When computing the average speed of a car over different time periods from
given GPS data, it is conventional wisdom that the maximal average speed over
all time intervals of fixed length decreases if the interval length increases.
However, this intuition is wrong. We investigate this phenomenon and make
rigorous in which sense this intuition is still true

Havenith, Thomas

Scholtes, Sebastian

English

arXiv

Scholtes, Sebastian Tobias

RWTH Publications

Comparing maximal mean values on different scales

Publikationsserver der RWTH Aachen University

COMPARING MAXIMAL MEAN VALUES ON DIFFERENT SCALESTHOMAS HAVENITH AND SEBASTIAN SCHOLTESAbstract. When computing the average speed of a car over different time periods from givenGPS data, it is conventional wisdom that the maximal average speed over all time intervals offixed length decreases if the interval length increases. However, this intuition is wrong. This canbe easily seen by the example in Figure 1 for the two time periods T < S. We investigate thisphenomenon and make rigorous in which sense this intuition is still true.Figure 1. Maximal average speeds for larger time periods are not necessarily smaller.In many applications, rates are measured as average velocities; for instance the speed monitoredby Speed Check Services SPECS in the UK, capacities of rides in amusement parks and the driftvelocity of electrons are obtained in this way [Smi12, Ahm97, Gri99]. In other fields mean valuesover different time intervals and their relation are important:• An athlete has a high heart rate during a competition and maybe even a very high heartrate for a small time during this contest, but having a similar heart rate for a wholeweek would result in serious health problems.• While approximately 10 percent of people that are exposed to a radiation dose of 1 Svover a short time period die from light radiation poisoning in the first 30 days [Har11,p.400], this is also the maximal dose allowed for NASA astronauts over their career,[NAS08, p.7].• Certain materials, for example a power line, can endure fairly high stresses, currents etc.for a small time interval, but the average load on a larger time scale has to be muchsmaller. Another illustration for the effect of the time scale is the phenomenon of creep[MC99].All these examples show that one expects maximal mean values to become smaller as theinterval size decreases. We study mean values (in the discrete as well as in the continuous setting)with regard to this property and find that it is not true in general (Lemmata 3 and 8), but thatit is true if one compares the maximal mean value over all regions of size T with those taken onregions of size d · T for d ∈ N (Lemmata 3 and 7).Date: January 27, 2015.2010 Mathematics Subject Classification. 46B25; 62P30, 46A45, 47A30.Key words and phrases. mean value, arithmetic mean, norm, Lp norm.1arXiv:1501.06391v1  [math.GM]  20 Jan 20152 THOMAS HAVENITH AND SEBASTIAN SCHOLTES1. Discrete settingLet `∞ be the space of bounded real sequences. For x ∈ `∞, p ∈ (0,∞) and n ∈ N we define‖x‖∞,p,n := supj∈N( 1nj+(n−1)∑i=j|xi|p) 1p.Lemma 1 (For p ∈ [1,∞) the mapping ‖·‖∞,p,n is an equivalent norm to ‖·‖∞).Let p ∈ [1,∞), n ∈ N. Then ‖·‖∞,p,n is a norm on `∞ withn− 1p ‖x‖∞ ≤ ‖x‖∞,p,n ≤ ‖x‖∞.Proof. Clearly ‖·‖∞,p,n is positively homogenous