0;$\\
$(iii) \quad (A_{\infty} )\ implies\ (A_{p})\ for\ some\ p<\infty .$\\
The determination of extremal values of exponents is perhaps of interest for
any of these properties. The upper bound for all $p$ in the proposition $(i)$
was obtained in [3]. The limiting value for $\varepsilon $
in $(ii)$ was determined in [4].
The present paper is devoted to the determination of the exact value of
$p$ in $(iii)$. As in [3] and [4] we consider here the one-dimensional case.\\
{ \bf 1. Main results.} Let $f$ be a non negative measurable function on an interval
$[\alpha , \beta ]$. The nondecreasing rearrangement of the function $f$ on a
measurable set $E \subset [\alpha ,\beta ]$ is defined by the equality
$$f_{*}^{(r)}(t) \equiv f_{*,E}^{(r)}(t)=\sup\{\alpha >0:\ \midJ\{ x\in E:\
f(x)<\alpha \} \mid \leq t \}, \quad 0~~From this follows
$$\varepsilon ^{-p+1}\frac {1}{\varepsilon }\int _0^\varepsilon f(u)du+
\int _\varepsilon ^su^{-p}f(u)du-
\int _{\varepsilon /a}^{s/a}u^{-p}f(u)du\leq s^{-p}\int _0^sf(u)du.
\eqno(3.9)$$
Assume that the integral (3.8) is convergent. Then $u^{-p}f(u)=o(1/u),$
that is $f(u)=o(u^{p-1})$ when $u\rightarrow 0.$ Therefore
$$\varepsilon ^{-p+1}\frac {1}{\varepsilon }\int _0^\varepsilon f(u)du\leq
\varepsilon ^{-p+1}f(\varepsilon )=o(1),\quad \varepsilon \rightarrow 0.$$
Taking into account this and the convergence of the integral (3.8),
going over to the limit by $\varepsilon \rightarrow 0+$ in (3.9),
for $p>1$ we find
$$s^{-p}\int _0^s f(u)du\geq
\int _0^su^{-p}f(u)du-
\int _{0}^{s/a}u^{-p}f(u)du= \int _{s/a}^su^{-p}f(u)du\geq $$
$$\geq f\left(\frac {s}{a}\right)\int _{s/a}^su^{-p}du=
f\left(\frac {s}{a}\right)\frac {s^{-p+1}}{p-1}(a^{p-1}-1).$$
>From this, by (3.6), it follows that
$$\frac {1}{s}\int _0^sf(u)du\geq \frac {pc-1}{p-1}f\left(\frac {s}{a}\right).
\eqno(3.10)$$
If $c>1$ then, by Remark 3, $p>1.$ Since in this case
$(pc-1)/(p-1)>c$ then (3.10) contradicts (3.5).
Let $c=1.$ If $p>1$ that is $1From this for $0~~~~From this it follows that the second term on the right is equal to zero,
which is the only possible in trivial case $f(x) \equiv 0,\ 0~~From this it follows that
$$\int_\varepsilon ^1u^{-q}f(u)du\leq
c\ q\ a^{-q+1}\left(\int_{\varepsilon /a}^{\varepsilon }u^{-q}f(u)du+
\int_{\varepsilon }^{1}u^{-q}f(u)du\right)+
\int _0^1f(u)du,$$
that is
$$\left(\frac {a^{q-1}}{c\cdot q}-1\right)
\int_{\varepsilon }^{1}u^{-q}f(u)du\leq
\int_{\varepsilon /a}^{\varepsilon }u^{-q}f(u)du+
\frac {a^{q-1}}{c\cdot q}\int _0^1f(u)du\leq $$\\[-5mm]
$$\leq f(\varepsilon )\frac {\varepsilon ^{-q+1}}{q-1}(a^{q-1}-1)+
\frac {a^{q-1}}{c\cdot q}\int _0^1f(u)du.$$\\
Since $q>p$ then we obtain
$$0<\frac {a^{q-1}}{c\cdot q}-1\leq
\frac {a^{q-1}-1}{q-1}\cdot \frac {\varepsilon ^{-q+1}f(\varepsilon )}
{\int_{\varepsilon }^{1}u^{-q}f(u)du}+
\frac {a^{q-1}}{c\cdot q}\cdot \frac{\int _0^1f(u)du}
{\int_{\varepsilon }^{1}u^{-q}f(u)du}.
\eqno (3.11)$$
Let us now make use of the fact that the integral (3.8) is divergent.
Then for $q>p$ the integral $\int_{0}^{1}u^{-q}f(u)du$ diverges and therefore
the second term on the right in (3.11) tends to zero when $\varepsilon \rightarrow 0+.$
Therefore $f(\varepsilon )\varepsilon ^{-q+1}\rightarrow \infty $
when $\varepsilon \rightarrow 0+$
and hence for $0<\varepsilon <\varepsilon _0=\varepsilon _0(q)$
the inequality $f(\varepsilon )\geq \varepsilon ^{q-1}$ holds.
>From this, since $r>q$ we obtain
$$\int_0^1f^{-1/(r-1)}(t)dt
=\int_0^{\varepsilon _0}f^{-1/(r-1)}(t)dt
+\int _{\varepsilon _0}^1f^{-1/(r-1)}(t)dtJ\leq $$\\[-5mm]
$$\leq \int_0^{\varepsilon _0}t^{-(q-1)/(r-1)}dt
+\int _{\varepsilon _0}^1f^{-1/(r-1)}(t)dt<\infty .$$
By this the proof of Proposition 2 is complete.\\
{\it Proof of Theorem} 4. $a)$ Fix any interval $I\subset [\alpha ,\beta ]$
and denote $g(t)=(w\mid I)_*(t),\ 0