On the behavior at infinity of an integrable function

Transcription

1 O the behavior at ifiity of a itegrable fuctio Emmauel Lesige To cite this versio: Emmauel Lesige. O the behavior at ifiity of a itegrable fuctio. The America Mathematical Mothly, 200, 7 (2), pp <hal v3> HAL Id: hal Submitted o Dec 2009 HAL is a multi-discipliary ope access archive for the deposit ad dissemiatio of scietific research documets, whether they are published or ot. The documets may come from teachig ad research istitutios i Frace or abroad, or from public or private research ceters. L archive ouverte pluridiscipliaire HAL, est destiée au dépôt et à la diffusio de documets scietifiques de iveau recherche, publiés ou o, émaat des établissemets d eseigemet et de recherche fraçais ou étragers, des laboratoires publics ou privés.

2 ON THE BEHAVIOR AT INFINITY OF AN INTEGRABLE FUNCTION EMMANUEL LESIGNE We deote by x a real variable ad by a positive iteger variable. The referece measure o the real lie R is the Lebesgue measure. I this ote we will use oly basic properties of the Lebesgue measure ad itegral o R. It is well kow that the fact that a fuctio teds to zero at ifiity is a coditio either ecessary or sufficiet for this fuctio to be itegrable. However, we have the followig result. Theorem. Let f be a itegrable fuctio o the real lie R. For almost all x R, we have () lim f(x) = 0. Remark. It is too much hope i Theorem for a result for all x because we cosider a itegrable fuctio f, which ca take arbitrary values o a set of zero measure. Eve if we cosider oly cotiuous fuctios, the result does ot hold for all x. Ideed a classical result, usig a Baire category argumet, tells us that if f is a cotiuous fuctio o R such that for all ozero x, lim f(x) = 0, the lim x ± f(x) = 0. Thus for a cotiuous itegrable fuctio f which does ot ted to zero at ifiity, property () is true for almost all x ad ot for all x. Remark 2. Let f be a itegrable ad oegative fuctio o R. We have f(x) dx = f(x) dx. Hece for ay oegative real sequece (ε ) such that ǫ / < +, we have ε f(x) dx < +, ad the mootoe covergece theorem (or Fubii s theorem) esures that the fuctio x ε f(x) is itegrable, hece almost everywhere fiite. I particular, for almost all x, we have lim ε f(x) = 0. This argumet is ot sufficiet to prove Theorem. Now we will state that, i a sese, Theorem gives a optimal result. The stregth of the followig theorem lies i the fact that the sequece (a ) ca ted to ifiity arbitrarily slowly. Theorem 2. Let (a ) be a real sequece which teds to +. There exists a cotiuous ad itegrable fuctio f o R such that, for almost all x, lim sup a f(x) = +.

3 2 EMMANUEL LESIGNE Moreover, there exists a itegrable fuctio f o R such that, for all x, lim sup a f(x) = +. Questio. Uder the hypothesis of Theorem 2, does there exist a cotiuous ad itegrable f such that, for all x, limsup a f(x) = +? We do ot kow the aswer to this questio, ad we propose it to the reader. However, the ext remark shows that the aswer is positive uder a slightly more demadig hypothesis. Remark 3. If the sequece (a ) is odecreasig ad satisfies a < +, the there exists a cotiuous ad itegrable fuctio f o R such that for all x, limsup a f(x) = +. Remark 4. I Theorem 2 we caot replace the hypothesis lim a = + by limsup a = +. Ideed, by a simple chage of variable we ca deduce from Theorem the followig result: for all itegrable fuctios f o R, lim f(2 x) = 0 for almost all x. (Apply Theorem to the fuctio x xf(x 2 ).) Thus the coclusio of Theorem 2 is false for the sequece (a ) defied by { if is a square of iteger, a = 0 if ot. I the remaider of this ote, we give proofs of the two theorems ad of Remark 3. Proof of Theorem. The fuctio f is itegrable o R. Let us fix ε > 0 ad deote by E the set of poits x > 0 such that f(x) ε. We kow that E has fiite measure. We are goig to show that, for almost all x [0, ], we have x E for oly fiitely may s. (If A is a measurable subset of R, we deote by A its Lebesgue measure.) For each iteger m, let E m := E (m, m]. Let us fix a (0, ). For each iteger, we cosider the set ( ) F := E [a, ) = E m [a, ) = (E m [a, )). We have = F = m = m= m E m [a, ). I this doubly idexed sum of positive umbers, we ca ivert the order of summatio. Moreover, oticig that E m [a, ) = if > m/a or m, we obtai = F = [m/a] m= =m E m [a, ) E m m= [m/a] =m.

