Lectures on Topics in Mean Periodic Functions And The Two-Radius Theorem

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1 Lectures n Tpics in Mean Peridic Functins And The Tw-Radius Therem By J. Delsarte Tata Institute f Fundamental Research, Bmbay 96

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3 Intrductin Three different subjects are treated in these lectures.. In the first part, an expsitin f certain recent wrk f J.L. Lins n the transmutatins f singular differential peratrs f the secnd rder in the real case, is given. (J.L. Lins- Bulletin sc. Math. de France, 84(956)pp. 9 95). The secnd part cntains the first expsitin f several new results n the thery f mean peridic functins F, f tw real variables, that are slutins f tw cnvlutin equatins: T F= T F= 0, in the case f cuntable and simple spectrum. These functins can be, at least frmally, expanded in a series f mean-peridic expnentials, crrespnding t different pints f the spectrum. Having determined the cefficients f this develpment, we prve its uniqueness and cnvergence when T and T are sufficiently simple. The result is btained by using an interplatin frmula, in C, which is analgus t the Mittag-Leffler expansin, in C. The expsitin and the prfs given here can prbably later, be simplified, imprved, and perhaps generalized. They shuld therefre be cnsidered as a preliminary accunt nly. 3. Finally, in the third part, I state and prve the tw-radius therem, which is the cnverse f Gauss s classical therem n the spherical mean fr harmnic functins. The prf is the same as that recently published, (Cmm. Math. Helvetici, 959) in cllabratin with J.L. Lins; it uses the thery f transmutatins f singular differential iii

4 iv Intrductin peratrs f the secnd rder, and the fundamental therem f meanperidic functins in R. J. Delsarte

5 Cntents Intrductin iii I Transmutatin f Differential Operatrs Riemann s Methd 3 Riemann s Methd fr the Cauchy prblem Prf fr the Riemann s methd Transmutatin f Differential Operatrs 3 Transmutatin in the Irregular Case 7 The peratr B p fr Rep> Cntinuatin f the peratr B p Cntinuatin f B p Transmutatin in the Irregular Case Cntinuatin f T p II Tpics In Mean-Peridic Functins 5 Expansin f a Mean-peridic Functin in Series 53 Determinatin f the cefficients in the frmal series.. 55 v

6 vi Cntents Examples Mean Peridic Functin in R 6 3 The Heuristic Methd The general frmula in R by the heuristic prcess The frmula (F ) fr a plynmial The fundamental therem f Mean III The Tw-Radius Therem Study f certain Cauchy-Prblems The generalized tw-radius therem Discussin f the hypthesis H

7 Part I Transmutatin f Differential Operatrs

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9 Chapter Riemann s Methd Riemann s Methd fr the Cauchy prblem Definitin. A functin is said t be (C, r) n a subset f A f R n if all its 3 partial derivatives upt the rder r exists and are cntinuus in A. Let D be a regin (pen cnnected set) in R, the (x, y) plane. Let a, b, c be three functins which are (C, ) in D and u a functin (C, ) in D and let L dente the differential peratr Lu= u x y + a u x + b u y + cu. Let v be a functin which is (C, ) in D and L be a differential peratr f the same type as L: then L v= v x y x (av) y (bv)+cv vl(u) ul (v)= M x + N y where M and N are (C, ) in D and are certain cmbinatins f u, v and 4 their partial derivatives. vl(u)=v v x y + av u x + bv u y + cuv 3 ()

10 4. Riemann s Methd = cuv+ x (auv)+ y (buv)+ ( v u ) x y u x (av) u y (bv) v u x y [ ] v = u x y x (av) y (bv)+cv + [ auv+v u ] x y y [ u v ] x buv 5 vl(u) ul (v)= M x + N y where M= auv+v u, N= buv u v y x. The right hand member f () des nt change if we replace M by auv+ (v u y u v y ) and N by buv+ (v u x u v ). We prefer t have x where M= y (uv) up(v) N= x (uv) uq(v) P(v)= v av, Q(v)= v y x bv. Let C be a clsed curve lying entirely in the regin D. By Green s frmula, ( ) M (λm+µn)ds= x + N dxdy y C whereλ,µdente the directin csines f the interir nrmal t C and A dentes the regin enclsed by C. In view f equatin (), (λm+µn)ds= (vl(u) ul (v))dxdy () C A A (i) (ii)