of degree one and ‖x‖∞,p,n ≥ 0 with equality ifand only if x = 0. From the triangle inequality of the p-norm in Euclidean n-space we obtain thetriangle inequality for ‖·‖∞,p,n. Additionally, we obtainn− 1p ‖x‖∞ =( 1n‖x‖p∞) 1p ≤ ‖x‖∞,p,n = supj∈N( 1nj+(n−1)∑i=j|xi|p) 1p ≤( 1nn‖x‖p∞) 1p= ‖x‖∞.This means that the spaces (`∞, ‖·‖∞,p,n) are non-reflexive Banach spaces with the sametopology as the standard `∞.Lemma 2 (Inequality for norms).Let αi ∈ N with n =∑dl=1 αl. Then for all p ∈ [1,∞) and x ∈ `∞ we have that‖x‖p∞,p,n ≤d∑l=1αln‖x‖p∞,p,αl .Proof. Let j ∈ N, ε > 0 such that ‖x‖p∞,p,n − ε ≤ 1n∑j+(n−1)i=j xpi . Then‖x‖p∞,p,n − ε ≤1nj+(n−1)∑i=jxpi =1nd∑l=1j+∑l−1k=1 αk+(αl−1)∑i=j+∑l−1k=1 αkxpi ≤1nd∑l=1αl‖x‖p∞,p,αl .Lemma 3 (Norms ‖·‖∞,p,dn of multiples of n are ordered).Let m,n ∈ N and n ≤ m.• If n is a factor of m, i.e. m = dn for d ∈ N then‖x‖∞,p,dn = ‖x‖∞,p,m ≤ ‖x‖∞,p,n for all x ∈ `∞.• If n is no factor of m then there are x, y ∈ `∞ such that‖x‖∞,p,n < ‖x‖∞,p,m and ‖y‖∞,p,m < ‖y‖∞,p,n.Proof. If we choose all αl = n in Lemma 2 we immediately obtain the first item. For the seconditem define x(n) ∈ `∞ by x(n)i = 1 if i ∈ {(d− 1)n+ 1 | d ∈ N} and x(n)i = 0 otherwise. Then‖x(n)‖p∞,p,n =1n<ddn− 1 ≤d(d− 1)n+ j = ‖x(n)‖p∞,p,(d−1)n+jfor j = 1, . . . , n− 1 and d ∈ N. Note that the natural numbers of the form m = (d− 1)n+ j forj = 1, . . . , n − 1, d ∈ N are exactly those which are either smaller than n or such that n is nofactor of m. This proves the proposition as we obtain the opposite inequality by interchangingthe roles of m and n. In the following Lemma we show that counterexamples, however, cannot be too bad.COMPARING MAXIMAL MEAN VALUES ON DIFFERENT SCALES 3Lemma 4 (Counterexamples cannot be too bad).Let p ∈ [1,∞), n < m. Then for all x ∈ `∞ we have that‖x‖p∞,p,m ≤(bmn c+ 1)nm‖x‖p∞,p,n ≤ 2‖x‖p∞,p,n.Proof. Choose d := bmn c and ε > 0. Then there is j ∈ N such that‖x‖p∞,p,m − ε ≤1mj+(m−1)∑i=j|xi|p ≤ 1mj+((d+1)n−1)∑i=j|xi|p≤ (d+ 1)nm‖x‖p∞,p,(d+1)n ≤(d+ 1)nm‖x‖p∞,p,naccording to Lemma 3. 2. Continuous settingLet f : RN → R be a function. For V > 0, p ∈ (0,∞) we define‖f‖p,V := supΩ⊂RN openΩ convex|Ω|=V( 1|Ω|∫Ω|f(x)|p dx) 1p.Note that this norm bears a resemblance to the Hardy-Littlewood maximal function from harmonicanalysis. If we denote the space of all functions f : RN → R with f = 0 a.e. by N we can defineLpV := {f : RN → R | ‖f‖p,V <∞}/N .By a slight abuse of notation we also want to consider ‖·‖p,V as a mapping from LpV to R in theobvious manner.Lemma 5 (For p ∈ [1,∞) the mapping ‖·‖p,V is a norm on LpV ).Let p ∈ [1,∞), V > 0. Then ‖·‖p,V is a norm on LpV .Proof. Clearly ‖·‖p,V is positively homogenous of degree one and ‖f‖p,V ≥ 0 with equality if andonly if f = 0 a.e.. Now, the triangle inequality for Lp proves the triangle inequality for ‖·‖p,V . Lemma 6 (Inequality for norms).Let αi > 0 with V =∑dl=1 αl. Then for all p ∈ [1,∞), V > 0 and f ∈ LpV holds‖f‖pp,V ≤d∑l=1αlV‖f‖pp,αl .Proof. Step 1 Let ε > 0 and Ω ⊂ RN be open and connected such that ‖f‖pp,V − ε ≤1|Ω|∫Ω|f(x)|p dx. According to Step 2 there is a hyperplane that cuts Ω into two convex sets Ω1and Ω˜1 with |Ω1| = α1 and |Ω˜1| =∑dl=2 αl. Now iterating this procedure with Ω˜1 we finallyobtain d open convex sets Ωi with |Ωi| = αi and Ω =⋃dl=1 Ωi ∪ A, where HN (A) = 0, as A iscontained in a finite union of