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Basis property of the Legendre polynomials in variable exponent Lebesgue spaces M. G. Magomed-Kasumovab, T. N. Shakh-Emirova, R. M. Gadzhimirzaeva a Daghestan Federal Research Centre of the Russian Academy of Sciences, Makhachkala, Russia b Vladikavkaz Scientific Centre of the Russian Academy of Sciences, Vladikavkaz, Russia
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     § 1. IntroductionAccording to Pollard, the Legendre polynomials form a basis of the Lebesgue space $L^p([-1,1])$ for $4/3<p<4$ and do not form a basis for $p\in[1,4/3)\cup(4,\infty)$ (see [1]). Newman and Rudin [2] supplemented this result by showing that the Legendre polynomials do not form a basis for $p \in \{4/3, 4\}$ either. The basis problem for the Legendre polynomials for variable exponent Lebesgue spaces was considered by Sharapudinov [3]. Let us discuss this problem in greater detail. We let $\mathscr P(-1,1)$ denote the class of variable exponents $p(x)$ satisfying the following conditions: (A) $p(x)>1$ for all $x \in[-1,1]$, and $p(x)$ satisfies the Dini–Lipschitz condition (2.8) (see below) for $x,y\in[-1,1]$; (B) the exponent $p=p(x)$ is constant in some neighbourhoods of the points $\pm 1$, that is, there exist numbers $l=l(p)$, $r=r(p)$, $\delta_1=\delta_1(p)$ and $\delta_2=\delta_2(p)$ such that $l,r > 1$, $0<\delta_1,\delta_2<1$, and
$$
\begin{equation}
p(x)= \begin{cases} l & \text{if } x\in[-1,-1+\delta_1], \\ r & \text{if } x\in[1-\delta_1,1]. \end{cases}
\end{equation}
\tag{1.1}
$$
The following theorem was proved in [3]. Theorem A. Let $p=p(x)\in\mathscr P(-1,1)$ and $p(\pm1)\in(4/3,4)$. Then the orthonormal system of Legendre polynomials (2.17) forms a basis of the variable exponent Lebesgue space $L^{p(\cdot)}([-1,1])$. As noted in [3], the quantities $\delta_1$ and $\delta_2$ can be arbitrarily small. This suggests the problem of whether one can drop condition (B) of the constancy of the variable exponent near the points $\pm1$. In the present paper we give an affirmative answer in this problem. Namely, the following theorem will be proved. Theorem 1.1. Let $p=p(x)>1$ satisfy the Dini–Lipschitz condition (2.8), and let $p(\pm1)\in(4/3,4)$. Then the orthonormal system of Legendre polynomials (2.17) forms a basis of $L^{p(\cdot)}([-1,1])$. § 2. Auxiliary results2.1. Variable exponent Lebesgue spacesLet $p(x)$ be a nonnegative measurable function defined on a measurable set $E$. By definition, the variable exponent Lebesgue space $L^{p(\cdot)}(E)$ consists of the functions $f$ such that For $E=[-1,1]$ we write for brevity $L^{p(\cdot)}= L^{p(\cdot)}[-1,1]$. We also define $p_+(E)=\operatorname*{ess\,sup}_{x\in E} p(x)$ and $p_-(E)=\operatorname*{ess\,inf}_{x\in E} p(x)$. For $1\leqslant p_-(E)\leqslant p(x)\leqslant p_+(E)<\infty$ the space $L^{p(\cdot)}(E)$ is normable (see [4]); one of the equivalent norms can be defined by
$$
\begin{equation}
\|f\|_{p(\cdot)}=\|f\|_{p(\cdot)}(E)=\inf\biggl\{\lambda>0\colon \int_{E}\biggl|\frac{f(x)}\lambda\biggr|^{p(x)}\,dx\leqslant1\biggr\}.
\end{equation}
\tag{2.2}
$$
In what follows we require some properties of these spaces. Let $1\leqslant p(x)\leqslant q(x)\leqslant q_+(E)<\infty$, $E \subset [-1,1]$. Then $L^{q(\cdot)}(E)\subset L^{p(\cdot)}(E)$, and, for $f\in L^{q(\cdot)}(E)$ (see [5]) where $r_{p,q}=1/(\mu_-(E))+1/\mu'(E)$ ($\mu(x)=q(x)/p(x)$, $1/\mu(x)+1/\mu'(x)=1$).If $p(x)>1$, $x \in E$ (the case $p_-(E)=1$ is not excluded), then a Hölder-type inequality holds for Lebesgue variable exponent spaces (see [4], inequality (8)):
$$
\begin{equation}
\int_E |f(x)|\,|g(x)|\,dx \leqslant c(p) \|f\|_{p(\cdot)}(E) \|g\|_{p'(\cdot)}(E),
\end{equation}
\tag{2.4}
$$
where $1/p(x)+1/p'(x)=1$, $c(p)\leqslant 1/p_-(E)+1/p_-'(E)$. Here and in what follows $c, c(\alpha), c(\alpha,\beta),\dots$ are positive constants (depending only on the above parameters), which can be different at different places. For arbitrary measurable sets $A\subset B$, We also use the equivalent norm (see [3])
$$
\begin{equation}
\|f\|^*_{p(\cdot)}=\sup_{\substack{g\in L^{p'(\cdot)}(E)\\ \|g\|_{p'(\cdot)}\leqslant1}} \int_{E}f(x)g(x)\,dx.
\end{equation}
\tag{2.6}
$$
Note that
$$
\begin{equation}
\|f\|_{p(\cdot)}\leqslant\|f\|^*_{p(\cdot)}\leqslant c(p)\|f\|_{p(\cdot)}.