11 . Riemann s Methd fr the Cauchy prblem 5 We shall cnsider this equatin in the case when C cnsists f tw straight lines AX, AY parallel t the axes f crdinates and a curveγ, mntnic in the sense f AX and AY, jining X and Y. Suppse that L(u)=0 and L (v)=0 then C (λm+µn)ds=0 i.e. X A Ndx A Y Mdy= (λm+µn)ds substituting fr M and N frm (i) and (ii) respectively, X Y and X A A Ndx= X [(uv) X (uv) A ] Y Mdy= [(uv) Y (uv) A ]+ A uq(v)dx A Y up(v)dy. If the functins u, v satisfy 6 Lu=0, L v=0, P(v)=0n AY and Q(v)=0 n AX (3) then we btain the Riemann s Reslutin frmula: (uv) A = [(uv) X+ (uv) Y ]+ X Y (λm+µn)ds. Let A be (x 0, y 0 ). Then v=g(x, y; x 0, y 0 ) satisfying the cnditins (3) is the Riemann s functin fr the equatin L(u) = 0. In the situatin L(u)= f (x, y), L v=0, P(v)=0 n AY, Q(v)=0n AX,

12 6. Riemann s Methd exactly similar cmputatin gives the frmula: (uv) A = X [(uv) X+ (uv) Y ]+ (λm+µn)ds+ v f (x, y)dxdy. Y A An imprtant prperty f the Riemann s functin. Let nw Γ cnsider f tw straight lines XB, Y B parallel t the axes f crdinates. Then X X B (λm+µn)ds= Mdy Ndx Y B Y 7 We can write M= y (uv)+vp (u) N= x (uv)+vq (u) where P (u)= u y + au, Q (u)= u x + bu. X B Mdy= X [(uv) X (uv) B ]+ B vp (u)dx B Ndx= Y [(uv) B (uv) Y ] B Thus in this case Riemann s reslutin frmula becmes (uv) A = [(uv) X+ (uv) Y ] [(uv) X (uv) B ] X + vp (u)dx+ B [(uv) B (uv) Y ] B X i.e., (uv) A = (uv) B vq (u)dx+ vp (u)dx B Y Y B If u=h(x, y; x, y ), (x, y ) being the pint B, is such that Lu=0, P (u)=0 n BX, Q (u)=0 n BY, Y vq (u)dx vq (u)dx

13 . Prf fr the Riemann s methd 7 we get (uv) A = (uv) B. 8 Chsing cnstant multipliers fr u = h(x, y; x, y ) and v = g (x, y; x 0, y 0 ) in such a way that u= at B and v=at A, we have h(x 0, y 0 ; x, y )=g(x, y ; x 0, y 0 ) (4) This shws that the Riemann s functin g, cnsidered as a functin f (x 0, y 0 ) satisfies the differential equatin Lu=0. Prf fr the Riemann s methd We have btained (4) under the hypthesis that there exists functins u=h(x, y; x, y ) and v=g(x, y; x 0, y 0 ) which are (C, ) in D and which satisfy Lu= u x y + a u x + b u y + cu=0 P (u)= u + au=0 n BX y Q (u)= u + bu=0 n BY and x L v= v x y + a v x + b v y + c v=0 where a = a, b = b, c = a x b y + c P(v)= v y + a (v)=0 n AY Q(v)= v x + b (v)=0 n AX. By change f ntatin and translatin f the rigin, the prblem fr 9 the existence f the Riemann s functin v=g(x, y) in D fr the pint

14 8. Riemann s Methd (x 0, y 0 ) D fr the differential equatin Lu=0 reduces t the slutin f the prblem : with the cnditins u x y + a u x + b u y + cu=0 u(0, y)=α(y) u(x, 0)=β(x) I α(0)=β(0)= 0 where a, b, c are (C, ) in D andα,β are (C, ) functins f ne real variable. The slutin f the mre general prblem : [ u x y =λ a u ] x + b u y + cu with the cnditins I will give frλ= the slutin f the prblem (). We shall nw prve the existence f the unique slutin fr the prblem by using Piccard s methd f successive apprximatins. Cnsider the series u 0 (x, y)+λu (x, y)+ +λ n u n (x, y)+ (5) where u i (x, y) are defined by the fllwing recurrence frmula: u 0 x y = 0 u 0(0, y)=α(y), u 0 (x, 0)=β(x) u n x y = a u n + b u n + cu n, u n (x, 0)=u n (0, y)=0 fr n. x y It suffices t take u 0 (x, y)=α(y)+β(x) and u n (x, y)= x 0 φ n (ξ,η)dξdη where φ n = a u n x + b u n y + cu n We shall nw prve the cnvergence f the series (5) by the prcess f majrisatin which is classical. Suppse that a, b, c are (C, 0) in D y 0

