The rearrangement-invariant space $\Gamma _{p,\phi }$

Abstract

Fix $b\in \mathbb R _+$ and $p\in (1,\infty )$. Let $\phi$ be a positive measurable function on $I_b:=(0,b)$. Define the Lorentz Gamma norm, $\rho _{p,\phi }$, at the measurable function $f:\mathbb R _+\rightarrow \mathbb R _+$ by $\rho _{{}_{p,\phi }}(f):=\left[ \int _0^bf^{**}(t)^p\phi (t)\,dt\right] ^{\frac{1}{p}}$, in which $f^{**}(t):=t^{-1}\int _0^tf^{*}(s)\,ds$, where $f^*(t):=\mu _f^{-1}(t)$, with $\mu _f(s):=|\{ x\in I_b: |f(x)|>s\}|$. Our aim in this paper is to study the rearrangement-invariant space determined by $\rho _{{}_{p,\phi }}$. In particular, we determine its Köthe dual and its Boyd indices. Using the latter a sufficient condition is given for a Caldéron–Zygmund operator to map such a space into itself.

Appendix

It is our purpose here to give an heuristic argument to motivate the choice of $\psi$ in (4.1) when $\phi$ is a non-trivial weight on $I_b$ satisfying $\int _{I_b}\phi =\infty$.

Now,

$$\begin{aligned} \rho _{{}_{p,\phi }}^\prime (g)=\sup _{f\in \mathfrak M _+(I_b)} \frac{\int _0^b f^*(t)g^*(t)\,dt}{\left[ \int _0^bf^{**}(t)^p\phi (t)\,dt\right] ^{\frac{1}{p}}}=:I(g), \quad g\in \mathfrak M _+(I_b). \end{aligned}$$

It suffices to consider $f(t)=\int _t^b h(s)\frac{ds}{s}$ for some $h\in \mathfrak M _+(I_b)$, $h\not = 0$ a.e.. Since, in that case,

$$\begin{aligned} \int _0^b f^*(t)g^*(t)=\int _0^b\int _t^bh(s)\frac{ds}{s}g^*(t)\,dt=\int _0^b h(t)g^{**}(t)\,dt \end{aligned}$$

and

$$\begin{aligned} f^{**}(t)=t^{-1}\int _0^t\int _s^bh(y)\frac{dy}{y}\,ds=t^{-1}\int _0^t h(s)\,ds+\int _t^bh(s)\frac{ds}{s}\approx (Sh)(t), \quad t\in I_b, \end{aligned}$$

we have

$$\begin{aligned} I(g)=\sup _{h\in \mathfrak M _+(I_b)} \frac{\int _0^b h(t)g^{**}(t)\,dt}{\left[ \int _0^b (Sh)(t)^p\phi (t)\,dt\right] ^{\frac{1}{p}}}. \end{aligned}$$

If $\overline{\phi }$ is such that

$$\begin{aligned} \int _0^b (Sh)^p\phi \le C\int _0^b h^p\overline{\phi }, \quad h\in \mathfrak M (I_b), \end{aligned}$$

(7.1)

then,

$$\begin{aligned} I(g)\ge C^{-1}\sup _{h\in \mathfrak M _+(I_b)}\frac{\int _0^bh(t)g^{**}(t)\,dt}{\left[ \int _0^bh(t)^p\overline{\phi }(t)\,dt\right] ^{\frac{1}{p}}}. \end{aligned}$$

This suggests we take $\psi (t)=\overline{\phi }(t)^{1-p^\prime }$ where $\overline{\phi }$ is, in some sense the smallest weight such that (7.1) holds. Andersen’s condition (3.5) for (7.1) leads us to solve for $\overline{\phi }(t)^{1-p^\prime }$ in the equation

$$\begin{aligned} \left[ \int _0^b \frac{\phi (s)}{(s+t)^{p}}\,ds\right] ^{\frac{1}{p}} \left[ \int _0^b \left( \frac{t}{s+t}\right) ^{p^\prime }\overline{\phi }(s)^{1-p^\prime }\,ds\right] ^{\frac{1}{p^\prime }}=1, \end{aligned}$$

(7.2)

or, what is equivalent,

$$\begin{aligned} \int _0^t \overline{\phi }(s)^{1-p^\prime }\,ds+t^{p^\prime }\int _t^b \overline{\phi }(s)^{1-p^\prime }s^{-p^\prime }\,ds =\left[ t^{-p}\int _0^t \phi (s)\,ds+ \int _t^b \phi (s)s^{-p}\,ds\right] ^{1-p^\prime }. \end{aligned}$$