\end{equation}
\tag{2.7}
$$
The Dini–Lipschitz condition
$$
\begin{equation}
|p(x)-p(y)| \leqslant \frac{c}{|\log |x-y||}, \qquad x,y \in E,
\end{equation}
\tag{2.8}
$$
plays a fundamental role in the theory of variable exponent Lebesgue spaces. For such spaces this condition was introduced by Sharapudinov [5], who studied the basis property of the Haar system in $L^{p(\cdot)}([0,1])$. We also require the following result (see Corollary 2.23 in [6]). Lemma 2.1. Let $A$ be a measurable set, let and let $1\leqslant p_-(A)\leqslant p(x)\leqslant p_+(A)<\infty$. If $\|f\|_{p(\cdot)}>1$, then If $0<\|f\|_{p(\cdot)}\leqslant1$, then The proof of our main theorem depends on several lemmas, which enable one to change from a variable exponent to a constant one. Lemma 2.2. Let $p(x)$ satisfy the Dini–Lipschitz condition (2.8) on $[a,b]$, $0\leqslant a < b \leqslant 1$. Then If $b=1$, then a lower estimate also holds: Proof. By the Dini–Lipschitz condition
$$
\begin{equation*}
\begin{aligned} \, &\biggl(\frac{1}{1-x} \biggr)^{p(x)-p(1)} \leqslant \biggl(\frac{1}{1-x} \biggr)^{|p(x)-p(1)|} \leqslant \biggl(\frac{1}{1-x} \biggr)^{c/|\log(1-x)|} \\ &\qquad =\exp\biggl(\log \biggl(\frac{1}{1-x} \biggr)^{c/|\log(1-x)|}\biggr) =\exp\biggl(\frac{c}{|\log(1-x)|} \log\biggl(\frac{1}{1-x} \biggr)\biggr) \\ &\qquad= \exp\biggl(\frac{-c\log(1-x)}{|\log(1-x)|} \biggr)=e^{c}, \qquad x \in [0,1). \end{aligned}
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
J=\int_{a}^{b} \biggl(\frac{1}{1-x} \biggr)^{p(1)} \biggl(\frac{1}{1-x} \biggr)^{p(x)-p(1)}\,dx \leqslant c(p)\int_{a}^{b} \biggl(\frac{1}{1-x} \biggr)^{p(1)}\,dx, \qquad a \geqslant 0.
\end{equation*}
\notag
$$
Let us prove the lower estimate. We set $E_+=\{ x \in [a,1]\colon p(x)-p(1) \geqslant 0 \}$, $E_-=\{ x \in [a,1]\colon p(x)-p(1) < 0 \}$ and split the integral into two:
$$
\begin{equation*}
\int_{a}^{1} \biggl(\frac{1}{1-x} \biggr)^{p(x)}\, dx= \int_{E_+} \biggl(\frac{1}{1-x} \biggr)^{p(x)}\, dx + \int_{E_-} \biggl(\frac{1}{1-x} \biggr)^{p(x)}\, dx.
\end{equation*}
\notag
$$
Since $1/(1-x) \geqslant 1$, we have
$$
\begin{equation*}
\int_{E_+} \biggl(\frac{1}{1-x} \biggr)^{p(x)} \,dx= \int_{E_+} \biggl(\frac{1}{1-x} \biggr)^{p(1)} \biggl(\frac{1}{1-x} \biggr)^{p(x)-p(1)}\!\! dx \geqslant \int_{E_+} \biggl(\frac{1}{1-x} \biggr)^{p(1)} \,dx.
\end{equation*}
\notag
$$
For $x \in E_{-}$, Hence which proves the lemma. We say that $p(x)$ satisfies the Dini–Lipschitz condition to the left of a point $a$ if for all $x$ in some left neighbourhood of $a$.Lemma 2.3. Let $p(x)$ satisfy the Dini–Lipschitz condition to the left of the point $1$. Then there exists a constant $c(p)$, which depends on $p(x)$, such that for all $y$ in some left neighbourhood of $1$ and all $x \in [y,1]$ If $p(1)>0$, then the reverse inequality also holds: Proof. We write $f(x,y)$ in the form
$$
\begin{equation}
f(x,y)=\biggl(\frac{1}{1-y}\biggr)^{p(1)}\biggl(\frac{1}{1-y}\biggr)^{p(x)-p(1)}.
\end{equation}
\tag{2.10}
$$
Let show that, under the conditions of the lemma, the second factor is a bounded function. Indeed, by the Dini–Lipschitz condition (2.9) and the inequality $x \geqslant y$, which follows from the conditions of the lemma, we have
The reverse inequality is proved similarly to Lemma 2.2. This completes the proof. Lemma 2.4. Let $0 \leqslant f(x) \leqslant c/(1-x)^\alpha$, $x \in [a,1]$, and let $p(x)$ satisfy the Dini–Lipschitz condition on $ [a,1]$. Then Proof. We have
$$
\begin{equation*}
f(x)^{p(x)}=f(x)^{p(1)} f(x)^{p(x)-p(1)} \leqslant f(x)^{p(1)} \biggl(\frac{c}{(1-x)^\alpha}\biggr)^{p(x)-p(1)}.
