%Authors: L. Grafakos, S. Montgomery-Smith, and O. Motrunich %Title: A sharp estimate for the Hardy-Littlewood maximal function %Filename: grafakosmontsmithmotrunichhardy.tex %TeX: AMSTeX %Length: ????? %Received Date N/A %SubjectClass: 42B25 %Abstract: The best constant in the usual $L^p$ norm inequality for the centered % Hardy-Littlewood maximal function on $\Bbb R^1$ is obtained for the % class of all peak-shaped'' functions. A positive function % on the line is called peak-shaped'' if it is positive and % convex except at one point. % The techniques we use include convexity and % an adaptation of the standard Euler-Langrange variational % method. %Citation: preprint. %Special character check block %32 space 33 ! exclam. pt. 34 " double quote 35 # sharp %36 $dollar 37 % percent 38 & ampersand 39 ' prime %40 ( left paren. 41 ) rt. paren. 42 * asterisk 43 + plus %44 , comma 45 - minus 46 . period 47 / divide %58 : colon 59 ; semi-colon 60 < less than 61 = equal %62 > greater than 63 ? question mark 64 @ at %91 [ left bracket 92 \ backslash 93 ] right bracket 94 ^ caret %95 _ underline 96  left single quote %123 { left brace 124 | vertical bar 125 } right brace 126 ~ tilda %Insert your TeX file starting here. \input amstex \documentstyle{amsppt} \magnification\magstep1 \NoRunningHeads %\hsize 6.5 truein \vsize 8.0 truein %general \define\k{\ref\key} \define\pp{\pages} \define\en{\endref} \define\nod{\noindent} \define\f{\frac} \define\tf{\tfrac} \define\q{\quad} \define\qq{\qquad} \define\qqq{\quad\qquad} \define\qqqq{\qquad\qquad} %blackboard bold \define\r1{\Bbb R^1} \define\rr{\Bbb R} \define\rp{\Bbb R^+} \define\rn{\Bbb R^n} \define\zz{\Bbb Z} \define\zp{\Bbb Z^+} \define\torus{\Bbb T^1} \define\tn{\Bbb T^n} \define\cc{\Bbb C^1} \define\cn{\Bbb C^n} \define\hn{\Bbb H^n} %calligraphic letters \define\cb{\Cal P} \define\cd{\Cal D} \define\ce{\Cal E} \define\cf{\Cal F} \define\cm{\Cal M} \define\cs{\Cal S} %math symbols \define\p{\partial} \define\ppr{p^\prime} \define\nf{\infty} \define\pnf{+\infty} \define\mnf{-\infty} \define\ci{C^\infty} \define\coi{C_0^\infty} \define\lo{ {L^1} } \define\lt{ {L^2} } \define\lp{ {L^p} } \define\li{ {L^\infty} } \define\half{\f{1}{2}} \define\dtt{\, \frac{dt}{t}} \define\dss{\, \frac{ds}{s}} \define\drr{\, \frac{dr}{r}} \define\pv{{\text{p.v.}}} \define\loc{ {\text{loc}} } \define\il{\int\limits} \define\ra{\rightarrow} %Greek \define\al{\alpha} \define\be{\beta} \define\ga{\gamma} \define\Ga{\Gamma} \define\de{\delta} \define\De{\Delta} \define\ve{\varepsilon} \define\ze{\zeta} \define\th{\theta} \define\Th{\Theta} \define\la{\lambda} \define\La{\Lambda} \define\si{\sigma} \define\Si{\Sigma} \define\vp{\varphi} \define\om{\omega} \define\Om{\Omega} \hsize=30truecc \baselineskip=16truept \topmatter \title A sharp estimate for the Hardy-Littlewood maximal function\endtitle \author Loukas Grafakos$^*$, Stephen Montgomery-Smith$^{**}$, and Olexei Motrunich$^{***}$\endauthor \affil University of Missouri, Columbia and Princeton University\endaffil \noindent \address Department of Mathematics, University of Missouri, Columbia, MO 65211, \newline Department of Mathematics, University of Missouri, Columbia, MO 65211, \newline Department of Physics, Princeton University, Princeton, NJ 08544 \endaddress \email \newline \noindent loukas\@math.missouri.edu, stephen\@math.missouri.edu, motrunich\@princeton.edu \endemail \thanks$\,\,\,\,\,\, ^*$Research partially supported by the University of Missouri Research Board. \endthanks \thanks$\,\,\, ^{**}$Research partially supported by the National Science Foundation. \endthanks \thanks$^{***}$Research partially supported by the University of Missouri-Columbia Research Council. \endthanks \abstract The best constant in the usual$L^p$norm inequality for the centered Hardy-Littlewood maximal function on$\Bbb R^1$is obtained for the class of all peak-shaped'' functions. A function on the line is called peak-shaped'' if it is positive and convex except at one point. The techniques we use include variational methods. AMS Classification (1991): 42B25 \endabstract \endtopmatter \document {\bf 0. Introduction.