Differentiation with respect to $t$ yields

$$\begin{aligned} t^{p^\prime -1}\int _t^b \overline{\phi }(s)^{1-p^\prime }s^{-p^\prime }\,ds = \int _0^t\phi (s)\,ds \left[ t^{-p}\int _0^t \phi (s)\,ds+ \int _t^b \phi (s)s^{-p}\,ds\right] ^{-p^\prime }. \end{aligned}$$

Differentiating again with respect to $t$ we get

$$\begin{aligned} \overline{\phi }(t)^{1-p^\prime }&= \frac{pp^\prime t^{pp^\prime -1}\int _0^t \phi (s)\,ds \int _t^b \phi (s)s^{-p}\,ds}{\left[ \int _0^t \phi (s)\,ds+t^{p} \int _t^b \phi (s)s^{-p}\,ds\right] ^{p^\prime +1}} \\&-\frac{t^{p^\prime }\phi (t)}{\left[ \int _0^t \phi (s)\,ds+t^{p} \int _t^b \phi (s)s^{-p}\,ds\right] ^{p^\prime }} \end{aligned}$$

It seems we essentially have

$$\begin{aligned} \overline{\phi }(t)^{1-p^\prime }=\frac{(P\phi )(t)(Q_{p}\phi )(t)}{\left[ (P\phi )(t)+(Q_{p}\phi )(t)\right] ^{p^\prime +1}}. \end{aligned}$$

(7.3)

The weight $\widehat{\phi }(t)$ given by

$$\begin{aligned} \widehat{\phi }^{1-p^\prime }(t)= \frac{t^{p^\prime }\phi (t)}{\left[ \int _0^t \phi (s)\,ds+t^{p} \int _t^b \phi (s)s^{-p}\,ds\right] ^{p^\prime }} = \frac{\phi (t)}{\left[ (P\phi )(t)+(Q_{p}\phi )(t)\right] ^{p^\prime }} \end{aligned}$$

is readily shown to satisfy Andersen’s condition (7.2) and, hence, so will

$$\begin{aligned} \frac{(P\phi )(t)(Q_{p}\phi )(t)}{\left[ (P\phi )(t)+(Q_{p}\phi )(t)\right] ^{p^\prime +1}}= \overline{\phi }(t)^{1-p^\prime }+\widehat{\phi }(t)^{1-p^\prime }. \end{aligned}$$

Now, $\overline{\phi }(t)$ will be better then $\widehat{\phi }(t)$ in (7.1) if

$$\begin{aligned} \int _0^b g^{**}(t)^{p^\prime }\widehat{\phi }(t)^{1-p^\prime }dt \le C \int _0^b g^{**}(t)^{p^\prime } \overline{\phi }(t)^{1-p^\prime }dt. \end{aligned}$$

One readily infers from Theorem 5.1 that this will be so if and only if

$$\begin{aligned} \int _0^t s^{p^\prime -1}\int _s^b \widehat{\phi }(y)^{1-p^\prime }y^{-p^\prime }\,dy\,ds\le C \int _0^t s^{p^\prime -1}\int _s^b \overline{\phi }(y)^{1-p^\prime }y^{-p^\prime }\,dy\,ds. \end{aligned}$$

But,

$$\begin{aligned} \int _s^b \overline{\phi }(y)^{1-p^\prime }y^{-p^\prime }\,dy&\approx \int _s^b y^{-p^\prime } \frac{(P\phi )(y)(Q_{p}\phi )(y)}{\left[ (P\phi )(y)+(Q_{p}\phi )(y)\right] ^{p^\prime +1}}\,dy\\&= \int _s^b \frac{y^{p-1}\int _0^y \phi (z)\,dz \int _y^b \phi (z)z^{-p}\,dz}{\left[ \int _0^y \phi (z)\,dz+y^{p} \int _y^b \phi (z)z^{-p}\,dz\right] ^{p^\prime +1}}\,dy\\&= -\frac{1}{p^\prime }\int _s^b \int _0^y \phi (z)\,dz \frac{d}{dz}\left[ \int _0^y \phi (z)\,dz\!+\!y^{p} \int _y^b\phi (z) z^{-p}\,dz\right] ^{-p^\prime }dy\\&= \left. -\frac{1}{p^\prime } \int _0^y \phi (z)\,dz \left[ \int _0^y \phi (z)\,dz+y^{p} \int _y^b\phi (z) z^{-p}\,dz\right] ^{-p^\prime }\right| _s^b\\&+\frac{1}{p^\prime } \int _s^b \phi (y)\left[ \int _0^y \phi (z)\,dz+y^{p} \int _y^b \phi (z)z^{-p}\,dz\right] ^{-p^\prime }\,dy\\&\ge \frac{1}{p^\prime }\int _s^b \widehat{\phi }(y)^{1-p^\prime }y^{-p^\prime }\,dy, \end{aligned}$$

if $\int _0^b\phi (z)\,dz=\infty$.