\end{equation*}
\notag
$$
For the second factor Lemma 2.3 shows that which proves the lemma. Proof. We have
$$
\begin{equation*}
\begin{aligned} \, \int_a^{2x-1} \frac{|f(y)|\,dy}{(1-y)^{3/4}} &\leqslant \frac{c}{(1-x)^{3/4}} \int_a^{2x-1} |f(y)|\,dy \\ &\leqslant\frac{c(p)}{(1-x)^{3/4}} \|f\|_{p(\cdot)} \leqslant \frac{c(p)}{(1-x)^{3/4}}, \end{aligned}
\end{equation*}
\notag
$$
and now the required assertion is secured by Lemma 2.4. This completes the proof. We also note the following relation, which follows easily, for example, from Hölder’s inequality for sums:
$$
\begin{equation}
\biggl(\sum_{k=1}^N a_k \biggr)^p \leqslant N^{p-1} \sum_{k=1}^N a_k^p, \qquad p \geqslant 1.
\end{equation}
\tag{2.11}
$$
2.2. The kernel $K(x,y)$Following formula (5.16) in [3], we set
$$
\begin{equation}
K(x,y)=\frac{1}{|x-y|}\biggl| \biggl(\frac{1-y^2}{1-x^2} \biggr)^{1/4} - 1 \biggr|.
\end{equation}
\tag{2.12}
$$
We need some properties of the kernel $K(x,y)$. Proof. It can easily be shown that, under the conditions of the lemma, It remains to note that since $(y+1)/2< x$. This proves the lemma. Lemma 2.7. The following inequalities hold: Lemma 2.8. Let $-1+\varepsilon < x < 1 - \varepsilon$, $0 < \varepsilon < 1$. Then the kernel $K(x,y)$ is uniformly bounded with respect to $y \in [-1,1]$: 2.3. Legendre polynomialsLegendre polynomials, which can be defined by Rodrigues’s formula are orthogonal with weight 1 on $[-1,1]$, that is,
$$
\begin{equation*}
\int_{-1}^1P_n(x)P_m(x)\,dx=\frac{2}{2n+1}\delta_{nm},
\end{equation*}
\notag
$$
where $\delta_{nm}$ is the Kronecker delta. Recall the following properties of Legendre polynomials (see [7], Theorems (7.3.3) and (7.33.3)):
$$
\begin{equation}
(1-x^2)^{1/4}n^{1/2}|P_n(x)|\leqslant\sqrt{\frac2\pi},
\end{equation}
\tag{2.15}
$$
$$
\begin{equation}
(1-x^2)^{1/4}(n+1)^{1/2}|P_n(x)-P_{n+2}(x)|\leqslant c,\ x\in[-1,1].
\end{equation}
\tag{2.16}
$$
Orthonormal Legendre polynomials are as follows: With an integrable function $f(x)$ on $[-1,1]$ we can associate the Fourier–Legendre series where are the Fourier–Legendre coefficients of $f$. Partial sums of the series (2.18) are as follows:Below we need the integral representation for Fourier–Legendre partial sums in terms of the Christoffel–Darboux kernels
$$
\begin{equation*}
K_n(x,t)=\sum_{k=0}^n\frac{2k+1}{2}P_k(x)P_k(t) =\frac{n+1}2\frac{P_{n+1}(x)P_n(t)-P_{n+1}(t)P_n(x)}{x-t}.
\end{equation*}
\notag
$$
2.4. The Hilbert transformLet $\mathbb{P}(-1,1)$ be the class of exponents $p(x)>1$ satisfying condition (2.8) on $[-1,1]$. Let $p(x)\in\mathbb{P}(-1,1)$. For $f\in L^{p(\cdot)}$ the Hilbert transform is defined by The integral on the right-hand side of (2.20) is understood in the sense of the Cauchy principal value. Note that the function $Hf(x)$ is finite for nearly all $x\in[-1,1]$. In [8] and [9] it was shown that§ 3. Proof of Theorem 1.1Consider the operators
$$
\begin{equation}
T_1(f)(x)=\int_{-1}^1 K(x,y)|f(y)|\,dy\quad\text{and} \quad T_2(f)(x)=\int_{-1}^1 K(y,x)|f(y)|\,dy.
\end{equation}
\tag{3.1}
$$
According to [3], the boundedness of these operators in $L^{p(\cdot)}$ plays an important role in the proof of the uniform boundedness of the Fourier–Legendre partial sums in $L^{p(\cdot)}$. Lemma 5.2 in [3] shows that these operators are bounded if $p(x)\in\mathscr P(-1,1)$. In the present paper we show that these operators are also bounded without constraint (B) that the exponent $p(x)$ should be constant near the endpoints of the intervals. Lemma 3.1 (main lemma). Let $p=p(x)$ satisfy the Dini–Lipschitz condition (2.8), and let $p(\pm1)\in(4/3,4)$. Then the operators $T_1(f)$ and $T_2(f)$ are bounded in the space $L^{p(\cdot)}$. The proof of this lemma, which underlies the following result, is presented in § 4. Lemma 3.2. Let $p=p(x)$ satisfy the Dini–Lipschitz condition (2.8), and let $p(\pm1)\in(4/3,4)$. Then for $f \in L^{p(\cdot)}$ The proof of Lemma 3.2 follows mainly the proof of the uniform boundedness of the $S_n$ (see [3], the proof of Theorem 5), except for the fact that here we use Lemma 3.1 in place of Lemma 5.2 in [3], and we also apply Lemma 2.2. However, we present the proof for completeness. Proof of Lemma 3.2. From (2.19) and Lemma 5.1 in [3] we obtain
