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 Ω is a bounded Lebesgue measurable subset of ${\mathbf{R}}^{n}$ with Lebesgue measure equal to 1 and origin lies in Ω.

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 Minkowski’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 Minkowski’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 [3–5], and in the problems of optimal embeddings [6–8]. 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

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

 □

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. □

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 $\alpha_{E}=\beta_{E}=a$. □

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 $\alpha_{H}=\beta_{H}=\alpha$. □

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 Minkowski’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 Minkowski’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 Minkowski’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. □

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 Minkowski’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 Minkowski’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. □

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 Minkowski’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 Minkowski’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. □