Boyd indices for quasi-normed function spaces with some bounds

Abstract

We calculate the Boyd indices for quasi-normed rearrangement invariant function spaces with some bounds. An application to Lorentz type spaces is also given.

1 Introduction

Let $$L_{\mathrm{loc}}$$ be the space of all locally integrable functions f on $${\mathbf{R}}^{n}$$ and $$M^{+}$$ be the cone of all locally integrable functions $$g\geq0$$ on $$(0,1)$$ with the Lebesgue measure.

Let $$f^{\ast}$$ be the decreasing rearrangement of f given by

$$f^{\ast}(t)=\inf \bigl\{ \lambda>0:\mu_{f}(\lambda)\leq t \bigr\} ,\quad t>0,$$

and $$\mu_{f}$$ be the distribution function of f defined by

$$\mu_{f}(\lambda)= \bigl\vert \bigl\{ x\in{\mathbf{R}}^{n}: \bigl\vert f(x) \bigr\vert >\lambda \bigr\} \bigr\vert _{n},$$

$$\vert \cdot \vert _{n}$$ denoting Lebesgue n-measure.

Also,

$$f^{\ast\ast}(t):=\frac{1}{t}\int_{0}^{t} f^{\ast}(s)\,ds.$$

We use the notations $$a_{1}\lesssim a_{2}$$ or $$a_{2}\gtrsim a_{1}$$ for nonnegative functions or functionals to mean that the quotient $$a_{1}/a_{2}$$ is bounded; also, $$a_{1}\approx a_{2}$$ means that $$a_{1}\lesssim a_{2}$$ and $$a_{1}\gtrsim a_{2}$$. We say that $$a_{1}$$ is equivalent to $$a_{2}$$ if $$a_{1}\approx a_{2}$$.

We consider rearrangement invariant quasi-normed spaces $$E \hookrightarrow L^{1}(\Omega)$$ such that $$\|f\|_{E}=\rho_{E}(f^{\ast})<\infty$$, where $$\rho_{E}$$ is a quasi-norm rearrangement invariant defined on $$M^{+}$$.

For simplicity, we assume that Î© is a bounded Lebesgue measurable subset of $${\mathbf{R}}^{n}$$ with Lebesgue measure equal to 1 and origin lies in Î©.

There is an equivalent quasi-norm $$\rho_{p}\approx\rho_{E}$$ that satisfies the triangle inequality $$\rho_{p}^{p}(g_{1}+g_{2})\leq\rho_{p}^{p}(g_{1})+\rho _{p}^{p}(g_{2})$$ for some $$p\in(0,1]$$ that depends only on the space E (see [1]). We say that the quasi-norm $$\rho_{E}$$ satisfies Minkowskiâ€™s inequality if for the equivalent quasi-norm $$\rho_{p}$$,

$$\rho_{p}^{p} \Bigl(\sum g_{j} \Bigr) \lesssim\sum\rho_{p}^{p}(g_{j}),\quad g_{j}\in M^{+}.$$

Usually we apply this inequality for functions $$g\in M^{+}$$ with some kind of monotonicity.

Recall the definition of the lower and upper Boyd indices $$\alpha_{E}$$ and $$\beta_{E}$$. Let $$g_{u}(t)=g(t/u)$$ if $$t<\min(1,u)$$ and $$g_{u}(t)=0$$ if $$\min(1,u)< t<1$$, where $$g\in M^{+}$$, and let

$$h_{E}(u)=\sup \biggl\{ \frac{\rho_{E}(g^{\ast}_{u})}{\rho_{E}(g^{\ast})}: g\in M^{+} \biggr\} , \quad u>0$$

be the dilation function generated by $$\rho_{E}$$. Suppose that it is finite. Then

$$\alpha_{E}:=\sup_{0< t< 1} \frac{\log h_{E}(t)}{\log t} \quad \mbox{and}\quad \beta_{E}:=\inf_{1< t< \infty} \frac{\log h_{E}(t)}{\log t}.$$

The function $$h_{E}$$ is sub-multiplicative, increasing, $$h_{E}(1)=1$$, $$h_{E}(u) h_{E}(1/u)\geq1$$ and hence $$0\leq\alpha _{E}\leq\beta_{E}$$. We suppose that $$0<\alpha_{E}=\beta_{E}\leq1$$.