$$
\begin{equation*}
\begin{aligned} \, S_n(f,x) &=\frac{n+2}{2n+3}\,\frac{n+1}2P_{n+1}(x)\int_{-1}^{1} \frac{P_n(y)-P_{n+2}(y)}{x-y}f(y)\,dy \\ &\qquad +\frac{n+2}{2n+3}\,\frac{n+1}2[P_{n+2}(x)-P_n(x)] \int_{-1}^{1}\frac{P_{n+1}(y)}{x-y}f(y)\,dy \\ &\qquad +\frac{n+1}2P_{n+1}(x)\int_{-1}^{1}P_{n+1}(y)f(y)\,dy. \end{aligned}
\end{equation*}
\notag
$$
The first two integrals on the right-hand side are understood in the sense of the Cauchy principal value. By the weighted estimate (2.15) and inequality (2.16) we have
$$
\begin{equation}
\begin{aligned} \, \notag |S_n(f,x)| &\leqslant c(n+1)^{1/2}(1-x^2)^{-1/4}\biggl|\int_{-1}^{1}\frac{P_n(y)-P_{n+2}(y)}{x-y}f(y)\,dy\biggr| \\ \notag &\qquad +c(n+1)^{1/2}(1-x^2)^{1/4}\biggl|\int_{-1}^{1}\frac{P_{n+1}(y)}{x-y}f(y)\,dy\biggr| \\ \notag &\qquad+c(n+1)^{1/2}(1-x^2)^{-1/4}\biggl|\int_{-1}^{1}P_{n+1}(y)f(y)\,dy\biggr| \\ &=Q_1(x)+Q_2(x)+Q_3(x). \end{aligned}
\end{equation}
\tag{3.2}
$$
For $Q_3$, applying the weighted estimate (2.15) and Hölder’s inequality (2.4) we find that
$$
\begin{equation*}
Q_3(x)\leqslant c(p)(1-x^2)^{-1/4}\|(1-y^2)^{-1/4}\|_{p'(\cdot)}(-1,1).
\end{equation*}
\notag
$$
The integral $\displaystyle\int_{-1}^{1}(1-y^2)^{-p'(y)/4}\,dy$, and so also the norm $\|(1- y^2)^{-1/4}\|_{p'(\cdot)}(-1,1)$, are finite by Lemma 2.2. Hence $Q_3(x)\leqslant c(p)(1-x^2)^{-1/4}$ and
Next, for $Q_1$ we have
$$
\begin{equation}
\begin{aligned} \, \notag Q_1(x) &=\biggl|\int_{-1}^{1}\biggl(\frac{1-y^2}{1-x^2}\biggr)^{1/4}\frac{A_n(y)}{x-y}f(y)\,dy\biggr| \\ \notag &\leqslant\biggl|\int_{-1}^{1}\biggl[\biggl(\frac{1-y^2}{1-x^2}\biggr)^{1/4}-1\biggr] \frac{A_n(y)f(y)}{x-y}\,dy\biggr|+ \biggl|\int_{-1}^{1}\frac{A_n(y)f(y)}{x-y}\,dy\biggr| \\ &=Q_{11}(x)+Q_{12}(x), \end{aligned}
\end{equation}
\tag{3.4}
$$
where In view of (2.16) the sequence $A_n$ is uniformly bounded on $(-1,1)$. Hence the function $A_n(x)f(x)$ lies in $L^{p(\cdot)}$. Now, from (2.21) we obtain
$$
\begin{equation}
\|Q_{12}\|_{p(\cdot)}(-1,1)\leqslant c(p)\|A_nf\|_{p(\cdot)}(-1,1)\leqslant c(p)\|f\|_{p(\cdot)}(-1,1)\leqslant c(p).
\end{equation}
\tag{3.5}
$$
Let us estimate $Q_{11}$. By (2.12) and (3.1)
$$
\begin{equation*}
Q_{11}(x)\leqslant \int_{-1}^{1}|A_n(y)K(x,y)f(y)|\,dy\leqslant c \int_{-1}^{1}K(x,y)|f(y)|\,dy=c T_1(f)(x).
\end{equation*}
\notag
$$
Hence, using Lemma 3.1, Now, from (3.4)–(3.6) we obtain
The quantity $Q_{2}$ is estimated similarly to $Q_{1}$, the only difference is that in place of estimate (2.16) and the boundedness of the operator $T_1$ one should use (2.15) and the boundedness of $T_2$. So we have Using estimates (3.3), (3.7), (3.8) and inequality (3.2) we find that which is the result required. This proves Lemma 3.2.For the proof of Theorem 1.1 it suffices to note that the set of algebraic polynomials is dense in $L^{p(\cdot)}$ and $S_n(p_k,x)=p_k(x)$ for $k\leqslant n$, where $p_k(x)$ is an algebraic polynomial of degree $k$. Hence by Lemma 3.2
$$
\begin{equation*}
\|S_{n}(f)-f\|_{p(\cdot)}\leqslant \|p_n-f\|_{p(\cdot)}+\|S_{n}(f-p_n)\|_{p(\cdot)}<(c(p)+1)\varepsilon.
\end{equation*}
\notag
$$
§ 4. Proof of Lemma 3.1Let $f\in L^{p(\cdot)}$, and let We set $\delta=(1/2)\min\{ 4 - p(1), p(1)-4/3 \}$. Since $p'(x)$ is uniformly continuous on $[-1,1]$, there is a natural number $N_0$ such that for $N \geqslant N_0$, for each $k \in \mathcal{I}=\{-N+1, \dots, N\}$ we have
$$
\begin{equation}
p'_+(\Delta_k) - p'_-(\Delta_k) < \frac{\delta}{4-\delta}, \qquad \Delta_k=\Delta_k^{(N)}= \biggl[\frac{k-1}{N},\frac{k}{N}\biggr].
