,1],|f|>s\}\right)$$

be its distribution function. The function

$$f^{*}(t)=inf\{s:m(s,f)=t\}$$

is a nonincreasing rearrangement of f.

We say that a function f belongs to the Lorentz space $L_{p,q}$ if f is measurable and

$${?f?}_{L_{p,q}}={\left({?}_0^{\mathrm{8}}{(t^{\frac{1}{p}}{f}^{*}(t))}^q\frac{dt}{t}\right)}^{\frac{1}{q}}<\mathrm{8},$$

for $1=q<\mathrm{8}$ and

$${?f?}_{L_{p,\mathrm{8}}}=\underset{t>0}{sup}{t}^{\frac{1}{p}}{f}^{*}(t)<\mathrm{8},$$

for $q=\mathrm{8}$.

Theorem 1 Let$1<p_0<2=p_1$, $1=q_1=q_0=\mathrm{8}$, $\frac{1}{r}=\frac{1}{p_0}-\frac{1}{p_1}$, $\frac{1}{s}=\frac{1}{q_1}-\frac{1}{q_0}$and$f?L_{r,s}[0,1]$. If$A?{\mathfrak{S}}_{p_0,q_0}$, then$A_f?{\mathfrak{S}}_{p_1,q_1}$and

$${?A_f?}_{{\mathfrak{S}}_{p_1,q_1}}=c{?f?}_{L_{r,s}}{?A?}_{{\mathfrak{S}}_{p_0,q_0}},$$

i.e. $L_{r,s}[0,1]?M_{p_0,q_0}^{p_1,q_1}$.

In the following theorem the cases $p=p_o=q_0$, $q=p_1=q_1$ are considered. The upper and the lower estimates of the norm ${?f?}_{M_p^q}$ ($M_p^q:=M_{p,p}^{q,q}$) are obtained.

Theorem 2 Let$1<p<2<q<\mathrm{8}$, $p^{\text{'}}=\frac{p}{p-1}$. Let G be a set of all segments Q from$[0,1]$, F be a set of all compacts from$[0,1]$, then

$$c_1\underset{Q?G}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|={?f?}_{M_p^q}=c_3\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(s)ds|,$$

where$|Q|$is the measure of a set Q.

We shall define the class of generalized monotone functions for which the upper and the lower estimates coincide.

We say that function f is a generalized monotone function, if there exists a constant $c>0$ such that for every $x?(0,1]$ the inequality

$$|f(x)|=\frac{c}{x}|{?}_0^{x}f(y)dy|$$

holds. The class of such functions is denoted by $\mathfrak{N}$.

Corollary 1 Let$1<p<2<q<\mathrm{8}$. If$f?\mathfrak{N}$, then$f?M_p^q$if and only if

$$\underset{t>0}{sup}{t}^{\frac{1}{p}-\frac{1}{q}}{f}^{*}(t)<\mathrm{8}.$$

Moreover, ${?f?}_{M_p^q}~{sup}_{t>0}{t}^{\frac{1}{p}-\frac{1}{q}}{f}^{*}(t)$.

In case parameters $p_0$, $p_1$ are both either less or greater than 2, we use the space of smooth functions.

Let $1=p<\mathrm{8}$, $a>0$. We denote by $B_{p,q}^a[0,1]$ the space of all measurable functions f on $[0,1]$ such that

$${?f?}_{B_{p,q}^a}={\left(\underset{k=0}{\overset{\mathrm{8}}{?}}{\left(2^{ak}{?{\mathrm{?}}_{k}f?}_p\right)}^q\right)}^{\frac{1}{q}}<\mathrm{8}$$

for $1=q<\mathrm{8}$, and

$${?f?}_{B_{p,\mathrm{8}}^a}=\underset{k}{sup}{2}^{ak}{?{\mathrm{?}}_{k}f?}_p<\mathrm{8}$$

for $q=\mathrm{8}$. Here ${\{a_m(f)\}}_{m?N}$ are the Fourier coefficients of the function f by trigonometric system ${\{e^{2pikx}\}}_{k?\mathbb{Z}}$, ${\mathrm{?}}_{k}f={\mathrm{?}}_{k}f(x)={?}_{[2^{k-1}]=|m|<2^k}{a}_m(f)e^{2pimx}$, and $[2^{k-1}]$ is the integer part of $2^{k-1}$.

This class is called the Nikol’skii-Besov space.

Theorem 3 Let$1<p_0=p_1<\mathrm{8}$, $2?[p_0,p_1]$, $1<q_0=q_1=\mathrm{8}$,

$$a=\underset{x?[\frac{1}{p_1},\frac{1}{p_0}]}{min}|\frac{1}{2}-x|,\frac{1}{r}=\underset{x?[\frac{1}{p_1},\frac{1}{p_0}]}{max}|\frac{1}{2}-x|,1-\frac{1}{s}=\frac{1}{q_0}-\frac{1}{q_1}$$

and$f?B_{r,s}^a[0,1]$.