If $$\beta_{E}<1$$, we have by using Minkowskiâ€™s inequality that $$\rho_{E}(f^{\ast})\approx\rho_{E}(f^{\ast\ast})$$. In particular, $$\|f\|_{E}\approx\rho_{E}(f^{\ast\ast})$$ if $$\beta_{E}<1$$. For example, consider the gamma spaces $$E=\Gamma^{q}(w)$$, $$0< q\leq\infty$$, w-positive weight, that is, a positive function from $$M^{+}$$, with a quasi-norm $$\|f\|_{\Gamma^{q}(w)}:=\rho_{E}(f^{\ast})$$, $$\rho_{E}(g):=\rho_{w,q}(\int_{0}^{1} g(tu)\,du)$$, where

$$\rho_{w,q}(g):= \biggl(\int_{0}^{1} \bigl[ g(t)w(t) \bigr]^{q} \,dt/t \biggr)^{1/q},\quad g\in M^{+}$$
(1.1)

and

$$\biggl(\int_{0}^{1} w^{q}(t)\,dt/t \biggr)^{1/q}< \infty.$$

Then $$L^{\infty}(\Omega)\hookrightarrow\Gamma^{q}(w)\hookrightarrow L^{1}(\Omega)$$. If $$w(t)=t^{1/p}$$, $$1< p<\infty$$, we write as usual $$L^{p,q}$$ instead of $$\Gamma^{q}(t^{1/p})$$. Consider also the classical Lorentz spaces $$\Lambda ^{q} (w)$$, $$0< q\leq\infty$$; $$f\in\Lambda^{q}(w)$$ if $$\|f\|_{\Lambda ^{q}_{w}}:=\rho_{w,q}(f^{\ast})<\infty$$, $$w(2t)\approx w(t)$$. We suppose that $$L^{\infty}(\Omega)\hookrightarrow\Lambda ^{q}(w)\hookrightarrow L^{1}(\Omega)$$.

The Boyd indices are useful in various problems concerning continuity of operators acting in rearrangement invariant spaces [2] or in optimal couples of rearrangement invariant spaces [3â€“5], and in the problems of optimal embeddings [6â€“8]. The main goal of this paper is to provide formulas for the Boyd indices with some bounds of rearrangement invariant quasi-normed spaces and to apply these results to the case of Lorentz type spaces.

2 Boyd indices for quasi-normed function spaces

Let $$\rho_{E}$$ be a monotone quasi-norm on $$M^{+}$$ and let E be the corresponding rearrangement invariant quasi-normed space consisting of all $$f\in L^{1}(\Omega)$$ such that $$\|f\|_{E}=\rho_{E}(f^{\ast})<\infty$$.

Theorem 2.1

Let

$$g_{u}(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} g(t/u) & \textit{if } 0< t< \min(1,u), \\ 0 & \textit{if } \min(1,u)\leq t< 1, \end{array}\displaystyle \displaystyle \displaystyle \displaystyle \right .$$

where $$g\in M^{+}$$, and let

$$h_{E}(u)=\sup \biggl\{ \frac{\rho_{E}(g^{\ast}_{u})}{\rho_{E}(g^{\ast})}: g\in M^{+} \biggr\} ,\quad u>0,$$

be the dilation function generated by $$\rho_{E}$$. Suppose that it is finite. Then the Boyd indices are well defined

$$\alpha_{E}:=\sup_{0< t< 1} \frac{\log h_{E}(t)}{\log t} \quad \textit{and}\quad \beta_{E}:=\inf_{1< t< \infty} \frac{\log h_{E}(t)}{\log t}$$

and they satisfy

\begin{aligned}& \alpha_{E}=\lim_{t\to0} \frac{\log h_{E}(t)}{\log t}, \end{aligned}
(2.1)
\begin{aligned}& \beta_{E}=\lim_{t\to\infty} \frac{\log h_{E}(t)}{\log t}. \end{aligned}
(2.2)

In particular, $$0\leq\alpha_{E}\leq\beta_{E}\leq\frac{\log h_{E}(2)}{\log2}$$.