15 . Prf fr the Riemann s methd 9 andα,β are (c, ) f ne real variable. Let K be a cmpact subset D cntaining the rectangle with sides parallel t the axes and (0, 0) and (x, y) as ppsite crners. Then there exists an M, A such that α(ξ)+ β(η) M, (α(ξ)+β(η) ) M, ξ and (α(ξ)+β(η) ) M, η fr (ξ,η) in K, and a, b, c, A in K. Then φ 0 (x, y) 3 A M. By the recurrence frmula fr n =, u (x, y) x y 0 0 φ 0 (ξ,η)dξdη 3AM x y u 3AM y and, u x y 3AM x. Hence u (x, y), u x, u y are each 3AM(+ x )(+ y ). Cmputingφ (x, y), we have immediately, ( ) ( ) + x + y u (x, y) 9A M!! ( ) u (x, y) + y x 9A M(+ x ) ( ) u (x, y) + x y 9A M (+ y ) In general u n (x, y), u n x, u n M[3A(+ x )(+ y )]n y, n! Cmparing with the expnential series, this majrizatin prves that the series (3) as als the series btained frm (3) by differentiating each term nce and twice are all cnvergent unifrmly n each cmpact subset f D and abslutely in D, s that (3) cnverges t a functin (C, ) in D. This functin u(x, y) is evidently the slutin f prblem and the prf fr the existence f Riemann s functins is cmplete.

16 0. Riemann s Methd Remark. Frm the recurrence frmula it is clear that the functin u(x, y) satisfies x y [ u(x, y)=α(y)+β(x) +λ a u ] x + b u y + cu dξ dη 0 0 x=ξ y=η This integral equatin is equivalent t the differential equatin f prblem with cnditins I. When a, b, c are (C, 0) andα,β are (C, ), the slutin u(x, y) f the integral equatin may nt be (C, ). But then it is a slutin f prblem in the sense f distributins.

17 Chapter Transmutatin f Differential Operatrs Let L and L dente tw differential peratrs n the real line R and 3 E m (x a) dente the space f functins m times cntinuusly differentiable in [a, ) furnished with the usual tplgy f unifrm cnvergence f functins tgether with their derivatives upt the rder m n each cmpact subset f [a, ). Let E be a subspace f the tplgical vectr space E m (x a). Definitin. A transmutatin f the differential peratr L int the differential peratr L in E is a tplgical ismrphism X f the tplgical vectr space E nt itself (i.e. a linear, cntinuus, ne-t-ne, nt map), such that XL = L X X is said t transmute the peratr L int the peratr L in E and L = XL X n E. Transmutatin in the regular case. Let L = D x q(x), L = D x where q satisfies certain cnditins f regularity. The cnstructin f the transmutatin peratr in this case depends n the cnsideratin f certain partial differential equatin. Prblem. T determineφ(x, y) in x a, y a satisfyingφ xx Φ yy

18 . Transmutatin f Differential Operatrs q(x)φ=0 with the bundary cnditins Φ(x, a)=0=φ(a, y);φ y (x, a)= f (x) 4 This mixed prblem is equivalent t the Cauchy prblem if we set u(x, y)=φ(x, y) fr x a = Φ(a x, y) fr x a We set withut prf the fllwing prpsitin. Prpsitin. (a) If q E (R), f E (x a) with f (a)=0, then prblem pssesses a unique slutin which is (C, ) in x a, y a and satisfies the differential equatin in the sense f distributins. (b) If q E (R) and if f E (x a) with f (a)=0, the slutin f prblem is (C, ) in x a, y a. In the regin y x(r x y), the slutin is (C, 3). (c) If q E (R) with q (a)=0 and f E 3 (x a) with f (a)= f (a)= 0, then the slutin u f the prblem is (C, 4) in y a. [Refer t E. Picard, Lecns sur quelques types simples d equatin aux derivces partielles, Paris, Gauthier-VIllars, 97.] With the help f this prpsitin we prve Prpsitin. If q E (R) with q (a)=0 and f E 4 (x a) with f (a)= f (a)=0 then D A f= AL f where L=D = q and A is defined by A f (y)= [Φ(a, y)], x 5 Φ(x, y) being the slutin f prblem. Letψ(x, y)=l x [Φ(x, y)]= Φ q(x)φ and g(x)=l x x f (x). AsΦ is (C, 4) by Prpsitin (c), and f E 4 (x a),ψ(x, y) is (C, ) and g E (x a) with g(a)=0. Replacing f by g in prpsitin (b), prblem pssesses a unique slutin. We verify belw that this unique slutin is Ψ(x, y). L x Ψ D yψ=l x L x Φ D yl x Φ=L x [L x Φ D yφ]=0;

