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# Abel and Euler Summation Formulas forSBVRBy Sergio VenturiniSubmitted: July 5th 2021Reviewed: September 9th 2021Published: October 19th 2021DOI: 10.5772/intechopen.100373

## Abstract

The purpose of this paper is to show that the natural setting for various Abel and Euler-Maclaurin summation formulas is the class of special function of bounded variation. A function of one real variable is of bounded variation if its distributional derivative is a Radom measure. Such a function decomposes uniquely as sum of three components: the first one is a convergent series of piece-wise constant function, the second one is an absolutely continuous function and the last one is the so-called singular part, that is a continuous function whose derivative vanishes almost everywhere. A function of bounded variation is special if its singular part vanishes identically. We generalize such space of special function of bounded variation to include higher order derivatives and prove that the functions of such spaces admit a Euler-Maclaurin summation formula. Such a result is obtained by deriving in this setting various integration by part formulas which generalizes various classical Abel summation formulas.

### Keywords

• Euler summation
• Abel summation
• bounded variation functions
• special bounded variation functions

## 1. Introduction

Abel and the Euler-Maclaurin summation formulas are standard tool in number theory (see e.g. [1, 2]).

The space of special functions of bounded variation(SBV) is a particular subclass of the classical space of bounded variation functions which is the natural setting for a wide class of problems in the calculus of variations studied by Ennio De Giorgi and his school: see e.g. [3, 4].

The purpose of this paper is to show that this class of functions (and some subclasses introduced here of function of a single real variable) is the natural settings for (an extended version of) the Euler-Maclaurin formula.

Let us describe now what we prove in this paper.

In Section 2 we obtain some “integration by parts”-like formulas for functions of bounded variations which imply the various “Abel summation” techniques (Propositions (0.6), (0.7), and the relative examples) and in Section 3 we give some criterion for the absolute summability of some series obtained by sampling the values of a bounded variations function.

The last section contains the proofs of the main result of this paper (Theorem (0.1)) that we will now describe.

We denote by C1R(resp. Ckab), L1Rand LRrespectively the space of continuously differentiable functions (resp. k-times differentiables functions on the closed interval ab), the space of Lebesgue (absolutely) integrable functions and the space of essentially bounded Borel functions on R.

Given f:RCand xRwe set

fx+=limh0+fx+h,E1
fx=limh0fx+h,E2
δfx=fx+fx.E3

We denote by BVRthe space of bounded variation complex functions on R; we refer to [5, 6] for the main properties of functions in BVR.

Any real function of bounded variation can be written as a difference of two non decreasing functions. It follows that if fBVRthen fx+, fxand δfxexist for each xRand the set xRδfx0is an arbitrary at most countable subset of R. Moreover, the derivative fxexists for almost all xRand fxL1R.

Let fBVR. We denote by dfthe unique Radon measure on Rsuch that for each open interval ]a,b[R

dfa,b=fafb+.E4

We recall that fis specialif for any bounded Borel function u

Ruxdfx=Ruxfxdx+xRuxδfx.E5

We denote by SBVRthe space of all special functions of bounded variation. We also say that fBVlocR(resp. fSBVlocR) if for each a,bR, with a<bthe function

fx=0ifx<aorx>b,fxifaxb,E6

is in BVR(resp. SBVR).

We define SBVnRinductively setting

SBV1R=SBVR,E7

and for each integer n>1

SBVnR=fSBVRfSBVn1RE8

We denote by Bnand Bnx, n=1,2,respectively the Bernoulli numbers and the Bernoulli functions. Let us recall that

B1x=0ifxZ,xx12ifxR\Z,E9

where xstands for the greatest integer less than or equal to xand Bnx, n=2,3,are the unique continuous functions such that

Bnx+1=Bnx,E10
Bnx=nBn1x,E11
01Bnxdx=0.E12

Moreover B2n+1=0for n>0and Bn=Bn0for n>1.

The main results of this paper is the following theorem.

Theorem 0.1 Let fSBVmR, m1and suppose f,,fmL1R. Then

nZfn++fn2=Rfxdx+xRk=1m1k1k!Bkxδfk1x+1m1m!RBmxfmxdx.E13

Remark.The sum “xR” in the right hand side of the above “Euler-Maclaurin formula” (13) is actually a sum over the subset of the xRsuch that some of the terms Bkxδfk1xdo not vanish. We point out that such a set can be an arbitrary at most countable subset of R.