\end{equation}
\tag{4.2}
$$
Let $N=\max\{N_1, N_2\}$. We choose $\varepsilon < 1/(2N)$ so that
$$
\begin{equation}
\frac 43+\delta < p_-(\mathcal{E}) \leqslant p_+(\mathcal{E}) \leqslant 4 - \delta, \qquad \mathcal{E}=[1-\varepsilon,1].
\end{equation}
\tag{4.3}
$$
This is possible by the continuity of $p(x)$ at $x=1$ and since $p(\pm 1) \in (4/3, 4)$. To prove our main lemma, Lemma 3.1, we need several auxiliary assertions. Lemma 4.1. Let $p(x)$ satisfy the Dini–Lipschitz condition (2.8), and let $4/3<p(1)<4$. Then for $\|f\|_{p(\cdot)}\leqslant1$ Proof. The relation $J(f,p(x))\leqslant c(p)J(f,p(1))$ readily follows from Lemmas 2.3 and 2.5. Let us prove the last inequality in (4.4).
We write $f(x)$ as the sum of two functions, $f(x)\chi_{E_1}(x)$ and $f(x)\chi_{E_2}(x)$, where $E_1=\{x\colon |f(x)| < 1\}$ and $E_2=\{x\colon |f(x)| \geqslant 1\}$. It is clear that $J(f) \leqslant c(p)\bigl(J(f\chi_{E_1})+J(f\chi_{E_2})\bigr)$. It is also easily verified (for example, using Lemma 2.2) that $J(f\chi_{E_1}) \leqslant c(p)$. So, in what follows we can assume that the function $|f(x)|$ is either equal to $0$, or $\geqslant1$ everywhere on $[1- 1/N,1]$. There are two cases to consider. 1. The case $p(x)\geqslant p(1)$. Let $1/p'(1)<\beta<3/4$. To estimate $J(f,p(1))$ we apply Hölder’s inequality to the inner integral. As a result,
$$
\begin{equation*}
\begin{aligned} \, J(f,p(1)) &=\int_\mathcal{E}\biggl[\frac1{(1-x)^{1/4}} \int_{1-1/N}^{2x-1}\frac{|f(y)|\,dy}{(1-y)^{3/4-\beta}(1-y)^{\beta}}\biggr]^{p(1)}\,dx \\ &\leqslant \int_\mathcal{E}\biggl[\frac1{(1-x)^{1/4}} \biggl(\int_{1-1/N}^{2x-1}\frac{|f(y)|^{p(1)}\,dy}{(1-y)^{(3/4-\beta)p(1)}} \biggr)^{1/p(1)} \\ &\qquad\times \biggl(\int_{1-1/N}^{2x-1}\frac{dy}{(1-y)^{\beta p'(1)}}\biggr)^{1/p'(1)} \biggr]^{p(1)}dx. \end{aligned}
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
\biggl(\int_{1-1/N}^{2x-1}\frac{dy}{(1-y)^{\beta p'(1)}}\biggr)^{1/p'(1)}\leqslant c(p)(1-x)^{1-1/p-\beta},
\end{equation*}
\notag
$$
and therefore
$$
\begin{equation*}
J(f,p(1))\leqslant c(p)\int_\mathcal{E}\frac1{(1-x)^{1-(3/4-\beta)p(1)}} \int_{1-1/N}^{2x-1}\frac{|f(y)|^{p(1)}\,dy}{(1-y)^{(3/4-\beta)p(1)}}\,dx.
\end{equation*}
\notag
$$
Hence, changing the order of integration and using (2.3), this gives
$$
\begin{equation}
\begin{aligned} \, \notag J(f,p(1)) &\leqslant c(p)\int_{1-1/N}^{1}|f(y)|^{p(1)}\frac1{(1-y)^{(3/4-\beta)p(1)}} \int_{(y+1)/2}^{1}\frac{dx}{(1\,{-}\,x)^{1-(3/4-\beta)p(1)}}\,dy \\ &=c(p)\int_{1-1/N}^{1}|f(y)|^{p(1)}\,dy=c(p)\|f\|_{p(1)}^{p(1)}\leqslant c(p)\|f\|_{p(\cdot)}^{p(1)}\leqslant c(p). \end{aligned}
\end{equation}
\tag{4.5}
$$
2. The case $p(x)\leqslant p(1)$. We write $f\in L^{p(\cdot)}$ in the form
$$
\begin{equation}
f(x)=g(x)\biggl(\frac1{1-x}\biggr)^\alpha, \qquad g(x)=(1-x)^\alpha f(x), \qquad \alpha=\frac6{p'(1)}.
\end{equation}
\tag{4.6}
$$
From the Dini–Lipschitz condition (2.8) we obtain
$$
\begin{equation}
|f(x)|^{p(1)}=|f(x)|^{p(x)}|f(x)|^{p(1)-p(x)}\leqslant c(p)|f(x)|^{p(x)}|g(x)|^{p(1)-p(x)}.
\end{equation}
\tag{4.7}
$$
Next, consider the sets $E_1=\{x\in[1-1/N,1]\colon g(x)\geqslant1\}$ and $E_2=\{x\in[1-1/N,1]\colon g(x)<1\}$. We have
$$
\begin{equation}
f(x)=f(x)\chi_{E_1}(x)+f(x)\chi_{E_2}(x)=f_1(x)+f_2(x).
\end{equation}
\tag{4.8}
$$
Arguing as in (4.5) we see that
$$
\begin{equation*}
J(f_2,p(1))\leqslant c(p)\int_{1-1/N}^{1}|f_2(y)|^{p(1)}\,dy.