} \smallskip Let $$(Mf)(x)= \sup_{\de >0} {1 \over 2 \de }\int_{x-\de}^{x+\de} |f(t)| \, dt\tag0.1$$ be the centered Hardy-Littlewood maximal operator on the line. This paper grew out from our attempt to find the operator norm of$M$on$\lp (\r1)$. Since$M$is a positive operator, we may restrict our attention to positive functions. Let$\cb$be the set of all positive functions$f$on$\r1$, which are convex except at one point (where we also allow$f$to be discontinuous). We call such functions `peak-shaped.'' We were able to find the best constant in the inequality $$\|Mf\|_{L^p} \le C(p) \|f\|_{L^p}\qqq \text{for f in \cb \cap \lp.}\tag0.2$$ for$11} {(\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 } $$for f in \cb \cap L^1 is C_1=1. This result is sharp and is analogous to our result when p=1. (In fact, this result is valid for the wider class of positive functions that are increasing on (-\infty,c), and decreasing on (c,\infty) for some number c.) It is still a mystery what happens for general functions f. It is conjectured in [DGS] that c_p is the operator norm of the Hardy-Littlewood maximal function on L^p(\Bbb R^1). Our methods will not work for arbitrary functions and we will point out during the proof where they break down. For general functions f \in L^1, the conjecture used to be that C_1=3/2. However, it has recently been shown by Aldaz [Al] that C_1 lies between 3/2 and n 2. This result tends to suggest that the value c_p given by our Theorem is not the best constant for general f\in L^p. The authors would like to thank the anonymous referee for many valuable comments, and for pointing out errors in the original version of the manuscript. \bigskip {\bf 1. Some preliminary Lemmas.} \smallskip Throughout this paper we fix a p with 10, and \xi_x(0) = f(x). It can be seen that \xi_x(t) is a C^\nf function of t> 0 (except at t = |x|, where it is merely continuous) and that it tends to zero as t\ra \nf. Furthermore, we see that$$ \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$$Then (Mf)'(q)  would be equal to either f'(q+\de(q)) or f'(q-\de(q)). By Lemma 1,  (Mf)'(q) is the slope of the line segment that joins (q-\de(q),f(q-\de(q))) to (q+\de(q),f(q+\de(q))). By the convexity conditions on f, this line would then necessarily lie on the graph of f. By (1.2), this would imply that (Mf)(q) \le f(q), a contradiction if condition~(3) is imposed upon f. Therefore Mf has no inflection points away from 0, hence it is either concave or convex there. Since (Mf)(x) looks like {1\over x} near \pm \nf, it follows that Mf is convex on (-\infty,0) and on (0,+\infty). We now show that if x < 0, then \de'(x) < -1. Let x_1 x_2+\de(x_2)  which proves that x+\de(x) is decreasing on (-\nf ,a). Therefore \de'(x) < -1 on (-\nf ,0). Likewise one can show that \de'(x) > 1 on (0 ,+\nf). This completes the proof of Lemma 2. QED. \bigskip {\bf 2. The variational functional.} \smallskip It will be convenient to have \de(x) \le 0 for x<0. To achieve this we define s(x) to be equal to \de(x) for x>0, to be equal to -\de(x) for x<0, and equal to 0 if x=0. We observe that (1.1), (1.2), and (1.3) remain valid for s(x). We also observe that s'(x)>1 for x \ne 0 and that \roster \item \lim\limits_{x\ra +\nf}x+s(x)= +\nf, \item \lim\limits_{x\ra -\nf}x+s(x)= -\nf, \item \lim\limits_{x\ra +\nf}x-s(x)= \inf (\text{support(f)}), \item \lim\limits_{x\ra -\nf}x-s(x)= \sup (\text{support(f)}). \endroster For simplicity, we denote by g=Mf the maximal function of f. It turns out that a suitable convex combination of the integrals of the functions (g(x)-g'(x)s(x))^p(s'(x)-1) and (g(x)+g'(x)s(x))^p(s'(x)+1) will give rise to a functional related to \|f\|_{L^p}^p. Our goal will be to minimize this functional by selecting a suitable s(x). To find such a minimizer, we solve the corresponding Euler-Langrange equations. By Lemma 1 we have that$$f(x+s(x))=g(x)+g'(x)s(x),\tag2.1$$and$$f(x-s(x))=g(x)-g'(x)s(x).