If$A?{\mathfrak{S}}_{p_0,q_0}$, then$A_f?{\mathfrak{S}}_{p_1,q_1}$and

$${?A_f?}_{{\mathfrak{S}}_{p_1,q_1}}=c{?f?}_{B_{r,s}^a}{?A?}_{{\mathfrak{S}}_{p_0,q_0}},$$

i.e., $B_{r,s}^a?M_{p_0,q_0}^{p_1,q_1}$.

In the case $p_0=p_1=q_0=q_1$, Karadzhov’s result (see [5]) follows from Theorem 3:

$$B_{r,1}^{\frac{1}{r}}?M_p=M_{p,p}^{p,p},\frac{1}{r}=|\frac{1}{p}-\frac{1}{2}|.$$

Now consider the case $1=q_1<q_0=\mathrm{8}$.

Theorem 4 Let$1<p_0<p_1<\mathrm{8}$, $1=q_1<q_0=\mathrm{8}$, $2?(p_0,p_1)$, $\frac{1}{r}-a=\frac{1}{p_0}-\frac{1}{p_1}$, $\frac{1}{s}=\frac{1}{q_1}-\frac{1}{q_0}$, $a>{min}_{x?[\frac{1}{p_1},\frac{1}{p_0}]}|\frac{1}{2}-x|$.

Then$B_{r,s}^a[0,1]?M_{p_0,q_0}^{p_1,q_1}$.

3 Properties of $M_{p_0,q_0}^{p_1,q_1}$ class

To prove the properties of $M_{p_0,q_0}^{p_1,q_1}$ class we need the following lemma. We first define a discrete Lorentz space. $l_{pq}$ is called a discrete Lorentz space whose elements are sequences of numbers $?={\{{?}_k\}}_{k=\mathrm{8}}^{\mathrm{8}}$ with the only limit point 0 such that

$${???}_{l_{pq}}={\left(\underset{m=1}{\overset{\mathrm{8}}{?}}{\left|{?}_m^{*}\right|}^q{m}^{\frac{q}{p}-1}\right)}^{\frac{1}{q}},1=q<\mathrm{8}$$

where ${\{{?}_m^{*}\}}_{m=1}^{\mathrm{8}}$ nonincreasing rearrangement of the sequence ${\{|{?}_k|\}}_{k=\mathrm{8}}^{\mathrm{8}}$.

For $q=\mathrm{8}$,

$${???}_{l_{p\mathrm{8}}}=\underset{m}{sup}{m}^{\frac{1}{p}}{?}_m^{*}.$$

Lemma 1 (See [6])

Let$1<r,p_0,p_1<\mathrm{8}$, $1=q_0,q_1,s=\mathrm{8}$. Then

$${?a*b?}_{l_{p_1,q_1}}=c{?b?}_{l_{r,s}}{?a?}_{l_{p_0,q_0}},$$

where$\frac{1}{p_1}+1=\frac{1}{r}+\frac{1}{p_0}$, $\frac{1}{q_1}=\frac{1}{s}+\frac{1}{q_0}$.

Let $\overline{X}=(X_0,X_1)$, where $X_0$, $X_1$ are Banach spaces, be a compatible pair. We define the functional $K(t,a)$ for $t>0$ and $a?X_0+X_1$ by the following formula:

$$K(t,a)=\underset{a=a_0+a_1}{inf}({?a_0?}_{X_0}+t{?a_1?}_{X_1}).$$

We denote by ${\overline{X}}_{?,q,k}$ the space $\{a?X_0+X_1:{?a?}_{?,q,k}={\mathrm{F}}_{?,q}(K(t,a))\}$, where ${\mathrm{F}}_{?,q}$ is a functional defined on nonnegative functions f by formula

$${\mathrm{F}}_{?,q}(f(t))={\left({?}_0^{\mathrm{8}}{(t^{-?}f(t))}^q\frac{dt}{t}\right)}^{\frac{1}{q}},1=q<\mathrm{8}$$

and

$${\mathrm{F}}_{?,\mathrm{8}}(f(t))=\underset{t>0}{sup}{t}^{-?}f(t),q=\mathrm{8}.$$

Let $X_{a_1^0,p_1^0}$ and $X_{a_2^1,p_2^1}$ be the spaces obtained by the method of real interpolation of Banach pairs of spaces $(X_0^1,X_1^1)$, $(X_0^2,X_1^2)$ respectively.