Remark.Let pand q, p<qbe two integers and let fbe a function of class Cmon the interval pq. Set fx=0when xis outside of the interval pq. Then the classical Euler-Maclaurin formula (see, e.g. Section 9.5 of [7])

k=pq1fk=pqfxdx+k=1mBkk!fk1qfk1p+1m1Bmm!pqBmxfmxdx,E14

follows easily from Theorem 0.1.

Remark.Any fBVRdecomposes uniquely as f=f1+f2+f3, where f1xcan be written in the form

f1x=n=1+φnxE15

where each φnxis a piece-wise constant function, f2xis an absolutely continuous function and f3xis a singular function, that is f3xis continuous and f3x=0for almost all xR. Then f=f1+f2+f3is special if, and only if, f3=0and in this case, for each bounded Borel function ux,

Ruxdf1x=xRuxδfx,E16
Ruxdf2x=Ruxfxdx.E17

In this paper we do not need of the existence of such a decomposition.

## 2. Integration by parts formulas

Our starting point is the following theorem:

Theorem 0.2 Let f,g:RCtwo complex function. Assume that fBVRL1Rand gBVlocRLR. Then

Rfx+dgx+Rgxdfx=0,E18
Rfxdgx+Rgx+dfx=0,E19
Rfx++fx2dgx+Rgx++gx2dfx=0.E20

Proof:Let a,bRwith a<b. Theorem 7.5.9 of [5] yields

]a,b[fx+dgx+]a,b[gxdfx=fbgbfa+ga+,E21
]a,b[fxdgx+]a,b[gx+dfx=fbgbfa+ga+.E22

Since fL1Rthen necessarily

limb+fb=limafa+=0.E23

Since gLRthen gx+and gxare bounded and we also have

limb+fbgb=limafa+ga+=0.E24

and hence one obtains the formulas (18) and (19) taking the limits as aand b+respectively in (21) and (22).

Formula (20) is obtained summing memberwise (18) and (19) and dividing by two. □

Next we prove:

Theorem 0.3 Let f,g:RCtwo complex function. Assume that fBVRL1Rand gSBVlocRLRand suppose that gLR. Then

Rfxgxdx+xR'fx+δgx+Rgxdfx=0,E25
Rfxgxdx+xR'fxδgx+Rgx+dfx=0,E26
Rfxgxdx+xR'fx++fx2δgx+Rgx++gx2dfx=0.E27

where

xR'limab+a<x<b.E28

Moreover, if the function falso is continuous then

Rfxgxdx+Rgxdfx=0,E29

Proof:Given a,bR, a<bset

gabx=0xa,gxa<x<b,0xb.E30

The function hx=gabxis in SBVRLR. Hence, formula (18) yields

Rfx+dhx+Rhxdfx=0.E31

Since hSBVRwe have

Rfx+dhx=abfx+gxdx+xRfx+δgabx.E32

But fx+=fxfor almost all xRand hence

Rfx+dhx=abfxgxdx+xRfx+δgabx,E33

which combined with (31) yields

abfxgxdx+xRfx+δgabx+Rgabxdfx=0.E34

Using the definition of gabxwe have

xRfx+δgabx=fa+ga++a<x<bfx+δgx,E35

and hence

a<x<bfx+δgx=fa+ga+abfxgxdxRgabxdfx.E36

As in the proof of the previous theorem we have

limafa+ga+=0.E37

Since fL1Rand gLRthen fgL1Rand hence

limab+abfxgxdx=Rfxgxdx.E38

The Radon measure dfxis bounded and the functions xgabxare equibounded with respect to aand b; by the Lebesgue dominated convergence we have

limab+Rgabxdfx=Rgxdfx.E39

From (36) it follows that

limab+a<x<bfx+δgx=xR'fx+δgx=RfxgxdxRgxdfxE40

which is equivalent to (25).

The proof of (26) is obtained in a similar manner using (19) instead of (18), and (27) is obtained summing memberwise (25) and (26) and dividing by two.

If the function gis continuous then gx+=gx=gxfor each xR,

xR'fx+δgx=0,E41

and (29) follows from, e.g., (25).