\end{equation*}
\notag
$$
Hence by (4.7)
$$
\begin{equation}
J(f_2,p(1))\leqslant c(p)\int_{1-1/N}^{1}|f_2(y)|^{p(1)}\,dy\leqslant \int_{1-1/N}^{1}|f_2(y)|^{p(y)}\,dy\leqslant c(p).
\end{equation}
\tag{4.9}
$$
Let us estimate $J(f_1,p(1))$. Note that for $x\in E_1$, from (4.7) we obtain
$$
\begin{equation*}
|f(x)|^{p(1)}\leqslant c(p)|f(x)|^{p(x)}|g(x)|^{p(1)-p(x)+\gamma}, \qquad \gamma=\frac{p(1)}{6},
\end{equation*}
\notag
$$
or, what is the same,
$$
\begin{equation*}
|f(x)|\leqslant c(p)|f(x)|^{p(x)/p(1)}|g(x)|^{(p(1)-p(x)+\gamma)/p(1)}.
\end{equation*}
\notag
$$
Using this relation and applying the Hölder-type inequality to $J(f_1)$, this gives
$$
\begin{equation}
\begin{aligned} \, \notag J(f_1,p(1))&\leqslant c(p)\int_\mathcal{E}\frac{1}{(1-x)^{p(1)/4}} \int_{1-1/N}^{2x-1}\frac{|f_1(y)|^{p(y)}\,dy}{(1-y)^{(3/4-1/p'(1))p(1)}} \\ &\qquad\times \biggl(\int_{1-1/N}^{2x-1}\frac{|g(y)|^{r(y)}}{1-y}\,dy\biggr)^{p(1)/p'(1)}\,dx, \end{aligned}
\end{equation}
\tag{4.10}
$$
where $r(y)=(p(1)-p(y)+\gamma)(p'(1)-1)$. Since $r(y)\leqslant p(y)$, we have
$$
\begin{equation}
I(x)=\int_{1-1/N}^{2x-1}\frac{|g(y)|^{r(y)}}{1-y}\,dy \leqslant \int_{1-1/N}^{2x-1}\frac{|f(y)|^{p(y)}\,dy}{(1-y)^{1-\alpha r(y)}}.
\end{equation}
\tag{4.11}
$$
For $\alpha$ and $\gamma$ as above we have
$$
\begin{equation*}
1-\alpha r(y)=1-\alpha (p(1)-p(y)+\gamma)(p'(1)-1)\leqslant0,
\end{equation*}
\notag
$$
and so $I(x)\leqslant c(p)$ by (4.11). Hence, using (4.10) and changing the order of integration we obtain the estimate
$$
\begin{equation}
\begin{aligned} \, \notag J(f_1,p(1)) &\leqslant c(p)\int_{1-1/N}^1\frac{|f_1(y)|^{p(y)}}{(1-y)^{(3/4-1/p'(1))p(1)}} \int_{(y+1)/2}^{1}\frac{dx}{(1-x)^{p(1)/4}}\,dy \\ &=c(p)\int_{1-1/N}^1\frac{|f_1(y)|^{p(y)}}{(1-y)^{(3/4-1/p'(1))p(1)+p(1)/4-1}}\,dy \notag \\ &=c(p)\int_{1-1/N}^1 |f_1(y)|^{p(y)}\,dy \leqslant c(p). \end{aligned}
\end{equation}
\tag{4.12}
$$
From (4.8), (4.9) and (4.12), for $p(x) \leqslant p(1)$ we finally obtain
Proceeding with the general case, we set $E_+=\{x\in[1-1/N,1]\colon p(x)\geqslant p(1)\}$ and $E_-=[1-1/N,1]\setminus E_+$. Now by (4.5) and (4.13) we have
$$
\begin{equation}
J(f,p(1))\leqslant c(p)(J(f\chi_{E_+},p(1))+J(f\chi_{E_-},p(1)))\leqslant c(p),
\end{equation}
\tag{4.14}
$$
which proves Lemma 4.1. Lemma 4.2. There exists a constant $c(p)$ such that for each $f$, $\|f\|_{L^{p(\cdot)}} \leqslant 1$, Proof. Using (2.14), for the inner integral we have
$$
\begin{equation*}
\begin{aligned} \, &\int_{x-(1-x)}^{x+(1-x)}K(x,y)|f(y)|\,dy \leqslant c(p)\frac{1}{(1-x)^{1/4}} \int_{x-(1-x)}^{x+(1-x)}\frac{|f(y)|}{(1-x)^{3/4}}\,dy \\ &\qquad\leqslant c(p)\frac{1}{2(1-x)} \int_{x-(1-x)}^{x+(1-x)}|f(y)|\,dy \leqslant c(p) Mf(x). \end{aligned}
\end{equation*}
\notag
$$
Hence, by Corollary 2.23 in [6] and since the Hardy–Littlewood maximal function is bounded in $L^{p(\cdot)}$ (see Theorem 3.16 in [6]), we have which proves the lemma. We continue the proof of the main lemma. First consider the operator $T_1$. We split the integral into three as follows:
$$
\begin{equation}
\begin{aligned} \, \notag &\int_{-1}^1 T_1(f)(x)^{p(x)}\,dx \\ &\qquad=\biggl(\int_{-1}^{-1+\varepsilon}+\int_{-1+\varepsilon}^{1-\varepsilon}+ \int_{1-\varepsilon}^{1}\biggr)T_1(f)(x)^{p(x)}\,dx=J_1+J_2+J_3, \end{aligned}
\end{equation}
\tag{4.15}
$$
where $\varepsilon$, which depends only on $p$, was fixed before (4.3). By Lemma 2.8 for $J_2$ we have
$$
\begin{equation*}
J_2\leqslant\int_{-1+\varepsilon}^{1-\varepsilon}\biggl(\frac c\varepsilon\int_{-1}^{1} |f(y)|dy\biggr)^{p(x)}\,dx\leqslant c(p)\int_{-1+\varepsilon}^{1-\varepsilon}\biggl(\int_{-1}^{1} |f(y)|dy\biggr)^{p(x)}\,dx.