Lemma 2 (See [7])

Let$0<a_i,{ß}_i<1$, $1=p_i,q_i=\mathrm{8}$, $i=0,1$, $a_0?a_1$, ${ß}_0?{ß}_1$. If T is a bilinear operator:

$$T:X_{a_0,p_0}\times Y_0?Z_{{ß}_0,q_0}$$

and

$$T:X_{a_1,p_1}\times Y_1?Z_{{ß}_1,q_1}$$

then

$$T:X_{a,p}\times Y_{?,r}?Z_{ß,q}.$$

Here$a=(1-?)a_0+?a_1$, $ß=(1-?){ß}_0+?{ß}_1$, $\frac{1}{p}+\frac{1}{r}>1$, $1+\frac{1}{q}=\frac{1}{p}+\frac{1}{r}+(1-?)(\frac{1}{q_0}-\frac{1}{p_0})+?{(\frac{1}{q_1}-\frac{1}{p_1})}_{+}$, $x_{+}=max(x,0)$.

Remark Since the s-numbers of convolution operator A coincide with the modules of the Fourier coefficients of the kernel K, the problem of estimating the s-numbers of “transformed” operator $A_f$ can be reduced to the study of the following inequality

$${?a*b?}_{l_{p_1,q_1}}=c{?a?}_{l{p}_0,q_0},$$

(1)

and we have to describe the class of those functions f with Fourier coefficients $b={\{b_m\}}_{m?Z}$, for which Inequality (1) holds.

Theorem 5

• (1) Let$1=p_0,p_1<\mathrm{8}$, $1=q_0,q_1=\mathrm{8}$, $\frac{1}{p_i}+\frac{1}{p_i^{\text{'}}}=\frac{1}{q_i}+\frac{1}{q_i^{\text{'}}}=1$, $i=0,1$. Then

  $$M_{p_0,q_0}^{p_1,q_1}=M_{p_1^{\text{'}},q_1^{\text{'}}}^{p_0^{\text{'}},q_0^{\text{'}}}.$$

• (2) Let$1<p_0<r_0<p_1^{\text{'}}<\mathrm{8}$, $\frac{1}{p_1}+\frac{1}{p_1^{\text{'}}}=1$, $\frac{1}{p_1}-\frac{1}{p_0}=\frac{1}{r_1}-\frac{1}{r_0}$, then

  $$M_{p_0,q_0}^{p_1,q_1}?M_{r_0,s}^{r_1,t},$$

where$\frac{1}{t}-\frac{1}{s}={(\frac{1}{q_1}-\frac{1}{q_0})}_{+}$.

Proof The proof of the first statement follows from Remark and from the fact that $?{(T_f)}^{*}?=?T_{\overline{f}}?$, where $\overline{f}$ is a complex conjugate of the function f. Now we prove (2).

Let $f?M_{p_0,q_0}^{p_1,q_1}$, then by (1) it follows that $f?M_{p_1^{\text{'}},q_1^{\text{'}}}^{p_0^{\text{'}},q_0^{\text{'}}}$, and

where $\frac{1}{p_i}+\frac{1}{p_i^{\text{'}}}=\frac{1}{q_i}+\frac{1}{q_i^{\text{'}}}=1$. According to Lemma 1, the operator $T(a,f)=a*b$

$$T:l_{p_0,q_0}\times M_{p_0,q_0}^{p_1,q_1}?l_{p_1,q_1}$$

is bounded. Using (1) we have

$$T:l_{p_1^{\text{'}},q_1^{\text{'}}}\times M_{p_0,q_0}^{p_1,q_1}?l_{p_0^{\text{'}},q_0^{\text{'}}}.$$

Further, applying the theorem on bilinear interpolation (Lemma 2) we find that the operator

$$T:l_{r_0,s}\times M_{p_0,q_0}^{p_1,q_1}?l_{r_1,t}$$

is also bounded, i.e., $M_{p_0,q_0}^{p_1,q_1}?M_{r_0,s}^{r_1,t}$, where

$$\frac{1}{r_1}=\frac{1-?}{p_1}+\frac{?}{p_0^{\text{'}}},\frac{1}{r_0}=\frac{1-?}{p_0}+\frac{?}{p_1^{\text{'}}},\frac{1}{t}-\frac{1}{s}={(\frac{1}{q_1}-\frac{1}{q_0})}_{+}$$

for every $0<?<1$. Eliminating ? from this equation, we obtain that

$$\frac{1}{p_1}-\frac{1}{p_0}=\frac{1}{r_1}-\frac{1}{r_0},$$

and the condition $0<?<1$ implies the condition $1<p_0<r_0<p_1^{\text{'}}<\mathrm{8}$, where $\frac{1}{p_1}+\frac{1}{p_1^{\text{'}}}=1$.

The proof is complete. ?

By (2), in particular, the following proposition follows.

Let $1<p<r<p^{\text{'}}<\mathrm{8}$, $\frac{1}{p}+\frac{1}{p^{\text{'}}}=1$, then

$$M_{p,q}?M_{r,t},$$

where $M_{p,q}=M_{p,q}^{p,q}$ and $q,t?[1,\mathrm{8}[$ are any.