19 3 Ψ(x, a)=l x [Φ(x, a)]=l x [0]=0; y Ψ(x, y)=d yl x Φ(x, y)=l x D y Φ(x, y) Ψ y (x, a)=l x D y Φ(x, a)=l x f (x)=g(x); Ψ(a, y)=d y[φ(a, y)]=0. s that Nw by definitin f A, AL[ f (y)]=a.g= Ψ(a, y) x Ψ x (x, y)=d x L x Φ= D y[φ x (x, y)] gives Ψ x (a, y) = D y [Φ x(x, y)]=d ya f (y). Hence we have prved that AL=D y A. Cmputatin f the slutin u(x, y), y a f prblem by using 6 Riemann s functin. Let K(x, y; x 0, y 0 ) be the Riemann s functin defined in the shaded part f the satisfying the cnditins K K x y q (x)k= 0 with q (x)=q(x) if x a = q(a x) if x<a

20 4. Transmutatin f Differential Operatrs and K n Mm = K n Mm =. In this case Riemann s methd gives u(x 0, y 0 )= x0 +y 0 a a y 0 +x 0 f (x)k(x, a; x 0, y 0 )dx where f (x)= f (x) if x a = f (a x) if x a 7 Hence A[ f (y 0 )]= u(a, y 0 ) x 0 = f (y 0 ) y0 a f (x) x 0 K(x, a; a, y 0 )dx. Prblem. T determine the functinφ(x, y) in x a, y a satisfying the cnditinsφ xx Φ yy q(x)φ=0;φ(a, y)=0;φ x (a, y)=g where g E (y a) with g(a) = 0, andφ(x, a) = 0. Prblem, is the same as Prblem if in the bundary cnditins the lines x=a, y=a are interchanged. IfΦis the slutin f Prblem, we define a g(x)= Φ(x, a). y Prpsitin 3. If q E (R) and f E (x a) with f (a)=0, then aa f= Aa f=f. LetΦbe the slutin f Prblem and let g(y)=a f (y)= Φ(a, y). x By Prpsitin (b), g is (c, ) in y a and g(a)=0. HenceΦis the slutin f Prblem and a g(x)= y Φ(x, a)= f (x). This shws that aa f=f. Similarly Aa f=f. Prpsitin 3 tgether with Prpsitin shws that if q E (R) with q (a)=0 the map A, which is bviusly linear is ne-t-ne f the space E= { f/f E 4 (x a), f (a)= f (a)=0 } nt itself and verifies AL f = D A f. Further in view f the frmula fr A f n page 3 in items f Riemann s functins, A is cntinuus n E with the tplgy induced by E 4 (x a). As E is a clsed subspace f the Frechet space

21 5 E 4 (x a), A is a tplgical ismrphism. Thus we have prved the existence f the transmutatin peratr A in E transmuting D q(x) int D when q is sufficiently regular. We nw cnsider the prblem f transmuting mre general differen- 8 tial peratrs L i = D + r i (x)d+ s i (x) (i=, ) int each ther when r i and s i are regular (e. g. r i, s E (R)). Prpsitin 4. There exists an ismrphism A L L f E which satisfies A L L L = L A L L. The prpsitin will be prved if we prve the existence f transmutatin X f the peratr L int the peratr D q. Fr then the same methd will give a transmutatin f L int L = D q and each f the peratrs D q i (i=, ) can be transmutated int the peratr D s that finally we btain the required transmutatins A L L by cmpsing several transmutatins. Let R (x)= x a r (ξ)dξ then the verificatin f the fllwing equatins is straight frward: ] L [e R(x) f (x) = e R(x) L [ f ]. Hence we have X f (x)=e R (x) f (x).l = D q where q (x)= 4 r (x)+ r (x) s (x) E. Applicatin f transmutatin t the Mixed Prblems f differential equatins. IfΛis an elliptic peratr inr n ( independent f the variable t which 9 crrespnds t time) we cnsider the prblem f finding a functin u(x, t) (x R n, t time) which satisfies the differential equatin ( ) Λ x (x, t)+ t + r(t) t + s(t) u(x, t)=0

22 6. Transmutatin f Differential Operatrs with the Cauchy data in a bunded dmainω R n fr t=0 and als n the hemicylinderω [t 0] whereω dentes the frntier fω. In this prblem the variable x which crrespnds t space and the variable t which crrespnds t time are strictly separated. Suppse that A t is a transmutatin in the variable t which transmutes t + r(t) t + s(t) int t and let V(x, t)=a tu(x, t) when u(x, t) is the slutin f the differential equatin. Then applying A t t the left hand member f the equatin we have A t Λ x u(x, t)+a t ( t + r(t) + s(t))u(x, t)=0 t i.e., Λ x v(x, t)+ v(x, t) t = 0 and v(x, t) satisfies Cauchy s data i.e., by means f the transmutatin, cnsideratin f the gives equatin is reduced t the cnsideratin f the wave equatin.