\tag2.2$$Raise both sides of (2.1) to the power p, multiply them by 1+s'(x), and integrate from -\nf to \nf to obtain$$ \align &\int_{-\nf}^{+\nf} (g(x)+g'(x) s(x))^p (s'(x)+1)\,dx \\ =& \int_{-\nf}^{+\nf} f(x+s(x))^p(s'(x)+1)\,dx \\ =& \int_{-\nf}^{+\nf} f(x)^p\,dx = \|f\|_{L^p}^p \tag2.3 \endalign $$Similarly, raise both sides of (2.2) to the power p, multiply them by s'(x)-1, and integrate from -\nf to \nf. We obtain$$ \align &\int_{-\nf}^{+\nf} (g(x)-g'(x) s(x))^p (s'(x)-1)\,dx \\ =& \int_{-\nf}^{+\nf} f(x-s(x))^p(s'(x)-1)\,dx \\ =& \int_{-\nf}^{+\nf} f(-x)^p\,dx = \|f\|_{L^p}^p \tag2.4 \endalign $$At this point, we remark that the calculation above will not work for general functions f, because in that case the function s(x) will have many discontinuities, and above formulae will have to include terms needed to account for these discontinuities. These discontinuities are generally rather unpredictable, and we have not been able to find a way to deal with this problem. Let \half <\al <1 be a real number to be selected later to depend on p only. Let F be the following function of three variables:$$ F(x,y,z)= \al (g(x)+g'(x)y)^p(z+1) + (1-\al) (g(x)-g'(x)y)^p(z-1).\tag2.5 $$The domain of F is the set of all (x,y,z) which satisfy \roster \item -\nf \half. \smallskip {\smc Proof.} To prove Lemma 3, rewrite (2.7) as$$[ \al(g(x)+g'(x)\phi(x))^{p-1}-(1-\al) p (g(x)-g'(x)\phi(x))^{p-1}] g''(x) \phi(x) =0.$$Then substituting for \phi = s_0, we obtain the result. QED. \smallskip We would like to to be able to directly deduce that I(s)\ge I(s_0). Unfortunately, general theorems from calculus of variations (see for example [Br]) are not directly applicable here since F(x,y,z) does not satisfy the usual convexity conditions needed. As it turns out, the desired inequality I(s)\ge I(s_0) will be a consequence of the key inequality below which is true because of the very specific structure of the function F(x,y,z). \proclaim{Lemma 4} For all x\ne0 we have$$ \leqalignno{ & F(x, s(x),s'(x)) - F(x,s_0(x),s_0'(x)) \cr &\ge (\p_2 F)(x,s_0(x),s_0'(x)) (s(x)-s_0(x)) \cr & \quad + (\p_3 F)(x,s_0(x),s_0'(x)) (s'(x)-s_0'(x)) &(2.9)\cr } $$\endproclaim {\smc Proof.} We observe that if h is a convex function on an interval J, then for all x,y in J we have$$ h(x)-h(y) \ge h'(y) (x-y)\tag2.10 $$irrespectively of the order of x and y. Next observe that for all x and all z> 1 the function F(x,y,z) is convex in y. This is because when z>1, (\p_2^2 F)(x,y,z) >0 for all x,y. Let x \ne 0. Then by Lemma 2, s'(x)=|\de'(x)|> 1 and$$ \leqalignno{ & F(x,s(x),s'(x))- F(x,s_0(x),s_0'(x)) \cr &= [F(x,s(x),s'(x))-F(x,s_0(x),s'(x))] \cr &\quad + [F(x,s_0(x),s'(x))-F(x,s_0(x),s_0'(x))] \cr &\ge (\p_2 F)(x,s_0(x),s'(x)) (s(x)-s_0(x)) \cr &\quad + (\p_3 F)(x,s_0(x),s_0'(x)) (s'(x)-s_0'(x)), &(2.11) \cr } $$by convexity of F in the second variable, (2.10), and linearity of F in the third variable. Calculation gives$$ (\p_3\p_2F)(x,y,z)=pg'(x)\big[\al (g(x)+g'(x)y)^{p-1} - (1-\al) (g(x)-g'(x)y)^{p-1}\big].\tag2.12 $$Setting y=s_0(x) in (2.12) we obtain that$$ (\p_3\p_2F)(x,s_0(x),z)=pg'(x)\big[\al (1-\be)^{p-1} - (1-\al) (1+\be)^{p-1}\big]=0, $$since by the definition of \be, it follows that$$ \al = { (1+\be)^{p-1} \over (1+\be)^{p-1} + (1-\be)^{p-1}}. $$We have now proved that the function (\p_2F)(x,s_0(x),z) is constant in z. The proof of (2.9) is now complete if we replace s'(x) by s_0'(x) in the first summand of (2.11). QED. \bigskip {\bf 3. The core of the proof.} \smallskip Next we have the following. \proclaim{Lemma 5} Both s and s_0 lie in the domain of the functional I. We have the equality$$ I(s_0) = r(\alpha) \|g\|_{L^p}^p ,\tag 3.1$$where r(\alpha) = \gamma_1(\alpha) + p \beta(\alpha) \gamma_2 (\alpha) with \gamma_1(\alpha) = \alpha(1-\beta)^p - (1-\alpha)(1+\beta)^p and \gamma_2(\alpha) = \alpha(1-\beta(\alpha))^p + (1-\alpha)(1+\beta(\alpha))^p. We also have the inequality$$ I(s) = \|f\|_{L^p}^p \ge I(s_0). \tag 3.2 \endproclaim {\smc Proof.} First, it is clear that $s$ is in the domain of $I$, by the calculations at the beginning of the previous section. Let us work with $s_0$. We see that if \$0