4 Proof of main results

For a given pair $\overline{X}=(X_0,X_1)$ we consider the space $G(\overline{X})$ consisting of all functions f bounded and continuous in the strip

$$S=\{z:0=Rez=1\}$$

with values in $X_0+X_1$. Moreover, f are analytic in the open strip

$$S_0=\{z:0<Rez<1\}$$

and such that the mapping $t?f(j+it)$ ($j=0,1$) is a continuous function on the real axis with values in $X_j$ ($j=0,1$) which tends to 0 for $|t|?\mathrm{8}$. It is clear that $G(\overline{X})$ is a vector space. We endow G with the norm

$${?f?}_G=max(\underset{t}{sup}{?f(it)?}_{X_0},\underset{t}{sup}{?f(1+it)?}_{X_1}).$$

The space ${\overline{X}}_{[?]}$$0=?=1$ consists of all elements $a?X_0+X_1$ such that $a=f(?)$ for some function $f?G(\overline{X})$. The norm on ${\overline{X}}_{[?]}$ is equal to

$${?a?}_{[?]}=inf\{{?f?}_G:f(?)=a,f?G\}.$$

In order to prove our main result, we need two lemmas in [8].

Lemma 3 (Bilinear interpolation, the complex method, see [8])

Let T be a bilinear operator such that

$$T:X_0\times Y_0?Z_0$$

and

$$T:X_1\times Y_1?Z_1.$$

Then

$$T:X_{[?]}\times Y_{[?]}?Z_{[?]},$$

where$X_{[?]}$, $Y_{[?]}$, $Z_{[?]}$are the spaces obtained by the method of complex interpolation of Banach pairs of spaces$(X_0,X_1)$, $(Y_0,Y_1)$, $(Z_0,Z_1)$respectively.

Lemma 4 (Bilinear interpolation, the real method, see [8])

Let T be a bilinear operator such that

$$T:X_0\times Y_0?Z_0$$

and

$$T:X_1\times Y_1?Z_1$$

with the norms$B_0$, $B_1$respectively. Then

$$T:X_{?,t_1}\times Y_{?,t_2}?Z_{?,s},$$

where$\frac{1}{s}+1=\frac{1}{t_1}+\frac{1}{t_2}$. Moreover,

$$?T?=c{B}_0^{1-?}{B}_1^{?}.$$

Proof of Theorem 1 First we prove the inequality:

$${?a*b?}_{l_{p_1,q_1}}=c{?f?}_{L_r}{?a?}_{l_{p_0,q_0}},$$

(2)

where $b={\{b_k\}}_{k?\mathbb{Z}}$ are Fourier coefficients of the function f.

If $r=2$, Inequality (2) follows by Lemma 1 and Hardy-Littlewood-Paley inequality [9]. Indeed, since $f?L_{rs}$, by the Hardy-Littlewood-Paley theorem, we have $b?l_{r^{\text{'}}s}$ and the following inequality holds

$${?b?}_{l_{r^{\text{'}}s}}=c{?f?}_{L_{rs}}.$$

Taking $s=r$, we get

$${?b?}_{l_{r^{\text{'}},r}}=c{?f?}_{L_r}.$$

Now let $2<r<\mathrm{8}$. Let $a?l_2$$f~{?}_{k?Z}{a}_k{e}^{2pikx}$, then by Parseval’s equality we get

$${?a*b?}_{l_2}={?ff?}_{L_2}={?f?}_{L_2}{?f?}_{L_{\mathrm{8}}}={?f?}_{L_{\mathrm{8}}}{?a?}_{l_2},$$

i.e., $M_2=L_{\mathrm{8}}$. From Lemma 1, using Parseval’s equality we have

$${?a*b?}_{l_{p_1,q_1}}=c{?f?}_{L_2}{?a?}_{l_{p_0,q_0}},$$

where $\frac{1}{p_1}+1=\frac{1}{2}+\frac{1}{p_0}$$\frac{1}{p_1}+\frac{1}{2}=\frac{1}{p_0}$$\frac{1}{q_1}=\frac{1}{q_0}+\frac{1}{2}$.

Thus, for the bilinear operator $T(a,f)=a*b$ we obtain

Applying the method of complex interpolation (Lemma 3), we obtain Inequality (2). Now we shall prove the inequality

$${?a*b?}_{l_{p_1,q_1}}=c{?f?}_{L_{r,s}}{?a?}_{l_{p_0,q_0}},$$

(3)

where $\frac{1}{s}=\frac{1}{q_1}-\frac{1}{q_0}$.