23 Chapter 3 Transmutatin in the Irregular Case Intrductin. Our aim in this chapter is t btain a transmutatin p- 0 eratr fr a differential peratr with regular cefficients. In rder t reduce the mixed prblem relative t the peratr Λ+ p+ t+ t, p real r cmplex, t whereλ= ( being the Laplacian in the variables x,..., x n ) t the mixed prblem relative t the peratr Λ+ t we shall cnstruct an peratr will transmute the peratr L p = D + p+ D x int the peratr D. The difficulty in this case arises due t the presence f the cefficient which has a singularity at x=0. If we seek the x slutin f the prblem in x a, where a>0, the methd f the preceding chapter is perfectly valid withut any change. But the imprtant case is precisely the ne in which a=0. 7

24 8 3. Transmutatin in the Irregular Case We shall determine ismrphisms B p, B p (f certain space which will be precisely specifies in the sequel ) which satisfy D B p = B p L p ; B p D = L p B p. The peratr B p fr <Rep< and the peratr B p fr Rep> are classical. B p is the Pissn s peratr and B p is the derivative f the Snine peratr. The peratr B p fr Rep> It can be frseen that the peratr B p is defined in terms f the slutin fr y>0 f the partial differential equatin Φ xx Φ yy + p+ Φ x = 0 x with the cnditinsφ(0, y)=g (y) being an even functin g (y)=g(y) fr y>0 and g (y)=g( y) fr y<0, andφ x (0, y)=0 Nw we define B p [g(x)]=φ(x, 0). Changing the variables (x, y) t the variables (s, t) be means f the frmulae s =y+ x and t ( =y x, we see by simple cmputatin that u(s, t)=φ s t, s+t ) is a slutin f the partial differential equatin u st p+ s t u s+ p+ s t u t= 0 with the cnditins u(s, s)=g (s ) (u s u t ) s=t = 0

25 . The peratr B p fr Rep> / 9 Frα= p+, Pissn s slutin has the frm u(s, t)= Γ(α) [Γ(α)] valid fr Reα>0 (r Rep> ). Then B[g(x)]=Φ(x, 0)=u( x, x ) 0 g { [s+(t s)ρ] } ( ρ) α ρ α dρ setting ρ=r, = Γ(α) [Γ(α)] 0 g [x( ρ)]( ρ) α ρ α dρ B p [g(x)]= Γ(α) [Γ(α)] α 0 = Γ(p+) πγ(p+ ) ( UsingΓ(α)= Γ(α)Γ(α+ )α π. Nte that B p [g(0)]=. 0 ( r ) α g (rx)dr ( t ) p g(tx)dt. Prpsitin. The mapping f B p f is a linear cntinuus map f 3 F = { f/f E (x 0), f (0)=0 } int itself and satisfies B p D f= L p B p f fr any f F. As fr Rep>, differentiatin { under sign } f integratin is permissible, d B p f E (x 0). Further dx B p[ f ] = f (0) k (k being sme x=0 cnstant), = 0. Hence B p f F. Evidently B p is linear. In rder t prve cntinuity, since F has a metrizable tplgy induced by that f E (x 0), it is

26 0 3. Transmutatin in the Irregular Case 4 5 sufficient t shw that if a sequence{ f n }, n=,,..., f n F. cnverges t 0 in F, then B p f n cnverges t zer in F. But f n 0 in F implies f n 0 unifrmly n each cmpact set, in particular n the cmpact set [0, ] frm which it fllws that B p f n 0. It remains t verify that B p satisfies the given cnditin. Writingβ p = Γ(p+) πγ(p+/) { Lp B p f B p D f } = β p β p + (p+)t x = p+ x = p+ x = { t ( t ) p f (tx) ( t ) p f (tx) ( t ) p f (tx) } dt 0 t( t ) p f (tx)dt ( t ) p+ f (tx)dt. 0 ( t ) p+ p+ f (tx) + ( t ) p+ f (tx)dt 0 x ( t ) p+ f (tx)dt (integrating the first integral by parts). Remark. Let J p (x) dente the classical Bessel functin and let j p (x)= p Γ(p+)x p J p (x) j p (x) is in F and is the unique slutin f the differential equatin d y ( ) dx + p+ dy dy x dx + s y=0 with the cnditins (y) x=0 = and = 0. dx x=0 Nw cs sx F. Let B p (cs sx)=g(x). Using L p B p = B p D, we get i.e., L p B p [cs sx]=l p [g(x)]=b p D (cs sx) L p g+ s g=0. = s B p [cs sx]= s g(x). Further g(0)=cs 0= and g (0)=0 since g F. This shws that B p (cs sx)=g(x) j p (sx).