4 ON THE BEHAVIOR AT INFINITY OF AN INTEGRABLE FUNCTION 3 By compariso of the discrete sum with a itegral, we see that, for all m, ( la). Thus we have [m/a] =m = F ( la) m= E m = ( la) E < +. This implies that almost every x belogs to oly fiitely may sets F. (This statemet is the Borel-Catelli lemma, which has a oe lie proof : ½F < + almost everywhere sice ½F(x) dx = ½ F(x) < +.) Returig to the defiitio of F, we coclude that, for almost all x [a, ], for all large eough, x / F, i.e. x / E. Sice a is arbitrary, we have i fact: for almost all x [0, ], for all large eough, x / E. We have proved that, for all ε > 0, for almost all x [0, ], for all large eough, f(x) ε. Sice we have to cosider oly coutably may ε s, we ca ivert for all ε > 0 ad for almost all x [0, ]. We coclude that, for almost all x [0, ], lim f(x) = 0. It is immediate, by a liear chage of variable (for example), that this result exteds to almost all x R. Proof of Theorem 2. We will utilize the followig theorem, a fudametal result i the metric theory of Diophatie approximatio [, Theorem 32]. Khichi s Theorem. Let (b ) be a sequece of positive real umbers such that the sequece (b ) is oicreasig ad the series b diverges. For almost all real umbers x, there are ifiitely may itegers such that dist(x, Z) < b. We will also make use of the followig lemma, which will be proved i the sequel. Lemma. Let (c ) be a sequece of oegative real umbers goig to zero. There exists a sequece of positive real umbers (b ) such that the sequece (b ) is oicreasig, b = +, ad b c < +. Let us prove Theorem 2. Replacig if ecessary a by if k a k, we ca suppose that the sequece (a ) is odecreasig. Applyig the precedig lemma to the the sequece c = / a, we obtai a sequece (b ) such that the sequece (b ) is oicreasig, b = +, ad b / a < +. The sequece (b ) teds to zero, ad we ca impose the additioal requiremet that b < /2 for all. We cosider the fuctio f defied o R by { / a if x b for a iteger, f (x) = 0 if ot. This fuctio is itegrable, due to the last coditio imposed o (b ). By Khichi s theorem, for almost all x > 0, there exist pairs of positive itegers (, k()), with arbitrarily large, such that x k() b. Let us cosider oe fixed such x i the iterval (0, ). We have lim k() = + ad, sice lim + b = 0, we have k() for all large eough. For such a