Let $q_0=\mathrm{8}$ and $p_0$ be fixed in Inequality (2). Taking $\frac{1}{q_1^i}=\frac{1}{r_i}$$i=0,1$, choose parameters $r_0$$r_1$$p_1^0$$p_1^1$ such that

$$\frac{1}{p_0}=\frac{1}{p_1^i}+\frac{1}{r_i},i=0,1.$$

(4)

Then from Inequality (2) we have

Using Marcinkiewicz-Calderón interpolation theorem (see [8]), we get

$${?a*b?}_{l_{p_1,s}}={\left(c_1{?a?}_{l_{p_0,\mathrm{8}}}\right)}^{?}{\left(c_2{?a?}_{l_{p_0,\mathrm{8}}}\right)}^{1-?}{?f?}_{L_{r,s}}=c{?a?}_{l_{p_0,\mathrm{8}}}{?f?}_{L_{r,s}},$$

(5)

where $\frac{1}{p_1}=\frac{1-?}{p_1^0}+\frac{?}{p_1^1}$$\frac{1}{r}=\frac{1-?}{r_0}+\frac{?}{r_1}$, i.e., $\frac{1}{p_0}-\frac{1}{p_1}=\frac{1}{r}$.

Now we apply Lemma 2 with fixed parameters r, s and parameters $p_1^i$, $p_0^i$, $i=0,1$ satisfying (4) and the inequality of type (5). We have:

$${(L_{r,s},L_{r,s})}_{?,1}\times {(l_{p_0^0,\mathrm{8}},l_{p_0^1,\mathrm{8}})}_{?,q_0}?{(l_{p_1^0,s},l_{p_1^1,s})}_{?,q_1}$$

or

$$T:L_{r,s}\times l_{p_0,q_0}?l_{p_1,q_1},$$

where $\frac{1}{q_1}-\frac{1}{q_0}=\frac{1}{s}-\frac{1}{\mathrm{8}}$, $\frac{1}{p_1}=\frac{1-?}{p_1^0}+\frac{?}{p_1^1}$, $\frac{1}{p_0}=\frac{1-?}{p_0^0}+\frac{?}{p_0^1}$, i.e., $\frac{1}{q_1}=\frac{1}{s}+\frac{1}{q_0}$, $\frac{1}{p_0}-\frac{1}{p_1}=\frac{1}{r}$.

Since the parameters $p_1^i$, $p_0^i$, $i=0,1$ are arbitrary in Inequality (5), it guarantees the arbitrary of the corresponding parameters in Inequality (4).

Thus, the following inequality holds:

$${?a*b?}_{l_{p_1,q_1}}=c{?a?}_{l_{p_0,q_0}}{?f?}_{L_{r,s}},$$

where $b={\{b_m\}}_{m?Z}$ are Fourier coefficients of the function f and $\frac{1}{r}=\frac{1}{p_0}-\frac{1}{p_1}$, $\frac{1}{s}=\frac{1}{q_1}-\frac{1}{q_0}$. According to Remark, this inequality is equivalent to the statement of Theorem 1. ?

Proof of Theorem 2 Let $f?M_p^q$ and Q be an arbitrary segment in $[0,1]$,

$$f_0(x)=\{\begin{array}{cc}1,\hfill & x?Q,\hfill \\ 0,\hfill & x?Q.\hfill \end{array}$$

Note that by Boas theorem [10] (see also [11]) we get

$${?\widehat{f_0}?}_{l_p}~{?f_0?}_{L_{p^{\text{'}},p}}={\left({?}_0^1{(t^{\frac{1}{p^{\text{'}}}}{f}_0^{*}(t))}^p\frac{dt}{t}\right)}^{\frac{1}{p}}={|Q|}^{\frac{1}{p^{\text{'}}}}.$$

(6)

Applying Theorem 5 from [12] and using (6), we obtain:

$$\begin{array}{rcl}{?f?}_{M_p^q}& =& \underset{f?0}{sup}\frac{{?\widehat{ff}?}_{l_q}}{{?\hat{f}?}_{l_p}}=\frac{{?\widehat{f_{0}f}?}_{l_q}}{{?\widehat{f_0}?}_{l_p}}\\ =& \frac{c}{{|Q|}^{\frac{1}{p^{\text{'}}}}}{?}_0^1{\left(t^{\frac{1}{q^{\text{'}}}}{\left(\underset{|W|=t,W?G}{sup}\frac{1}{|W|}\right|{?}_W{f}_0(x)f(x)dx\left|\right)}^q\frac{dt}{t}\right)}^{\frac{1}{q}}\\ =& \frac{c}{{|Q|}^{\frac{1}{p^{\text{'}}}}}\underset{t>0}{sup}{t}^{\frac{1}{q^{\text{'}}}}\left(\underset{|W|=t,W?G}{sup}\frac{1}{|W|}\right|{?}_{WnQ}f(x)dx\left|\right)\\ =& \frac{c_1}{{|Q|}^{\frac{1}{p^{\text{'}}}}}{|Q|}^{\frac{1}{q^{\text{'}}}-1}|{?}_{Q}f(x)dx|=\frac{c_1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|.\end{array}$$

Since the interval Q is arbitrary, we get

$${?f?}_{M_p^q}=c_1\underset{Q?G}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|,$$

where constants c and $c_1$ depend only on parameters p and q.