Proof

We have

$$g_{uv}=(g_{u})_{v} \quad\mbox{if } u< v.$$
(2.3)

Indeed, since $$\min(1,uv)\leq\min(1,v)$$ for $$u< v$$, we find $$(g_{u})_{v}(t)=g_{u}(t/(uv))$$ if $$0< t<\min(1,uv)$$ and $$(g_{u})_{v}(t)=0$$ if $$\min (1,uv)\leq t<1$$. Thus (2.3) is proved. This implies that the function $$h_{E}$$ is sub-multiplicative.

Further, the function $$\omega(x)=\log h_{E}(e^{x})$$ is sub-additive increasing on $$(-\infty,\infty)$$ and $$\omega(0)=0$$. Hence, by [2], Lemma 5.8, (2.2) is satisfied and evidently $$\beta _{E}\leq \frac{\log h_{E}(2)}{\log2}$$.

Since $$h_{E}(1)=1$$ and $$h_{E}$$ is sub-multiplicative, therefore

$$h_{E}(u_{1} u_{2})\leq h_{E}(u_{1})h_{E}(u_{2}).$$

Replacing $$u_{2}$$ by $$\frac{1}{u_{1}}$$, we get

$$h_{E}(1)\leq h_{E}(u_{1})h_{E} \biggl(\frac{1}{u_{1}} \biggr),$$

which implies that

$$1\leq h_{E}(u_{1})h_{E} \biggl(\frac{1}{u_{1}} \biggr); \quad\mbox{because } h_{E}(1)=1,$$

it follows that $$1\leq h_{E}(u) h_{E}(1/u)$$.

We have

$$\alpha_{E}\leq\beta_{E}.$$

Indeed

$$\log \bigl(h_{E}(u) \bigr)\geq\log \biggl(\frac{1}{h_{E}(\frac{1}{u})} \biggr),$$

if $$u>1$$, then

$$\frac{\log(h_{E}(u))}{\log u}\geq\frac{\log (\frac{1}{h_{E}(\frac {1}{u})} )}{\log u}=\frac{\log (h_{E}(\frac{1}{u}) )}{\log\frac{1}{u}},$$

which implies that

$$\lim_{u\rightarrow\infty}\frac{\log(h_{E}(u))}{\log u}\geq\lim_{u\rightarrow\infty} \frac{\log (h_{E}(\frac{1}{u}) )}{\log \frac{1}{u}}.$$

Since $$\beta_{E}$$ is finite, therefore $$\alpha_{E}$$ is also finite. Since $$h_{E}(1)=1$$ and we know that $$h_{E}$$ is increasing function, so

$$h_{E}(u)\leq1 \quad\mbox{for } 0< u< 1,$$

which implies that

$$\log \bigl(h_{E}(u) \bigr)\leq0,$$

which implies that

$$\frac{\log(h_{E}(u))}{\log u}\geq0,$$

which implies that

$$\alpha_{E}=\sup_{0< u< 1}\frac{\log(h_{E}(u))}{\log u}\geq0,$$

and hence

$$0\leq\alpha_{E}\leq\beta_{E}.$$

â€ƒâ–¡

Let $$\rho_{H}$$ be a monotone quasi-norm on $$M^{+}$$ and let H be the corresponding quasi-normed space, consisting of all locally integrable functions on $$(0,1)$$ with a finite quasi-norm $$\|g\|_{H}=\rho_{H}(|g|)$$.

