1} {(\tau +1)^{p-1\over p} + (\tau -1)^{p-1\over p} \over 2\tau {p-1\over p}},\tag0.4 $$ that is, the constant in (0.3). \endproclaim One may ask the corresponding question when $p=1$. It was communicated to us by Jos\'e Barrionuevo [Ba] that the best constant $C_1$ in the weak type inequality $$ |\{x: (Mf)(x)>\la \}| \le C_1 { \|f\|_{L^1} \over \la } $$ for $f$ in $\cb \cap L^1$ is $C_1=1$. This result is sharp and is analogous to our result when $p=1$. (In fact, this result is valid for the wider class of positive functions that are increasing on $(-\infty,c)$, and decreasing on $(c,\infty)$ for some number $c$.) It is still a mystery what happens for general functions $f$. It is conjectured in [DGS] that $c_p$ is the operator norm of the Hardy-Littlewood maximal function on $L^p(\Bbb R^1)$. Our methods will not work for arbitrary functions and we will point out during the proof where they break down. For general functions $f \in L^1$, the conjecture used to be that $C_1=3/2$. However, it has recently been shown by Aldaz [Al] that $C_1$ lies between $3/2$ and n $2$. This result tends to suggest that the value $c_p$ given by our Theorem is not the best constant for general $f\in L^p$. The authors would like to thank the anonymous referee for many valuable comments, and for pointing out errors in the original version of the manuscript. \bigskip {\bf 1. Some preliminary Lemmas.} \smallskip Throughout this paper we fix a $p$ with $1

0$, and
$\xi_x(0) = f(x)$. It can be
seen that $\xi_x(t)$ is a $C^\nf$ function of $t> 0$
(except at $t = |x|$, where it is merely continuous)
and that it tends to zero
as $t\ra \nf$.
Furthermore, we see that
$$ \xi_x'(t) = {{1\over2}(f(x+t)+f(x-t)) - \xi_x(t) \over t} .$$
Convexity shows us that $\xi_x'(t) \ge 0$ for $t \in (0,|x|)$, and
the third condition on $f$ shows us that $\xi_x'(t) > 0$ for $t$ close
to $|x|$. Thus we see that
$\xi_x(t)$ is non-decreasing for $t$ in some open neighborhood
of $(0, |x|]$.
The global maximum of $\xi_x$ over $[0,\nf)$ is
equal to $(Mf)(x)$. This maximum
is attained on some set of real numbers $B_x=\{ t: \xi_x(t) =
\sup_{u\ge 0}\xi_x(u)\}$. Set $\de(x)= \max B_x$. Since $B_x$ is
a closed set, it contains $\de(x)$. Note that $\de(x) > |x|$ for
$x\ne0$.
Thus
$$
(Mf)(x)= {1 \over 2 \de(x) }\int_{x-\de(x)}^{x+\de(x)}
f(t) \, dt.\tag1.1
$$
Since $\de(x)$ is a critical point of $\xi_x$, it follows that
$\xi_x'(\de(x))=0$. A simple calculation and (1.1)
give formula (1.2) below.
Now fix $x_0 \ne 0$.
By the Implicit Function Theorem, the equation
$\xi_x'(\de)=0$ can be solved for $\de$ as a smooth
function of $x$ in the vicinity of any point $(x_0,\de(x_0))$,
as long as $\dfrac{ \p \xi_x'(\de)}{ \p \de} \neq 0$ at $(x_0,\de(x_0))$.
This condition is equivalent to
$f'(x_0+\de(x_0))\neq f'(x_0-\de(x_0))$, which follows from the fact
that $x_0+\de(x_0)$ and $x_0-\de(x_0)$ lie on opposite sides of
the origin and that
$f$ has different kind of monotonicity on each side.
Therefore $\de$ coincides with a smooth function in the
neighborhood of every point $x_0 \ne 0$, which implies that
$\de(x)$ is a smooth function of $x \ne 0$.
As a consequence $(Mf)(x)$ is also smooth for $x \ne 0$.
We notice that for sufficiently small $|x|$ that
$\de(x) = (1+\tau_{2p}) |x|$ for a fixed value $\tau_{2p}$, and
that $(Mf)(x) = c_{2p} |x|^{-1/2p}$.
Thus $\de(x)$ is a continuous function of $x$.
\proclaim{Lemma 1} For $x\ne 0$, we have
$$
(Mf)(x)= {f(x+\de(x))+f(x-\de(x))\over 2},\tag1.2
$$
and
$$
(Mf)'(x)= {f(x+\de(x))-f(x-\de(x))\over 2\de(x)}.\tag1.3
$$
\endproclaim
{\smc Proof.} $\,$ (1.2) is proved as indicated above.
To prove (1.3), differentiate the identity (1.1) and use (1.2).
This completes the proof of Lemma 1. $QED$.
\medskip
Formula (1.3) indicates that the points $x+\de(x)$ and
$x-\de(x)$ are the $x$-coordinates of some two points of
intersection of the graph of $f$ with the
tangent line to the graph of $Mf$ at $(x,f(x))$.
\proclaim{Lemma 2} If $x>0$ then $\delta'(x) > 1$, and if
$x<0$ then $\delta'(x) < -1$. Moreover $Mf$ is in $\cb$ with
its maximum at $0$.
\endproclaim
{\smc Proof.} $\,$
We begin by showing that $Mf$ has no inflection points away
from $0$. Differentiating (1.2) and (1.3) we obtain
that for $x\ne 0$, we have
$$
\align
(Mf)'(x) &= f'(x+\de(x)){ (1+\de'(x))\over 2} +
f'(x-\de(x)){ (1-\de'(x))\over 2 } \tag1.4 \\
(Mf)'(x)\de'(x) +\de(x)(Mf)''(x) &=
f'(x+\de(x)){ (1+\de'(x))\over 2} -
f'(x-\de(x)){ (1-\de'(x))\over 2 }. \tag1.5
\endalign
$$
If $q \ne 0$ were an inflection point, then $(Mf)''(q)=0$,
and by (1.4) and (1.5) it would follow that
$$\align
f'(q+\de(q)) (1+\de'(q))&=(1+\de'(q))(Mf)'(q)\\
\noalign{\noindent or}
f'(q-\de(q)) (1-\de'(q))&=(1-\de'(q))(Mf)'(q).
\endalign$$
Then $(Mf)'(q) $ would be equal to either
$f'(q+\de(q))$ or $f'(q-\de(q))$. By
Lemma 1, $ (Mf)'(q)$ is the slope of the line
segment that joins $(q-\de(q),f(q-\de(q)))$ to
$(q+\de(q),f(q+\de(q)))$. By the convexity conditions
on $f$, this line would then necessarily lie on
the graph of $f$. By (1.2), this would imply that
$(Mf)(q) \le f(q)$, a contradiction if condition~(3) is imposed
upon $f$.
Therefore $Mf$ has no inflection points away from $0$, hence it is either concave or
convex there.
Since $(Mf)(x)$ looks like ${1\over x}$ near $\pm \nf$, it follows that
$Mf$ is convex on $(-\infty,0)$ and on $(0,+\infty)$.
We now show that if $x < 0$, then $\de'(x) < -1$.
Let $x_1