27 . The peratr B p fr Rep> / Hence we btain the classical frmula, j p (sx)= Γ(p+) πγ(p+ ) 0 ( t ) p cs(stx)dt. The peratr B p was cnsidered by Pissn in this particular questin f the transfrmatin f the cnsine int the functin j p. The peratr B p fr <Rep< Fr f E 0 (x 0), the definitin f B p is B p [ f (x)]=b p x x 0 (x y ) p 3 y p+ f (y)dy = b p t p+ ( t ) p 3/ f (tx)dt 0 where /b p = ( π Γ(p+)Γ p ) Prpsitin. Fr <Rep< and f F, f B p f is a linear 6 cntinuus map f F int itself satisfying D B p f= B p L p f. The prf f the fact that B p is a linear cntinuus map f F int itself is analgus t the ne we have given in Prpsitin. We verify that B p satisfies the given cnditin, again as in Prpsitin, by integratin by parts { } Bp L p D B p f (x)= t p+ ( t ) p 3/ b p 0 { f (tx)+ p+ } f (tx) t f (tx) dt tx = = t p+ ( t ) p ) f (tx)+ p+ 0 x [ ] t p+ ( t ) p f (tx) 0 0 f (tx) x 0 d dt t p ( t ) p 3/ ) f (tx)dt

28 3. Transmutatin in the Irregular Case t p+ dt+ p+ ( t ) p+ x t p 0 f (tx) ( t ) p+3/ dt=0. 7 The Snine peratr B p fr <Rep<. B p is defined fr every f E 0 (x 0)by B p f= x b p 0 t p+ ( t ) p+ f (tx)dt x = b p y p+ (x y ) p f (y)dy 0 π where b p = Γ(p+)Γ( p+ ). The integral cnverges if <Rep< and we can differentiate under the sign f integratin d dx B p [( f (x))]=b p [ f (x)] fr every f E 0 (x 0) Relatin between B p and B p When < Rep<, bth B and B p are defined and it is easy t prve by direct cmputatin Abel s functinal equatin B p B p [ f (x)]= In fact x 0 f (y)dy. x y B p B p [ f (x)]= b p y p+ (x y ) p βp y p f (z)(y z ) p dz 0 y = b p β p dy 0 y x = b p β p [ f (z) 0 0 x y(x y ) p f (z)dz z (x y ) p (y z ) p ydy]dz 0

29 . Cntinuatin f the peratr B p 3 8 Setting x sin θ+z cs θ=y, we have x y = (x z ) cs θ, y z = (x z ) sin θ ydy=(x z ) sinθcsθdθ x z (x y ) p (y z ) p ydy π = cs p θ sin p θdθ= ( 0 B p+, p+ ) = Γ ( ( ) p+ ) Γ p+ Γ() Hence B p B p [ f (x)]= x 0 f (z)dz s that D B p B p [ f (x)]= f (x). Cntinuatin f the peratr B p Fr any f E (x 0)=E (x 0), we define In general T [ f (t, x)]=d t [t f (tx)], T [ f (t, x)]=d t [t 3 T { f (t, x)}]. T n [ f (t, x)]=d t {t 3 T n [ f (t, x)]}. Lemma. T n [ f (t, x)] = t n g n (t, x) where g n (t, x) is an indefinitely 9 differentiable functin in [0, ] [0, ). The prf f the lemma is trivial and is based n indicatin n n. Fr n=, we have nly t set g (t, x)=d t [t f (tx)]. Assume that the lemma is true fr n s that Define T n [ f (t, x)]=t n 4 g n (t, x). g n (t, x)=3g n (t, x)+(n 4)g (n ) (t, x)+td t g n (t, x).

30 4 3. Transmutatin in the Irregular Case By definitin T n [ f (t, x)]=3t T n [ f (t, x)]+t 3 D t T n [ f (t, x)] = 3t t n 4 g n (t, x)+t 3{ (n 4)t n 5 g n (t, x)+t n 4 D t g n (t, x) }. = t n g n (t, x). n 3 Tn f (x, t)dt cn- Crllary. The integral 0 tp (n 3) ( t ) verges fr <Rep<n. p+ 30 The crllary is immediate since the integral can be written as 0 tp+ ( t ) p+n 3 g n (t, x). Prpsitin. Fr <Rep<, and fr f E (x 0), B p [ f (x)]= ( ) n b p (p+)(p ) (p n 3) 0 t p (n 3) ( t n 3 ) p T n f (t, x)dt The prf is based n inductin n n and the fllwing frmula which is bvius: d t p λ = (p λ)tp λ dt ( t ) p λ ( t ) p λ/+ () fr any λb p [ f (x)]=b p t p+ ( t ) p 3/ f (tx)dt Let n=. If p+ 3 = p λ +, i.e., λ= t p d t ( t ) p+3/= p+ (p+) dt ( t ) p+ 0