Theorem 2.2

Let

$$(\Psi_{u}g) (t)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} g(ut), & \textit{if } 0< t< \min(1,\frac{1}{u}), \\ g(1), & \textit{if } \min(1,\frac{1}{u})\leq t< 1, \end{array}\displaystyle \displaystyle \displaystyle \displaystyle \right .$$

where $$g\in M^{+}$$, and let

$$h_{H}(u)=\sup \biggl\{ \frac{\rho_{H}(\Psi_{u} g)}{\rho_{H}(g)}: g \in G_{a} \biggr\} ,\quad u>0,$$

be the dilation function generated by $$\rho_{H}$$. Suppose that it is finite, where

$$G_{a}:= \bigl\{ g\in M^{+}: t^{-a/n}g(t) \textit{ is decreasing} \bigr\} ,\quad a>0.$$

Then the Boyd indices are well defined

$$\alpha_{H}:=\sup_{0< t< 1} \frac{\log h_{H}(t)}{\log t} \quad \textit{and}\quad \beta_{H}:=\inf_{1< t< \infty} \frac{\log h_{H}(t)}{\log t}$$

and they satisfy

\begin{aligned}& \alpha_{H}=\lim_{t\to0} \frac{\log h_{H}(t)}{\log t}, \end{aligned}
(2.4)
\begin{aligned}& \beta_{H}=\lim_{t\to\infty} \frac{\log h_{H}(t)}{\log t}. \end{aligned}
(2.5)

In particular, $$\frac{\log h_{H}(1/2)}{\log1/2}\leq\alpha_{H}\leq\beta _{H}\leq a/n$$.

Proof

We have

$$\Psi_{uv}g=\Psi_{u}(\Psi_{v} g) \quad\mbox{if } u< v.$$
(2.6)

Indeed, since $$\min(1,1/(uv))\leq\min(1,1/u)$$ for $$u< v$$, we find $$\Psi_{u}(\Psi_{v} g)(t)= g(t/(uv))$$ if $$0< t<\min(1,1/(uv))$$ and $$\Psi_{u}(\Psi _{v} g)(t)=g(1)$$ if $$\min(1,1/(uv))\leq t<1$$. Thus (2.6) is proved. This implies that the function $$h_{H}$$ is sub-multiplicative. Since the function $$u^{-a/n}h_{H}(u)$$ is decreasing, it follows that the function $$u^{a/n} h_{H}(1/u)$$ is increasing and sub-multiplicative. Hence we can apply the results from Theorem 2.1. This establishes Theorem 2.2.â€ƒâ–¡

Example 2.3

If $$E=\Lambda^{q} (t^{a} w)$$, $$0\leq a\leq1$$, $$0< q\leq\infty$$, where w is slowly varying, then $$\alpha_{E}=\beta_{E}=a$$.

Proof

We give a proof for $$0< q<\infty$$, the case $$q=\infty$$ is analogous. We have, for $$g\in M^{+}$$,

$$\rho_{E} \bigl(g^{\ast}_{u} \bigr)= \biggl(\int _{0}^{1} \bigl[g^{\ast}_{u}(t)t^{a} w(t) \bigr]^{q} \,dt/t \biggr)^{1/q}= \biggl(\int _{0}^{\min(1,u)} \bigl[g^{\ast}(t/u)t^{a} w(t) \bigr]^{q} \,dt/t \biggr)^{1/q}$$

and by a change of variables,

$$\rho_{E} \bigl(g^{\ast}_{u} \bigr) \leq \biggl(\int_{0}^{1} \bigl[g^{\ast}(t) (tu)^{a} w(tu) \bigr]^{q} \,dt/t \biggr)^{1/q}.$$
(2.7)

From the definition of a slowly varying function it follows that for every $$\varepsilon>0$$, $$t^{-\varepsilon}w(t)\approx d(t)$$, where d is a decreasing function. Then, for $$u>1$$, we have $$d(tu)\leq d(t)$$, thus

$$(tu)^{-\varepsilon}w(tu)\lesssim d(t u) \lesssim t^{-\varepsilon}w(t),$$

which implies that

$$w(tu)\lesssim u^{\varepsilon}w(t),\quad u>1.$$
(2.8)