The proof of obtaining an upper estimate follows from Theorem 1 and the embedding $l_{p,p}?l_{p,q}$, for $p<q$.

Indeed, from Theorem 1 it follows

$$L_{r,\mathrm{8}}?M_p^q,$$

i.e.,

$$\begin{array}{rcl}{?f?}_{M_p^q}& =& c_2\underset{t>0}{sup}{t}^{\frac{1}{r}}{f}^{*}(t)=c_3\underset{t>0}{sup}\frac{1}{t^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}{?}_0^t{f}^{*}(s)ds\\ =& c_3\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}{?}_Q|f(x)|dx~\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|.\end{array}$$

?

Proof of Corollary 1 Let Q be an arbitrary compact from F.

From the condition of generalized monotonicity of the function f we have

$$\begin{array}{rcl}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(y)dy|& =& \frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}{?}_Q\frac{c}{y}|{?}_0^{y}f(x)dx|dy\\ =& \frac{c}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}\underset{A?G}{sup}\frac{1}{{|A|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{A}f(x)dx|{?}_Q\frac{dy}{y^{1-\frac{1}{q}-\frac{1}{p^{\text{'}}}}}\\ =& c_1\underset{A?G}{sup}\frac{1}{{|A|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{A}f(x)dx|.\end{array}$$

Taking into account that $Q?F$ is arbitrary, we have

$$\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|=c\underset{Q?G}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|.$$

Thus, from Theorem 2 we get

$${?f?}_{M_p^q}~\underset{Q?F}{sup}\frac{1}{{|Q|}^{\frac{1}{p^{\text{'}}}+\frac{1}{q}}}|{?}_{Q}f(x)dx|~\underset{t>0}{sup}{t}^{\frac{1}{p}-\frac{1}{q}}{f}^{*}(t).$$

?

Proof of Theorem 3 Let $2<p_0=p_1<\mathrm{8}$. For a sequence of numbers $a={\{a_m\}}_{m?Z}$ and a function $f?L_1[0,1]$ we consider the mapping T of the form $T(a,f)=a*b$, where $b={\{b_m\}}_{m?Z}$ is the sequence of Fourier coefficients on the trigonometric system of functions f. This map is bilinear and from Karadzhov’s theorem [5] and Remark it follows that it is bounded from $l_1\times B_{2,1}^{\frac{1}{2}}$ to $l_1$.

Since $M_2=L_{\mathrm{8}}$, the mapping

$$T_f:l_2\times L_{\mathrm{8}}?l_2$$

is also bounded. Thus, for the operator T, the following is true

Then, by Lemma 4 on bilinear interpolation, we get

$${(l_1,l_2)}_{?,q}\times {(B_{2,1}^{\frac{1}{2}},L_{\mathrm{8}})}_{?,1}?{(l_1,l_2)}_{?,q},$$

i.e., the operator T is bounded from $l_{p,q}\times {(B_{2,1}^{\frac{1}{2}},L_{\mathrm{8}})}_{?,1}$ to $l_{p,q}$. In the paper [5] it is shown that $B_{r,1}^{\frac{1}{r}}?{(B_{2,1}^{\frac{1}{2}},L_{\mathrm{8}})}_{?,1}$, where $\frac{1}{r}=\frac{1-?}{2}$. Thus, taking into account Theorem 5, we will get

$$T:l_{p,q}\times B_{r,1}^{\frac{1}{r}}?l_{p,q},$$

(7)

for $2<p<\mathrm{8}$$1=q=\mathrm{8}$$\frac{1}{r}=\frac{1}{2}-\frac{1}{p}$.

From Minkowski’s inequality and Parseval’s equality we get

$$T:l_1\times L_2?l_2.$$

Thus, for the operator T, the following is true

Then, by Lemma 3 we get

$${(l_{p,q},l_1)}_{[?]}\times {(B_{r,1}^{\frac{1}{r}},L_2)}_{[?]}?{(l_{p,q},l_2)}_{[?]}$$

i.e., T is a bounded mapping from $l_{p_0,q_0}\times B_{r,s}^a$ to $l_{p_1,q_1}$, where $2<p_0=p_1<\mathrm{8}$, $\frac{1}{r}=\frac{1}{2}-\frac{1}{p_1}$, $a=\frac{1}{2}-\frac{1}{p_0}$. The arbitrary choice of parameters guarantees the arbitrary of the parameters available in the theorem. ?

The case $1<p_0<p_1<2$ follows from the statements proved above and the property $M_{p_0,q_0}^{p_1,q_1}=M_{p_1^{\text{'}},q_1^{\text{'}}}^{p_0^{\text{'}},q_0^{\text{'}}}$.