31 . Cntinuatin f the peratr B p 5 s that B p [ f (x)]= b p p+ = b p p+ 0 0 d t p+ dt ( t ) p+ t p+ t f (tx)dt [ ] T ( t ) p+ f (t, x) dt (integrating by parts, the integrated part being zer since Rep+ < 0 3 and Rep+ >0). Thus the frmula t be prved hlds fr n=. Assuming it fr n, we establish it fr n B p [ f (x) ] = ( ) n b p (p+)(p )...(p n+5) 0 t p (n 5) ( t n 5 ) p T n [ f (t, x)]dt Using () with p n 5 = p λ + i.e., λ=n 3, the integral n the right hand side equals p n+3 0 = = d dt t p n+3 ( t n 3 ) p p n+3 p n+3 0 t 3 T n [ f (t, x)]dt t p+ ( t n 3 ) p 0 g n (t, x) t p n+3 ( t n 3 ) p t p n+3 ( t n 3 ) p 0 T n [ f (t, x) ] dt T n [ f (t, x) ] dt, the integrated part being zer since <Rep<. We write 3 B n p [ f (x) ] = b (n) p 0 t p (n 3) [ ] T ( t n 3 n f (t, x) dt () ) p

32 6 3. Transmutatin in the Irregular Case where b (n) p = ( ) n b p (p+)(p )(p 3) (p n+3) (3) The integral is cnvergent fr < Rep < n s that under this cnditin, we can differentiate under the sign f integratin and B n p f E. We btain fr each n a functin which assigns t each p in <Rep<n, a map Bn p f E int itself which cincides with B p if 33 <Rep<. It is easy t see that Bn p is a linear map f E int itself. In rder t shw that it is cntinuus, as E is metrizable, it is enugh t prve that if a sequence { } f j j=,,... tends t zer in E, then Bn p f j tends t 0 in E. We have where D r xb n [ p f j (x) ] b (n) t p (n 3) p M(r, x, j) dt 0 ( t n 3 ) p M(r, x, j)= sup D r x T [ n f j (t, x) ] t [ Nw T n f f (t, x) ] is a plynmial in t, x with cefficients which are derivatives f rder n f f j and f j tgether with all its derivatives cnverge t zer n each cmpact subset. Hence M(r, x, j) 0 as j unifrmly fr x n each cmpact subset i.e. B n p f j 0 in E. Thus we btain a functin p B n p n <Rep< with values in L (E,E ), the space f linear cntinuus maps f E int itself. We intend t prve that this functin is analytic and can be cntinued in the whle cmplex plane int a functin which ia analytic in the half plane Rep > and mermrphic in Rep < with a sequence f ples lying n the real axis. Befre prving this cntinuatin therem we give first the definitin f a vectr valued analytic functin and sme f its prperties which fllw immediately frm the definitin. Definitin. Let 0 be an pen subset f the cmplex plane and E a lcally cnvex vectr space. A functin f : 0 E is called analytic if fr every e in the tplgical dual E f E (i.e. the space f linear cntinuus frms n E) the functin z < f (z), e > is analytic in 0 where<,> dentes the scalar prduct between E and E.

33 . Cntinuatin f the peratr B p 7 Lemma. If E is lcally cnvex vectr space in which clsed cnvex envelpe f a cmpact set is cmpact, a functin f : 0 E is analytic if f is cntinuus and fr every e in a ttal set M f E, z < f (z), e > is analytic in 0. Let C be any simple clsed curve lying entirely in 0, enclsing re- 34 gin ( pen cnnected set ) cntained in 0. Fr e M, the functin z < f (z), e > is analytic in 0, s that c < f (z), e > dz=0 i. e. < c f (z)dz, e >= 0. f (z)dz is the integral f the cntinuus E-valued c functin f ver the cmpact set C and is an element f E since E has the prperty that the clsed cnvex envelpe f any cmpact subset is cmpact. ( Fr integratin f a vectr valued functin, refer t N. Burbaki, Elements de Mathematique, Integratin, Chapter III). Hence c f (z)dz= 0 since M is ttal, s that< c f (z)dz, e >= c < f (z), e > dz=0 fr every e E. Als as f is cntinuus it fllws that z < f (z), e > is cntinuus. This prves that z < f (z), e > is analytic fr every e E since the chice f C was arbitrary. Prpsitin. Suppse that <Rep<n. Then a) B n p f E (x 0) fr every f E (x 0) b) The mapping f B n p f is linear cntinuus f E int itself. c) The functin p B n p n the strip < Rep < n with values in L s (E,E ) is analytic where L s (E,E ) is the space f linear cntinuus maps f E int E endwed with the tplgy f simple cnvergence. We have t prve nly (c). We first bserve that any linear cntinuus frm n L s (E, F) is given by finite linear cmbinatin f frms 35 f the type u < ue, f > with e E and f F. The therem will be prved if we shw that the map p < B n p f, T> is analytic in <Rep<n where f is any element f E and T any element f E ; i.e. fr fixed f we have t shw that the map p B n p f is analytic with values in E. Nw E ia a Hausdrff cmplete lcally cnvex vectr

