Given a smooth, compactly supported hypersurface S in ℝn
that does not pass through the origin, and denoting by tS the
surface dilated by a factor t>0, we can
consider the averaging operator defined for functions f∈[Sscr ],
the Schwartz class of functions, by
formula here
where dσ is Lebesgue measure on S. We can
now define a maximal average operator,
formula here
In the case where S is Sn−1,
the sphere of unit radius in ℝn, we are looking at Stein's
spherical maximal function, as treated in his paper [10].
Stein proved that M is a bounded operator on the Lp
spaces if and only if p>n/(n−1)
when n>3. Subsequently, Bourgain [2]
showed that if S is any compactly supported smooth
curve with non-vanishing Gaussian curvature, then M will be bounded on
Lp, if and only if p>2, thus dealing
with the case of the circular maximal function in the plane.
(For related results, see also [6] and [7]).
In the case where S is a curve whose curvature
vanishes to order at most m−2 at a single point,
Iosevich [4] showed that M is
bounded on Lp for p>m,
and unbounded if p = m. If we study curves given by
Γ(s) = (s, γ(s)+1),
s∈[0, 1], for some suitably smooth γ, where
γ(0) = γ′(0) = … =
γ(m−1)(0) ≠ γ(m)(0)>0,
then we can reinterpret his results as follows. Define
formula here
for Schwartz functions f. Iosevich proved that
formula here
for p>2. If we note that κ(s), the curvature
of the curve Γ(s) is approximately
2−k(m−2)
whenever s∈[2−k, 21−k],
then we have that the operator
formula here
is bounded on Lp for some p>2,
if σ is sufficiently large, since
formula here
which is finite so long as
σ>(m/p−1) (m−2)−1.
If we want to choose σ independent of m>2, the type of the curve,
such that Mσ is bounded on Lp
for some fixed p>2, then clearly we can take σ = 1/p.
In this paper we show that Mσ will be bounded
on Lp for p>max{σ−1, s}
for a class of infinitely flat, convex curves in the plane.
Counterexamples will show that this is the best possible result, in that there exist flat
curves for which Mσ is unbounded for
2<p[les ]σ−1.