Inserting this estimate in (2.7), we arrive at

$$\rho_{E} \bigl(g^{\ast}_{u} \bigr)\lesssim u^{a+\varepsilon}\rho_{E} \bigl(g^{\ast}\bigr),\quad u>1,$$

which yields $$h_{E}(u)\lesssim u^{a+\varepsilon}$$, $$u>1$$. Then it follows that $$\beta_{E}\leq a+\varepsilon$$. Analogously, $$\alpha_{E}\geq a-\varepsilon$$. Since $$\varepsilon>0$$ is arbitrary and $$\alpha_{E}\leq\beta_{E}$$, we obtain $$\alpha_{E}=\beta_{E}=a$$.â€ƒâ–¡

Example 2.4

If $$H=L^{q}_{\ast}(w(t)t^{-\alpha})$$, $$0\leq\alpha< a/n$$, $$0< q\leq\infty$$, where w is slowly varying, then $$\alpha_{H}=\beta_{H}=\alpha$$.

Proof

We give a proof for $$0< q<\infty$$, the case $$q=\infty$$ is analogous. We have, for $$g\in G_{a}$$,

\begin{aligned} \rho_{H}(\Psi_{u} g)&= \biggl(\int_{0}^{1} \bigl[\Psi_{u} g(t)t^{-\alpha} w(t) \bigr]^{q} \,dt/t \biggr)^{1/q} \\ &= \biggl(\int_{0}^{\min(1,1/u)} \bigl[g(tu)t^{-\alpha} w(t) \bigr]^{q} \,dt/t \biggr)^{1/q}+I(u), \end{aligned}

where $$I(u)= (\int_{\min(1,1/u)}^{1} [t^{-\alpha} w(t)]^{q} \,dt/t)^{1/q}g(1)$$. Note that $$I(u)=0$$ for $$0< u<1$$. Since for every $$\varepsilon>0$$ we have $$w(t)\lesssim t^{\varepsilon}$$, it follows that $$I(u)\lesssim u^{\alpha+\varepsilon} g(1)$$, $$u>1$$. Also, $$g(1)\rho _{H}(t^{a/n})\leq\rho_{H}(g)$$ and $$\rho_{H}(t^{a/n})<\infty$$ due to $$\alpha< a/n$$.

On the other hand, by a change of variables,

$$\rho_{H}(\Psi_{u} g)\lesssim \biggl(\int _{0}^{1} \bigl[g(t) (t/u)^{-\alpha} w(t/u) \bigr]^{q} \,dt/t \biggr)^{1/q}+u^{\alpha+\varepsilon} \rho_{H}(g).$$

As in the proof of the previous example, we have

$$w(t/u)\lesssim u^{\varepsilon}w(t),\quad u>1,$$

therefore

$$\rho_{H}(\Psi_{u} g)\lesssim u^{\alpha+\varepsilon} \rho_{H}(g),\quad u>1, g\in G_{a}.$$

Hence $$h_{H}(u)\lesssim u^{\alpha+\varepsilon}$$, $$u>1$$. Then it follows that $$\beta_{H}\leq\alpha+\varepsilon$$. Analogously, $$\alpha_{H}\geq\alpha -\varepsilon$$. Since $$\varepsilon>0$$ is arbitrary and $$\alpha_{H}\leq\beta_{H}$$, we obtain $$\alpha_{H}=\beta_{H}=\alpha$$.â€ƒâ–¡

3 Basic inequalities

Here we prove a few inequalities, which are of independent interest.

Theorem 3.1

If $$\alpha<\alpha_{H}$$, then

$$\rho_{H} \biggl(t^{\alpha}\int_{0}^{t} s^{-\alpha}g(s)\frac{ds}{s} \biggr)\lesssim\rho_{H}(g), \quad g \in G_{a}$$

and if $$\beta_{H}<\beta$$, then

$$\rho_{H} \biggl(t^{\beta}\int_{t}^{1} s^{-\beta}g(s)\frac{ds}{s} \biggr)\lesssim \rho_{H}(g), \quad g \in G_{a}.$$

Proof

We are going to use Minkowskiâ€™s inequality for the equivalent p-norm of $$\rho_{H}$$. To this end, first we replace the integrals by sums using monotonicity properties of $$g\in G_{a}$$.