\end{equation*}
\notag
$$
Applying Hölder’s inequality to the inner integral and using (4.1) we obtain
$$
\begin{equation}
J_2\leqslant c(p)\int_{-1+\varepsilon}^{1-\varepsilon}\|1\|_{p'(\cdot)}^{p(x)}\|f\|_{p(\cdot)}^{p(x)}\,dx\leqslant c(p)\int_{-1+\varepsilon}^{1-\varepsilon}\|1\|_{p'(\cdot)}^{p(x)}\,dx\leqslant c(p).
\end{equation}
\tag{4.16}
$$
Now we estimate $J_3$. According to (3.1),
$$
\begin{equation*}
J_3=\int_{\mathcal{E}} \biggl(\int_{-1}^1 K(x,y)|f(y)|\,dy \biggr)^{p(x)}\,dx.
\end{equation*}
\notag
$$
Setting $\displaystyle I_k(x)=\int_{\Delta_k} K(x,y)|f(y)|\,dy$, splitting the inner integral in $J_3$ into the parts corresponding to the intervals $\Delta_k$ in (4.2) (where $N$ was defined before (4.3)) and using (2.11) we have
$$
\begin{equation}
J_{3}=\int_{\mathcal{E}} \biggl(\sum_{k=-N+1}^{N} I_k(x) \biggr)^{p(x)}\,dx \leqslant c(p) \sum_{k=-N+1}^{N} \int_{\mathcal{E}} I_k(x)^{p(x)}\,dx.
\end{equation}
\tag{4.17}
$$
Using the Hölder-type inequality (2.4) and (4.1), for $I_k(x)$ we have
$$
\begin{equation}
I_k(x) \leqslant c(p)\|K(x,\cdot)\|_{L^{p'(x)}(\Delta_k)}.
\end{equation}
\tag{4.18}
$$
By Lemma 2.1
$$
\begin{equation}
\|K(x,\cdot)\|_{L^{p'(x)}(\Delta_k)} \leqslant \max \biggl\{1, \biggl(\int_{\Delta_k} K(x,y)^{p'(y)}\,dy \biggr)^{1/p'_-(\Delta_k)} \biggr\}.
\end{equation}
\tag{4.19}
$$
To estimate the integral on the right-hand side of the above expression for ${k \!\in\! \mathcal{I} \!\setminus\! \{\mkern-1mu N\mkern-1mu\}}$ and $x \in \mathcal{E}$, we use (2.13). We have
$$
\begin{equation*}
\begin{aligned} \, &\int_{\Delta_k} K(x,y)^{p'(y)}\,dy\leqslant 2^{5p'_+(\Delta_k)/4} \biggl(\frac{1}{1-x}\biggr)^{p'_+(\Delta_k)/4} \int_{\Delta_k} \biggl(\frac{1}{1-y}\biggr)^{3p'_+(\Delta_k)/4} \,dy \\ &\qquad\leqslant2^{5p'_+(\Delta_k)/4} \biggl(\frac{1}{1-x}\biggr)^{p'_+(\Delta_k)/4} N^{3p'_+(\Delta_k)/4-1}= c(p)\biggl(\frac{1}{1-x}\biggr)^{p'_+(\Delta_k)/4}. \end{aligned}
\end{equation*}
\notag
$$
Hence by (4.18) and (4.19),
$$
\begin{equation*}
I_k(x) \leqslant c(p) \biggl(\frac{1}{1-x}\biggr)^{(p'_+(\Delta_k))/(4p'_-(\Delta_k))}, \qquad k \in \mathcal{I} \setminus \{ N \}.
\end{equation*}
\notag
$$
Substituting this estimate into (4.17), we find that
$$
\begin{equation}
J_3 \leqslant c(p)\biggl(\sum_{k \in \mathcal{I} \setminus \{N\}} \int_\mathcal{E} \biggl(\frac{1}{1-x}\biggr)^{p_+(\mathcal{E}) (p'_+(\Delta_k))/(4p'_-(\Delta_k))}\,dx+ \int_\mathcal{E} I_N(x)^{p(x)}\,dx \biggr).
\end{equation}
\tag{4.20}
$$
Note that in view of (4.2), (4.3) and the condition $p(x)>1$, for all $k \in \mathcal{I}$ we have
$$
\begin{equation*}
\frac14 p_+(\mathcal{E})\frac{p'_+(\Delta_k)}{p'_-(\Delta_k)} < \frac14 (4-\delta) \frac{p'_-(\Delta_k)+\delta/(4-\delta)}{p'_-(\Delta_k)} < 1.
\end{equation*}
\notag
$$
Therefore, by (4.20)
$$
\begin{equation}
J_3 \leqslant c(p)\biggl(1+\int_\mathcal{E} I_N(x)^{p(x)}\,dx \biggr).
\end{equation}
\tag{4.21}
$$
To estimate the integral in (4.21) we write $I_N(x)$ as the sum
$$
\begin{equation*}
I_N(x)=\biggl(\int_{1-1/N}^{x-(1-x)}+\int_{x-(1-x)}^1 \biggr) K(x,y)|f(y)|\,dy= I_N^{(1)}(x)+I_N^{(2)}(x).