Example.This example shows that in the hypoteses of Theorem (0.3) the series

xR'fx+δgxE42

is not, in general, absolutely convergent. Indeed, set

fx=0ifx1/2,1/x2ifx>1/2,E43

and

gx=1if2n1<x2n,nZ,0if2n<x2n+1,nZ.E44

Then the integral

Rfxgxdx=n=1+2n12ndfx=n=1+12n2n1E45

is absolutely convergent, but the series

xR'fx+δgx=n=1+1nnE46

is convergent but not absolutely convergent.

We also have the following theorem.

Theorem 0.4 Let f,g:RCtwo complex function. Assume that fSBVRL1Rand gSBVlocRLRand suppose that gLR. Then

Rfxgxdx+Rfxgxdx+xRδfxgx++xR'fxδgx=0,E47
Rfxgxdx+Rfxgxdx+xRδfxgx+xR'fx+δgx=0,E48
Rfxgxdx+Rfxgxdx+xRgx++gx2δfx+xR'fx++fx2δgx=0,E49

where

xR'limab+a<x<bE50

If the function galso is continuous then

Rfxgxdx+Rfxgxdx+xRgxδfx=0,E51

Proof:Let fand gbe as in the theorem. By formula (26) we have

Rfxgxdx+xR'fxδgx+Rgx+dfx=0.E52

Since fSBVR, using the fact that gx+=gxfor almost all xR, we obtain

Rgx+dfx=Rgxfxdx+xRgx+δfx.E53

Then (52) and (53) yield (47). Formulas (48) and (49) are obtained in a similar manner using respectively Formulas (25) and (27) instead of (26).

If the function gis continuous then gx+=gx=gxfor each xR,

xR'fx+δgx=0,E54

and (51) follows from, e.g., (47).□

Theorem 0.4 generalizes to high order derivatives.

Theorem 0.5 Let f,g:RCtwo complex function. Let m>0be a positive integer. Assume that fSBVmRwith f,,fmL1Rand gSBVlocmRwith g,,gmLR. Then

1m1Rfmxgxdx+Rfxgmxdx+xRk=1m1k1δfk1xgmkx++xR'k=1m1k1fk1xδgmkx=0,E55
1m1Rfmxgxdx+Rfxgmxdx+xRk=1m1k1δfk1xgmkx+xR'k=1m1k1fk1x+δgmkx=0,E56
1m1Rfmxgxdx+Rfxgmxdx+xRk=1m1k1δfk1xgmkx+gmkx+2+xR'k=1m1k1fk1x+fk1x+2δgmkx=0,E57

Proof:We prove first the formula (55). The proof is by induction on m. When m=1(55) reduces to (47). Assume that (55) holds for m1, that is

1m2Rfm1xgxdx+Rfxgm1xdx+xRk=1m11k1δfk1xgmk1x++xR'k=1m11k1fk1xδgmk1x=0.E58

Replacing fwith f, kwith k+1and changing the sign we obtain

1m1RfmxgxdxRfxgm1xdx+xRk=2m1k1δfk1xgmkx++xR'k=2m1k1fk1xδgmkx=0.E59

Replacing gwith gm1in (47) we obtain

Rfxgm1)xdx+Rfxgmxdx+xRδfxgm1x++xR'fx+δgm1x=0.E60

Summing (59) and (60) we obtain (55).

The proofs of (56) and (57) are similar.

We say that a function fSBVlocRis a step functionif fx=0for almost every xR.

The following propositions are easy consequences of Theorem (0.4).

Proposition 0.6 Let uvRbe a bounded closed interval and let fbe an absolutely continuous function on the closed interval uv. Let gSBVlocRbe a step function. Then

uvfxgxdx=fvgvfugu+u<x<vfxδgx.E61

Proof:First we extend the functions fas zero outside of the interval uv. We may also assume that the function gis zero outside of a bounded open interval containing the closed interval uv. Observe that then fu+=fu, fv=fvand fu=fv+=0and therefore δfu=fu, δfv=fvand δfx=0for xu,v. By (47), we have