Thus

\begin{aligned} t^{\alpha}\int_{0}^{t} s^{-\alpha}g(s) \frac{ds}{s}&=\int_{0}^{1} v^{-\alpha }g(tv)\frac{dv}{v} \\ &=\sum_{l=-\infty}^{0}\int_{2^{l-1}}^{2^{l}} v^{-\alpha}g(tv)\frac{dv}{v} \\ &\lesssim\sum_{l=-\infty}^{0} 2^{-l\alpha}g \bigl(t2^{l} \bigr). \end{aligned}

Applying Minkowskiâ€™s inequality, we get

\begin{aligned} \rho^{p}_{H} \biggl(t^{\alpha}\int _{0}^{t} s^{-\alpha}g(s)\frac{ds}{s} \biggr) &\lesssim\sum_{l=-\infty}^{0} 2^{-lp\alpha}\rho^{p}_{H} \bigl(g \bigl(t2^{l} \bigr) \bigr) \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=-\infty}^{0} 2^{-p\alpha l} h^{p}_{H} \bigl(2^{l} \bigr) \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=-\infty}^{0} 2^{-p\alpha l} 2^{{lp(\alpha_{H}-\varepsilon)}} \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=-\infty}^{0} 2^{{lp(\alpha _{H}-\varepsilon-\alpha)}}. \end{aligned}

The above series is convergent if we choose $$\varepsilon>0$$ such that $$\varepsilon< \alpha_{H}-\alpha$$, so we have

$$\rho_{H} \biggl(t^{\alpha}\int_{0}^{t} s^{-\alpha}g(s)\frac{ds}{s} \biggr)\lesssim\rho_{H}(g).$$

On the other hand, for $$g\in G_{a}$$,

\begin{aligned} t^{\beta}\int_{t}^{1} s^{-\beta}g(s) \frac{ds}{s}&=\int_{1}^{\infty}\chi _{(0,1)}(tv) v^{-\beta}g(tv)\frac{dv}{v} \\ &=\sum_{l=0}^{\infty}\int_{2^{l}}^{2^{l+1}} \chi_{(0,1)}(tv) v^{-\beta }g(tv)\frac{dv}{v} \\ &\lesssim\sum_{l=0}^{\infty}2^{-l\beta}g \bigl(t2^{l} \bigr)\chi _{(0,1)} \bigl(t2^{l} \bigr). \end{aligned}

Again applying Minkowskiâ€™s inequality, we get

\begin{aligned} \rho^{p}_{H} \biggl( t^{\beta}\int _{t}^{1} s^{-\beta}g(s)\frac{ds}{s} \biggr) &\lesssim\sum_{l=0}^{\infty}2^{-l\beta p} \rho^{p}_{H} \bigl(g \bigl(t2^{l} \bigr)\chi _{(0,1)} \bigl(t2^{l} \bigr) \bigr) \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=0}^{\infty}2^{-l\beta p} h^{p}_{H} \bigl(2^{l} \bigr) \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=0}^{\infty}2^{-l\beta p} 2^{pl(\beta _{H}+\varepsilon)} \\ &\lesssim\rho^{p}_{H}(g)\sum_{l=0}^{\infty}2^{{lp(\beta _{H}+\varepsilon-\beta)}}. \end{aligned}

The above series is finite if we choose a suitable $$\varepsilon>0$$ such that $$\varepsilon< \beta-\beta_{H}$$. The proof is finished.â€ƒâ–¡

Theorem 3.2

If $$\beta_{E}< a$$, then

$$\rho_{E} \biggl(t^{-a}\int_{0}^{t} s^{a}g(s)\frac{ds}{s} \biggr)\lesssim\rho _{E}(g), \quad g \in D_{0},$$

where $$D_{0}:=\{g \in M ^{+} : g(t) \textit{ is decreasing and } g(t)=0 \textit{ for } t \geq1\}$$.