Proof of Theorem 4 Let $1<p_0<p_1=2$. Let us consider the bilinear mapping $T(a,f)=a*b$, where $b={\{b_m\}}_{m?Z}$ is the sequence of Fourier coefficients of the function f. The mapping

$$T:l_{p_0,q_0}\times L_2?l_{p_1,q_1}$$

(8)

is bounded according to Theorem 1. Here $\frac{1}{p_0}-\frac{1}{p_1}=\frac{1}{2}$, $\frac{1}{q_1}-\frac{1}{q_0}=\frac{1}{2}$, $1<q_1<2<q_0$, $1<p_0<2<p_1$. The result of Theorem 3, in the case $q_0=q_1=1$, $p_0=p_1=p$ can be written as

$$T:l_{p,1}\times B_{t,1}^{1/t}?l_{p,1},\frac{1}{t}=\frac{1}{p}-\frac{1}{2}.$$

(9)

Applying Lemma 3 on the bilinear interpolation to (8) and (9), and taking into account the properties of the embedding of the spaces $l_{p,q}$ and $B_{p,q}^a$, we have:

$$T:l_{p_0,1}\times B_{r,1}^a?l_{p_1,\mathrm{8}},$$

(10)

where parameters r, a, $p_0$, $p_1$ satisfy the following conditions:

$$1<p_0<p_1=2,\frac{1}{r}-a=\frac{1}{p_0}-\frac{1}{p_1},a>\frac{1}{p_1}-\frac{1}{2}.$$

(11)

Let in (11) parameter r be fixed. Using Lemma 4 on bilinear interpolation and taking into account that

$${(B_{r,1}^{a_0},B_{r,1}^{a_1})}_{?,h}=B_{r,h}^a,\mathtt{\text{with}}a=(1-?)a^0+a^1,$$

we get

$$T:l_{p_0,h_1}\times B_{r,h_2}^a?l_{p,h_3},$$

where $\frac{1}{h_1}+1=\frac{1}{h_2}+\frac{1}{h_3}$, $a>\frac{1}{p_1}-\frac{1}{2}={min}_{x?[\frac{1}{p_1},\frac{1}{p_0}]}|\frac{1}{2}-x|$, $\frac{1}{r}-a=\frac{1}{p_0}-\frac{1}{p_1}$.

Therefore, with fixed $a?l_{p_0,\mathrm{8}}$ and r we obtain that

$$P_a:B_{r,1}^{a_i}?l_{p_1^i,\mathrm{8}},$$

and

$${?P_a?}_{B_{r,1}^{a_i}?l_{p_1^i,\mathrm{8}}}=c_i{?a?}_{l_{p_0,\mathrm{8}}},$$

where $\frac{1}{r}-a_i=\frac{1}{p_0}-\frac{1}{p_1^i}$, $a^i>\frac{1}{p_1^i}-\frac{1}{2}$, $i=0,1$.

Using Marcinkiewicz-Calderón interpolation theorem we have

$$P_a:B_{r,s}^a?l_{p_1,s},$$

and

$${?P_a?}_{B_{r,s}^a?l_{p_1,s}}=c{?a?}_{l_{p_0,\mathrm{8}}}.$$

Thus

$$T:l_{p_0,\mathrm{8}}\times B_{r,s}^a?l_{p_1,s}.$$

To complete the proof we fix the function f and the parameters r, s, a and we choose the parameters $p_0^i$, $p_1^i$, $i=0,1$ satisfying (11). We use Lemma 2 to get $B_{r,s}^a[0,1]?M_{p_0,q_0}^{p_1,q_1}$. ?

The case $2=p_0<p_1<\mathrm{8}$, as in the proof of Theorem 3, will follow from $M_{p_0,q_0}^{p_1,q_1}=M_{p_1^{\text{'}},q_1^{\text{'}}}^{p_0^{\text{'}},q_0^{\text{'}}}$.

5 Examples demonstrating the sharpness of the results

Proposition 1 Let$1<p_0<2=p_1$, $\frac{1}{r}=\frac{1}{p_0}-\frac{1}{p_1}$, $\frac{1}{s}={(\frac{1}{q_1}-\frac{1}{q_0})}_{+}$. If$q_1<q_0$, then for any$e>0$there exists$f_1?L_{r,s+e}$such that$f_1?M_{p_0,q_0}^{p_1,q_1}$, if$q_1=q_0$there exists$f_2?L_{r-e,\mathrm{8}}$such that$f_2?M_{p_0,q_0}^{p_1,q_1}$.