5 4 EMMANUEL LESIGNE, we have x k() b k() ad hece f (x) = ak(). (We used here the fact that the sequece (b ) is oicreasig.) Thus, for arbitrarily large, we have a f (x) = a a k(). ak() (We used here the fact that the sequece (a ) is odecreasig.) This proves that limsup a f (x) = +. This argumet applies to almost all x betwee 0 ad. For each iteger m, let us deote by f m the fuctio f m (x) = f (x/m). This fuctio f m is oegative ad itegrable o R. It is locally a step fuctio. For almost all x betwee 0 ad m, we have lim sup a f m (x) = +. From this, it is ot difficult to costruct a cotiuous ad itegrable fuctio f o R such that, for all m > 0, there exists A m > 0 with f f m o [A m, + ). (For example, we ca choose a icreasig sequece of umbers (A m ) such that + A m f (x) + f 2 (x) + + f m (x) dx m 2 ; the we defie g = f + f f m o the iterval [A m, A m+ ). Sice m Am+ A m f (x) + f 2 (x) + + f m (x) dx <, this fuctio g is itegrable. The we just have to fid a cotiuous ad itegrable fuctio f which domiates g; this ca be achieved sice the fuctio g is locally a step fuctio: choose f to be zero o (, 0] ad cotiuous o R such that g f ad, for all m > 0, m m f(x) g(x) dx /m2, so that + 0 f(x) g(x) dx < +.) For almost all x 0, we have limsup a f(x) = +. A symmetrizatio procedure exteds this property to almost all real umbers. The first part of Theorem 2 is proved. The secod part is a direct cosequece. We cosider the fuctio f costructed above, ad we deote by F the set of x such that the sequece (a f(x)) is bouded. The set {x x F, N} has zero measure. We modify the fuctio f o this set, choosig for example the value. The ew fuctio is itegrable ad satisfies, for all x, limsup a f(x) = +. Proof of Lemma. The sequece (c ) is give, ad it goes to zero. We will costruct by iductio a icreasig sequece of itegers ( k ) ad a oicreasig sequece of positive umbers (d k ), ad we will defie b = d k / for k < k. The umbers d k will be chose so that k i= k b i = ; thus we require that d k := k i i= k.

6 ON THE BEHAVIOR AT INFINITY OF AN INTEGRABLE FUNCTION 5 We start from 0 =, ad the we choose > 0 such that, for all, c /2. I the ext step, we choose 2 > such that d 2 d ad, for all 2, c /4. More geerally, if, 2,..., k have bee costructed, we choose k > k such that d k d k ad, for all k, c 2 k. (Of course, this is possible because lim + ( i= k i ) = 0.) This defies the sequece (b ) by blocks. The sequece (b ) is oicreasig ad, for all k, we have k i= k b i = ad k i= k b i c i 2 k. This guaratees that b = + ad b c < +. The lemma is proved. About Remark 3. Dirichlet s lemma i Diophatie approximatio (based o the pigeo-hole priciple) cocers the particular case b = / i Khichi s theorem ad it gives a result for all x. Lemma 2 (Dirichlet s Lemma). For all real umbers x, there exist ifiitely may itegers such that dist(x, Z). Now, we justify Remark 3. We cosider a odecreasig sequece of positive real umbers (a ) such that < +. a We claim that there exists a sequece of positive real umbers (b ) such that b a + ad b < +. Here is a proof of this claim: for each k, there exists (k) such that a k 2. We have (k) card{k (k) } = a k (k) a < +, ad we ca defie b := card{k (k) }/a. Give this sequece (b ), we cosider the fuctio f defied o R by { b k if x k /k, k a iteger, k 2, f(x) = 0 if ot. This fuctio is itegrable. Usig Dirichlet s lemma, we have the followig: for each fixed x i (0, ), there exist pairs of positive itegers (, k()), with arbitrarily large, such that x k() /. We have lim k() = + ad, for all large eough, k(). Hece there exist ifiitely may s such that x k() k() ad so f(x) = b k().

7 6 EMMANUEL LESIGNE For such a, we have a f(x) = a b k() a k() b k(). (We used here the fact that the sequece (a ) is odecreasig.) This proves that limsup a f(x) = +. This result obtaied for all umbers x betwee 0 ad exteds to all real umbers by the same argumet as the oe used i the proof of Theorem 2. We ca also replace the local step fuctio by a cotiuous oe as we did before. Theorem aswers a questio asked by Aris Dailidis. Refereces [] A. Ya. Khichi, Cotiued Fractios, Dover, Mieola, NY, 997; reprit of (tras. Scripta Techica, Ic.) Uiversity of Chicago Press, 96; reprit of 3rd Russia ed., State Publishig House of Physical-Mathematical Literature, Moscow, 96. Laboratoire de Mathématiques et Physique Théorique, Uiversité Fraçois-Rabelais Tours, Fédératio Deis Poisso - CNRS, Parc de Gradmot, Tours, Frace emmauel.lesige@lmpt.uiv-tours.fr