Proof

We are going to use Minkowskiâ€™s inequality for the equivalent p-norm of $$\rho_{E}$$. To this end, first we replace the integral by sums using monotonicity properties of $$g\in D_{0}$$.

Thus

\begin{aligned} t^{-a}\int_{0}^{t} s^{a}g(s) \frac{ds}{s}&=\int_{0}^{1} v^{a}g(tv)\frac{dv}{v} \\ &=\sum_{l=-\infty}^{0}\int_{2^{l}}^{2^{l+1}} v^{a}g(tv)\frac{dv}{v} \\ &\lesssim\sum_{l=-\infty}^{0} 2^{{al}}g \bigl(t2^{l} \bigr). \end{aligned}

Applying Minkowskiâ€™s inequality, we get

\begin{aligned} \rho^{p}_{E} \biggl(t^{-a}\int _{0}^{t} s^{a}g(s)\frac{ds}{s} \biggr) &\lesssim\sum_{l=-\infty}^{0} 2^{{pal}}\rho^{p}_{E} \bigl(g \bigl(t2^{l} \bigr) \bigr) \\ &\lesssim\rho^{p}_{E}(g)\sum_{l=-\infty}^{0} 2^{{pal}}\ h^{p}_{E} \bigl(2^{1} \bigr) \\ &\lesssim\rho^{p}_{E}(g)\sum_{l=-\infty}^{0} 2^{{pal}} \ 2^{{-1p(\beta_{E}+\varepsilon)}} \\ &\lesssim\rho^{p}_{E}(g)\sum_{l=-\infty}^{0} 2^{{lp(a-\beta _{E}-\varepsilon)}}. \end{aligned}

The above series is finite if we choose $$\varepsilon>0$$ such that $$\varepsilon< a-\beta_{E}$$, and this concludes the proof.â€ƒâ–¡

Theorem 3.3

If $$\alpha_{E}>0$$, then

$$\rho_{E} \biggl(\int_{t}^{1} g(u) \frac{du}{u} \biggr)\lesssim\rho _{E}(g),\quad g\in D_{0}.$$

Proof

We are going to use Minkowskiâ€™s inequality for the equivalent p-norm of $$\rho_{E}$$. To this end, first we replace the integral by sums using monotonicity properties of $$g\in D_{0}$$.

Thus

\begin{aligned} \int_{t}^{1} g(u)\frac{du}{u} & \lesssim\int _{1}^{\infty}\chi_{(0,1)}(tv)g(tv) \frac{dv}{v} \\ &= \sum_{l=0}^{\infty}\int_{2^{l}}^{2^{l+1}} \chi_{(0,1)}(tv)g(tv)\frac {dv}{v} \\ &\lesssim\sum_{l=0}^{\infty}\chi_{(0,1)} \bigl(t2^{l} \bigr)g \bigl(t2^{l} \bigr). \end{aligned}

Applying Minkowskiâ€™s inequality, we get

\begin{aligned} \rho^{p}_{E} \biggl(\int_{t}^{1} g(u)\frac{du}{u} \biggr) &\lesssim\sum_{l=0}^{\infty}\rho^{p}_{E} \bigl(\chi_{(0,1)} \bigl(t2^{l} \bigr)g \bigl(t2^{l} \bigr) \bigr) \\ &\lesssim\rho^{p}_{E}(g)\sum_{l=0}^{\infty}h^{p}_{E} \bigl(2^{-l} \bigr) \\ &\lesssim\rho^{p}_{E}(g)\sum_{l=0}^{\infty}2^{-l(\alpha _{E}-\varepsilon)}. \end{aligned}

Choosing $$\varepsilon>0$$ such that $$\alpha_{E}>\varepsilon$$, we conclude the proof.â€ƒâ–¡

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Acknowledgements

The authors are thankful to the editor and the referees for their valuable suggestions in improving the final version of the article.

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Correspondence to Shin Min Kang.

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Nazeer, W., Mehmood, Q., Nizami, A.R. et al. Boyd indices for quasi-normed function spaces with some bounds. J Inequal Appl 2015, 235 (2015). https://doi.org/10.1186/s13660-015-0754-9