hyperplanes. Therefore‖f‖pp,V − ε ≤1|Ω|∫Ω|f(x)|p dx = 1|Ω|d∑l=1∫Ωi|f(x)|p dx ≤ 1|Ω|d∑l=1αl‖f‖pp,αl .Step 2 Let Ω be an open convex set and α+ β = |Ω| for α, β > 0. If we defineΩs := {x ∈ Ω | x1 = s}we know by Cavalieri’s Principle (a special case of Fubini’s Theorem) that the function ϕ(t) :=∫ t−∞|Ωs|ds measures the volume of the set Ω intersected with the open half-space below (in e1direction) the hyperplane {x ∈ RN | x1 = t}. Additionally limt→−∞ ϕ(t) = 0, limt→∞ ϕ(t) = |Ω|and ϕ is continuous. Therefore the intermediate value theorem yields the existence of a t0 suchthat Ω1 =⋃t∈(−∞,t0) Ωt and Ω˜1 =⋃t∈(t0,∞) Ωt are open convex sets (as intersection of two suchsets) and |Ω1| = α and |Ω˜1| = |Ω| − α = β. 4 THOMAS HAVENITH AND SEBASTIAN SCHOLTESLemma 7 (Norms ‖·‖p,d·V of multiples of V are ordered).Let p ∈ [1,∞), V > 0 and d ∈ N. Then‖f‖p,d·V ≤ ‖f‖p,V for all f ∈ LpV .Proof. If we chose all αl = V in Lemma 6 we directly obtain the result. Lemma 8 (For N = 1, T < S and T no factor of S the norms are not ordered).Let N = 1, p ∈ [1,∞) and T, S > 0, T < S such that T is no factor of S. Then there aref, g ∈ LpT such that‖f‖p,T < ‖f‖p,S and ‖g‖p,S < ‖g‖p,T .Proof. Let T < S such that T is not a factor of S, i.e. ST 6∈ N. Write d := bST c. Then it is possibleto choose ε > 0 such that (d+ 1)ε < S − dT . Let ψε(x) be a bump function of Lp mass 1 withsupport in (0, ε). Then f(x) =∑di=0 ψε(x− i(T + ε)) is an LpT function and‖f‖pp,T =1T<d+ 1S= ‖f‖pp,S ,because ST < bST c+ 1. On the other hand, g = ψε ∈ LpT yields‖g‖pp,S =1S<1T= ‖g‖pp,T .Corollary 9 (Counterexamples cannot be too bad).Let p ∈ [1,∞), T < S. Then for all f ∈ LpS holds‖f‖pp,S ≤(bST c+ 1)TS‖f‖pT ≤ 2‖f‖pT .Proof. Choose d := bST c and ε > 0 such that ‖f‖pp,S − ε ≤ 1S∫Ω|f(t)|p dt. Then we can extend Ωconvexly to an open Ω˜ with Ω ⊂ Ω˜ and |Ω˜| = (d+ 1)T . This can for example be done by takingthe convex hull Ω˜y of Ω and a point y moving along a normal at a fixed point y ∈ ∂Ω. By anargument similar to that in the proof of Lemma 6 Step 2, we see that x can be chosen in such away that |Ω˜y| = (d+ 1)T . Now‖f‖pp,S − ε ≤1S∫Ω|f(t)|p dt ≤ 1S∫Ω˜|f(t)|p dt ≤ (d+ 1)TS‖f‖p(d+1)T ≤(d+ 1)TS‖f‖pTby Lemma 7. References[Ahm97] Reza H. Ahmadi, Opportunities for simulation in amusement parks, Oper. Res. 45 (1997), no. 1, 1–13.[Gri99] David Griffiths, Introduction to electrodynamics, Prentice-Hall, 1999.[Har11] Gavin D. J. Harper, Green Health: An A-to-Z Guide, ch. Radiation Sources, pp. 374–401, SAGE, 2011.[MC99] M. A. Meyers and K. K. Chawla, Mechanical Behavior of Materials, Cambridge University Press, 1999.[NAS08] NASA, Space Faring: The Radiation Challenge Introduction and Module 1: Radiation, Educator GuideEP-2008-08-116-MSFC, NASA, 2008.[Smi12] B.D.V. Smith, Vehicle speed monitoring system, US Patent 8,189,048, 2012.Thomas Havenith, Infrastrukturplanung (I.ETZ 1), DB Energie GmbH, Pfarrer-Perabo-Platz 2,60326 Frankfurt am Main, GermanyE-mail address: thomas.havenith@deutschebahn.comURL: http://www.dbenergie.de/Sebastian Scholtes, Institut für Mathematik, RWTH Aachen University, Templergraben 55, 52062Aachen, GermanyE-mail address: sebastian.scholtes@rwth-aachen.deURL: http://www.instmath.rwth-aachen.de/~scholtes/home/

Comparing maximal mean values on different scales

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