Rfxgxdx+Rfxgxdx+xR'fx+δgx+xRgxδfx=0.E62

Since gis a step function then gx=0for almost all xRand hence it follows that

Rfxgxdx=xR'fx+δgxxRgxδfx.E63

The function fby construction has compact support, and hence, as fv+=0, we have

xR'fx+δgx=fu+gu+gu+u<x<vfx+δgx=fugu+fugu+u<x<vfx+δgx,E64

and

xRgxδfx=guδfu+gvδfv=fugufvgv.E65

Summing memberwise the last two formulas we obtain

xR'fx+δgx+xRgxδfx=fvgv+fugu++u<x<vfx+δgx,E66

as desired.□

Proposition 0.7 Let f,gSBVlocRbe two step function. Let uvRbe a bounded closed interval. Then

u<x<vgx+δfx=fvgvfu+gu+u<x<vfxδgx.E67

Proof:Set both the functions fand gto zero outside the closed interval uv. Then formula (47) yields

xRfx+δgx+xRgxδfx=0.E68

But then

xRfx+δgx=fu+gu++u<x<vfx+δgx,E69

and

xRgxδfx=fvgv+u<x<vgxδfx;E70

hence

fu+gu++u<x<vfx+δgxfvgv+u<x<vgxδfx=0,E71

which is equivalent to (67).□

Example 1.(Abel summation I) Let an, nZbe a sequence of complex numbers such that an=0for n<<0. Then the function

Ax=n<xanE72

is a step function in SBVlocR. If fC1uvthen Proposition (0.6) yields

uvfxAxdx=fvAvfuAu+u<n<vfnan.E73

Example 2.(Abel summation II) Let an,bn, nZbe two sequence of complex numbers. Let f,gCbe defined respectively setting fx=anand gx=bnwhen nx<n+1, nZ. Clearly f,gSBVlocRand they are two step functions. Let be given two integers pand q, p<q. Set u=pand v=q+1. Then it is easy to show that

u<x<vgx+δfx=n=p+1qbnanan1E74

and

u<x<vfxδgx=n=p+1qan1bnbn1;E75

hence, Proposition (0.7) yields

n=p+1qbnanan1=aqbqapbpn=p+1qan1bnbn1.E76

## 3. Sampling estimates

In this section we give some conditions which ensures the absolute convergence of series of the form xEfx+fx+/2where fis a function absolutely integrable of bounded variation and Eis a countable subset of R.

The basic estimate is given in the following lemma.

Lemma 0.8 Let ARbe an open subset and let FAbe a finite subset of A. Assume that there exist a>0such that

x1,x2F,x1x2x1x2a,xF,yR\Axya/2.E77

Then, for any complex functionfBVRL1Rwe have

Proof:Let define

gx=0,ifx<1/2orx=0orx1/2,x+1/2,if1/2x<0,x1/2,if0x<1/2,E79

and set

Gx=yFgxya.E80

For each xRwe have

Gx+Gx+2=GxE81

By Eq. (27)

xR'fx++fx2δGx=RfxGxdx+RGxdfx.E82

We also have

δGx=1ifxF,0ifxR\F,E83

which implies

xR'fx++fx2δGx=xFfx+fx+2.E84

Set

E=xF]xa,x+a[.E85

Then FEAand

Gx=1/aifxE,0ifxR\E,E86

and hence

RfxGxdx=1aEfxdx.E87

Moreover we have Gx=0if xR\Eand hence

xFfx+fx+2=RfxGxdx+RGxdfx=1aEfxdx+EGxdfx.E88

Taking modules, and observing that Gx1/2for each xE, we obtain

as required.

Corollary 0.9 Let fBVRL1Rand let ERbe a countable subset. If there exists a real constant a>0such that for each pair of distinct x1,x2Ewe have x1x2athen

xEfx+fx+2<+.E90

Proof:It suffices to choose A=R; lemma (0.8) yields easily the assertion.

## 4. Proof of Theorem 0.1

Inserting Bmxinstead of gmxin formula (57) of Theorem 0.5 we easily obtain

nZ'fn++fn2=Rfxdx+xRk=1m1k1k!Bkxδfk1x+1m1m!RBmxfmxdx.E91

By Corollary 0.9 it follows that

nZ'fn++fn2=nZfn++fn2E92

is an absolutely convergent series, and hence Theorem 0.1 follows.

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Sergio Venturini (October 19th 2021). <article-title xmlns:mml="http://www.w3.org/1998/Math/MathML">Abel and Euler Summation Formulas for <inline-formula id="I1"><mml:math id="m1"><mml:mi mathvariant="italic">SBV</mml:mi><mml:mfenced open="(" close=")"><mml:mi mathvariant="double-struck">R</m [Online First], IntechOpen, DOI: 10.5772/intechopen.100373. Available from: