GAZETA MATEMATICËA - SSMR

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$Bull. Math. Soc. Sci. Math. Roumanie Tome 49(97) No. 4 ... - SSMR$

212 Note Matematice şi MetodiceThus, there exists a sequence t n > 0 such thatwhich is equivalent to:∫ t n0h(x)φ(x)dxφ(t n )≥ c>0,∫ t n0h(x)φ(x)dx ≥ cφ(t n ) > 0.On the other hand, using the continuity and the fact that φ is nondecreasingwe obtain:0
R. Gologan and C. Lupu, An Olympiad problem 213Now, by letting ε → 0, if we assume that∫1˜H(1)φ(1) =0∫ 10˜H(x)d 1 φ .h(x)dx = 0, we get:This implies that the function ˜H(x) cannot be of constant sign on (0, 1).Thus there is c ∈ (0, 1) such that ˜H(c) = 0. In the particular case when:we clearly have∫ 1h(t) =f(t)∫ 1c ∈ (0, 1) such that∫ 1000∫ 1g(x)φ(x)dx − g(t)0f(x)φ(x)dx,h(x)dx = 0, so by the considerations above there exists∫ c0∫ cf(x)dxh(x)φ(x)dx = 0 which is equivalent to:0g(x)φ(x)dx =∫ 10∫ cg(x)dx0f(x)φ(x)dx.To formulate a consequence, denote by C([0, 1]) the Banach space ofcontinuous functions on [0, 1] and by C null the subspace of functions havingzero integral.Theorem 1.4. Let φ : [0,1] → R a nondecreasing function continuousat 0 and such that φ(0) = 0, and consider the Volterra operatorV φ : C([0, 1]) → C([0, 1]) given by V φ (f)(x) =φ(t)f(t)dt. Then, all functionsin V φ (C null ) have at least one zero in (0, 1).∫ x0Bibliografie[1] T.M. Flett, A mean value problem, Mathematical Gazette 42(1958), 38–39.[2] T.L. Rădulescu, V.D. Rădulescu, T. Andreescu, Problems in Real Analysis: asvancedcalculus on the real axis, Springer Verlag, 2009.[3] T. Lupu, Probleme de Analiză Matematică: Calcul Integral, GIL Publishing House,1996.[4] C. Lupu, T. Lupu, Problem 11290, American Mathematical Monthly, no. 4/2007.
Page 1 and 2: GAZETA MATEMATICĂSERIA AANUL XXVII
Page 3 and 4: V. Pop, Metoda etichetării binare
Page 11 and 12: A. Reisner, Teorema Beatty, şiruri
Page 13: A. Reisner, Teorema Beatty, şiruri
Page 23 and 24: M. Olteanu, Asupra rafinării unor
Page 36 and 37: 208 Articolerazele sferei înscrise
Page 38 and 39: 210 Note Matematice şi Metodice0
Page 42 and 43: 214 Note Matematice şi MetodiceIne
Page 44 and 45: 216 Examene şi concursuriii) Inega
Page 46 and 47: 218 Examene şi concursuriconstat
Page 48 and 49: 220 Examene şi concursuri2. Proble
Page 50 and 51: 222 Examene şi concursuri=[ (x1a 1
Page 52 and 53: 224 Examene şi concursuri∀ x ∈
Page 54 and 55: 226 Examene şi concursuri∫Fie 0
Page 56 and 57: 228 Didactica MatematiciiProiect di
Page 58 and 59: 230 Didactica Matematicii2) repreze
Page 60 and 61: 232 ProblemeÎn realizarea sarcinil
Page 62 and 63: 234 ProblemeSOLUŢIILE PROBLEMELOR
Page 64 and 65: 236 ProblemeDacă im(αu a + βu b
Page 66 and 67: 238 Problemeschimbarea, deci q să
Page 68 and 69: 240 Problemepentru orice x>0, este
Page 70 and 71: 242 Problemeunde:După efectuarea c
Page 72 and 73: 3) 1 α + 1 β + 1 γ = 2 a . Danie
Page 74 and 75: 246 Problemeşi:Rezultă:( )ω2sign
Page 76 and 77: 248 Istoria Matematiciişi soluţia
Page 78 and 79: 250 Istoria MatematiciiRevista s-a
Page 80 and 81: 252 Istoria Matematiciicaute o alt
Page 82 and 83: 254 Istoria Matematiciiachitării d
Page 84 and 85: 256 Istoria MatematiciiSocietăţii
Page 86 and 87: 258 Istoria Matematiciirezolvarea C
Page 88 and 89: 260 Din viaţa societăţiitrecere
Page 90 and 91:
262 Din viaţa societăţiipreponde
Page 92 and 93:
264 Din viaţa societăţii44. Radu
Page 94 and 95:
266 Din viaţa societăţii(22) Adr
Page 96:
268 RecenziiADRIANA DRAGOMIR, LUCIA
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