Proof Let e be an arbitrary positive number, and numbers ${ß}_1$, ${ß}_2$ be such that

$${ß}_1>\frac{1}{s+e},{ß}_2>\frac{1}{q_0},{ß}_1+{ß}_2<\frac{1}{s}+\frac{1}{q_0}=\frac{1}{q_1}.$$

Let

and

$$f_1~\underset{k=-\mathrm{8}}{\overset{+\mathrm{8}}{?}}{b}_k{e}^{2pikx}.$$

Then for $m?0$

Thus, ${(a*b)}_m=c{(|m|+1)}^{-(\frac{1}{r^{\text{'}}}+\frac{1}{p_0})+1}|ln(|m{|+2)|}^{-{ß}_1-{ß}_2}$. Since

$$\underset{m=0}{\overset{+\mathrm{8}}{?}}\left({\left({(m+1)}^{-(\frac{1}{r^{\text{'}}}+\frac{1}{p_0})+1}{|ln(|m|+2)|}^{-{ß}_1-{ß}_2}\right)}^{q_1}{(|m|+1)}^{(\frac{q_1}{p_1}-1)}\right)=\mathrm{8},$$

$a*b?l_{p_1,q_1}$, and therefore $f_1?M_{p_0,q_0}^{p_1,q_1}$. Since Fourier coefficients of $f_1$ are the sequence ${\{b_k\}}_{k?Z}$ it follows that $f_1?L_{r,s}$.

To prove the second part of the proposition, we take $s=\mathrm{8}$. Let numbers $a_1$ and $a_2$ be such that

$$a_1>\frac{1}{{(r-e)}^{\text{'}}}=1-\frac{1}{r-e},a_2>\frac{1}{p_0},a_1+a_2<1-\frac{1}{r}+\frac{1}{p_0}$$

(note that the last inequality does not contradict the previous two). Choosing

$$b_k=\frac{1}{{(|k|+1)}^{a_1}},a_k=\frac{1}{{(|k|+1)}^{a_2}},f_2~\underset{k=-\mathrm{8}}{\overset{+\mathrm{8}}{?}}{b}_k{e}^{2pikx},$$

we can show that

$$a*b~{\left\{{(|k|+1)}^{-a_1-a_2+1}\right\}}_{k?Z}.$$

Hence $a*b?l_{p_1,q_1}$, and therefore $f_2?M_{p_0{q}_0}^{p_1{q}_1}$. At the same time taking into account the monotonicity of the sequence ${\{b_k\}}_{k?\mathbb{Z}}$ and Hardy-Littlewood theorem, we have that $f_2?L_{r-e,\mathrm{8}}$. The statement is proved. ?

Theorem 6 Let$1<p_0<p_1<2$, $1<q_1=q_0$, $\frac{1}{r}-a=\frac{1}{p_0}-\frac{1}{p_1},\frac{1}{s}=\frac{1}{q_1}-\frac{1}{q_0}$. Then for any$e>0$there exist$f_1?B_{r,\mathrm{8}}^{a-e}n{B}_{r-e,\mathrm{8}}^a$and$f_2?B_{r,s+e}^a$such that$f_1?M_{p_0,q_0}^{p_1,q_1}$, $f_2?M_{p_0,q_0}^{p_1,q_1}$.

Proof Let $s<\mathrm{8}$ and numbers ${ß}_1$, ${ß}_2$ be such that

$${ß}_1>\frac{1}{s+e},{ß}_2>\frac{1}{q_0},{ß}_1+{ß}_2<\frac{1}{s}+\frac{1}{q_0}=\frac{1}{q_1}.$$

Let $b={\{b_k\}}_{k?Z}$ and $a={\{a_k\}}_{k?Z}$, where

It is obvious that $a?l_{p_0,q_0}$, and $f_2~{?}_{k=-\mathrm{8}}^{+\mathrm{8}}{b}_k{e}^{2pikx}$ belongs to $B_{r,s+e}^a$.

It is easy to show that

$${(a*b)}_m=c{(|m|+1)}^{a-\frac{1}{r}+\frac{1}{p_0}}{(ln(|m|+2))}^{{ß}_1+{ß}_2}$$

and consequently, $a*b?l_{p_1,q_1}$. Therefore $f_2?M_{p_0,q_0}^{p_1,q_1}$.

To construct the function $f_1$, it is sufficient to consider the sequences

$$b={\left\{\frac{1}{{(|m|+1)}^{{?}_1}}\right\}}_{m?Z},a={\left\{\frac{1}{{(|m|+1)}^{{?}_2}}\right\}}_{m?Z},$$

where

$${?}_1>max(a-e-\frac{1}{r}+1,a-\frac{1}{r+e}+1),{?}_2>\frac{1}{p_0}$$

and

$${?}_1+{?}_2<a-\frac{1}{r}+\frac{1}{p_0}.$$

$f_1~{?}_{k=-\mathrm{8}}^{+\mathrm{8}}{b}_k{e}^{2pikx}$. The proof that $f_1?B_{r,\mathrm{8}}^{a-e}n{B}_{r-e,\mathrm{8}}^a$, $f_1?M_{p_0,q_0}^{p_1,q_1}$ is similar to the proof of the first part. ?