\end{equation*}
\notag
$$
In view of (2.11) the integral is estimated as follows:
$$
\begin{equation}
\int_\mathcal{E} I_N(x)^{p(x)}\,dx \leqslant c(p)\biggl(\int_\mathcal{E} I_N^{(1)}(x)^{p(x)}\,dx+\int_\mathcal{E} I_N^{(2)}(x)^{p(x)}\,dx\biggr).
\end{equation}
\tag{4.22}
$$
From Lemma 4.2 we readily obtain
$$
\begin{equation}
\int_\mathcal{E} I_N^{(2)}(x)^{p(x)}\,dx\leqslant c(p).
\end{equation}
\tag{4.23}
$$
Consider now the quantity $I_N^{(1)}(x)$. By (2.14),
$$
\begin{equation*}
I_N^{(1)}(x) \leqslant \int_{1-1/N}^{x-(1-x)} \overline{K}(x,y)|f(y)|\,dy\leqslant \frac{c}{(1-x)^{1/4}}\int_{1-1/N}^{2x-1}\frac{f(y)\,dy}{(1-y)^{3/4}}, \qquad x\in \mathcal{E},
\end{equation*}
\notag
$$
and so, by Lemma 4.1,
$$
\begin{equation}
\int_\mathcal{E} I_N^{(1)}(x)^{p(x)}\,dx\leqslant c(p)\int_\mathcal{E}\biggl[\frac1{(1-x)^{1/4}} \int_{1-1/N}^{2x-1}\frac{f(y)\,dy}{(1-y)^{3/4}}\biggr]^{p(x)}\,dx\leqslant c(p).
\end{equation}
\tag{4.24}
$$
Now from (4.21) and (4.22)–(4.24) we eventually obtain To evaluate $J_1$, we change in succession to $x=-t$ and $y=-u$ and use the obvious fact that $K(-x,-y)=K(x,y)$. As a result,
$$
\begin{equation*}
\begin{aligned} \, J_1&=\int_{-1}^{-1+\varepsilon}\biggl(\int_{-1}^{1}K(x,y)|f(y)|\,dy\biggr)^{p(x)}\,dx \\ &=\int_{1-\varepsilon}^{1}\biggl(\int_{-1}^{1}K(t,u)|f(-u)|\,du\biggr)^{p(-t)}\,dt= \int_{1-\varepsilon}^{1}\biggl(\int_{-1}^{1}K(t,u)|g(u)|\,du\biggr)^{q(t)}\,dt. \end{aligned}
\end{equation*}
\notag
$$
Next, we note that $g\in L^{q(\cdot)}$ and Indeed, by the definition of the norm (2.2) we have
$$
\begin{equation*}
\begin{aligned} \, \|f\|_{p(\cdot)} =\inf\biggl\{\lambda>0\colon \int_{-1}^{1}\biggl|\frac{f(x)}\lambda\biggr|^{p(x)}\,dx &=\int_{-1}^{1}\biggl|\frac{f(-x)}\lambda\biggr|^{p(-x)}\,dx \\ &=\int_{-1}^{1}\biggl|\frac{g(x)}\lambda\biggr|^{q(x)}\,dx\leqslant1\biggr\}=\|g\|_{q(\cdot)}. \end{aligned}
\end{equation*}
\notag
$$
Thus, $J_1$ for $f\in L^{p(\cdot)}$ coincides with $J_3$ for the function $g\in L^{q(\cdot)}$, which is bounded by the above, since the exponent $q(x)$ satisfies the condition in the lemma, and $\|g\|_{q(\cdot)}\leqslant1$. Therefore, From (4.16), (4.25) and (4.26) we finally obtain Let us now examine the boundedness of the operator $T_2$ in $L^{p(\cdot)}$. We follow the arguments in [3]. Using the norm (2.6) we have
$$
\begin{equation*}
\begin{aligned} \, \|T_2(f)\|^*_{p(\cdot)} &=\sup_{\substack{\|g\|_{p'(\cdot)}\leqslant1 \\ g(x)\geqslant0}}\int_{-1}^{1}g(x)\,dx \int_{-1}^{1}K(y,x)|f(y)|\,dy \\ &=\sup_{\substack{\|g\|_{p'(\cdot)}\leqslant1 \\ g(x)\geqslant0}}\int_{-1}^{1}|f(y)|\,dy \int_{-1}^{1}K(y,x)g(x)\,dx \\ &=\sup_{\substack{\|g\|_{p'(\cdot)}\leqslant1 \\ g(x)\geqslant0}}\int_{-1}^{1}|f(y)|T_1(g)(y)\,dy. \end{aligned}
\end{equation*}
\notag
$$
Since both $p'(x)$ and $p(x)$ satisfy the conditions of the lemma, by the above we have Therefore, we also have $\|T_1(g)\|_{p'(\cdot)} < c(p')$. Hence, using (2.7), (2.4) and (4.1) we finally obtain
$$
\begin{equation*}
\|T_2(f)\|_{p(\cdot)} \leqslant \sup_{\substack{\|g\|_{p'(\cdot)}\leqslant1 \\ g(x)\geqslant0}} c(p)\|f\|_{p(\cdot)}\|T_1(g)\|_{p'(\cdot)} \leqslant c(p)c(p')\|f\|_{p(\cdot)} < c(p).
\end{equation*}
\notag
$$
This proves Lemma 3.1.
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\Bibitem{MagShaGad24} |
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