34 8 3. Transmutatin in the Irregular Case space and therefre clsed cnvex envelpe f each cmpact subset f E is cmpact ( refer t N. Burbaki, Espaces Vectriels Tplgiques, Ch.II, 4, Prp.). Further it is easy t see that p B n p f is cntinuus and that the set { δ x }x 0, whereδ x is the Dirac measure with supprt at x, is ttal in E. Applying the lemma, we see that in rder t prve analyticity f the functin p B n p f we have nly t prve that the functin p < B n p f,δ x> i.e. the functin p B n p f (x) (fr f and x fixed ) is analytic in <Rep<n. Observing that b p = π Γ(p+)Γ( p ) is an entire functin with zers at p=,, 3,... and p=,, 3,... we see that bn p in (3) is an entire functin. The integral in () n the ther hand cnverges fr <Rep<n and therefre is analytic in the same regin, Hence p B n p f (x) is analytic in the strip <Rep<n. 36 Crllary. The functins B m p and Bn p where m and n are tw distinct psitive integers are identical in the intersectin f their dmains f definitin. We have in fact tw analytic functins B m p and B n p which cincide with B p in < Rep < which is cmmn t their dmains f definitins and therefre the tw functins cincide everywhere in the dmain which is the intersectin f their dmains f definitin due t analyticity. It fllws frm the crllary that we have a unique analytic functin B p defined fr Rep>. Remark. We have B / = identity. Fr n=, B p ( f )= b p p+ 0 t p+ ( t ) p+ d[ ] t f (tx) dt dt

35 . Cntinuatin f the peratr B p 9 and if p=, b p p+ = and B [ f (x) ] = 0 d[ ] t f (tx) dt= f (x). dt Cntinuatin f B p fr Rep<. We define U [ f (t, x) ] = Dt [ ( t ) f (tx) ]. U [ f (t, x) ] = Dt [ ( t ) 3/ f (tx) ]. In general U n [ f (t, x) ] = Dt [ ( t ) 3/ U n f (tx) ]. 37 Lemma 3. Fr every f ine(x 0)=E we have U n [ f (t, x) ] = ( t ) 3/ D t U n f (t, x) 3t( t ) Un [ f (t, x) ] and U n [ f (t, x) ] = ( t ) n hn (t, x) (5) where h n (t, x) is indefinitely differentiable in [ 0, ] [ 0, ). Relatin (4) is evident. (5) can be prved by inductin n n. It is true fr n= if we set h (t, x)= t f (tx)+ x( t ) f (t, x). Assuming it fr n, it is easy t verify that (5) hlds fr n if h n (t, x)= nth n (t, x)+( t )D t h n (t, x) (4) Crllary. The integral 0 n < Rep<. t p+n+ ( t p+ n+ ) U n [ f (x, t) ] dt cnverges fr Prpsitin 3. If <Rep<,

36 30 3. Transmutatin in the Irregular Case B p [ f (x) ] = ( ) n b p (p+)(p+3) (p+n+) 0 t p+n+ ( t n+ ) p+ U n [ f (t, x) ] dt The prf f this prpsitin is analgus t that f Prpsitin. 38 We prve it by inductin n n and by using frmula (). Fr <Rep<, 39 [ ] B p f (x) = bp t p+ ( t ) p 3 f (tx)dt Using () with p+=p λ i. e.λ=, we have, s that t p+ ( t ) p+= p+ [ ] b p B p f (x) = p+ = b p (p+) d dt ( t p+ ( t ) p+ ) d dt ( t p+ ( t ) p+ )( t ) f (tx)dt t p+ ( t ) p+ U [ ] f (t, x) dt, the integrated part being zer since <Rep<. The frmula is prved fr n=. We assume it fr n B p [ f (x) ] = ( ) n b p (p+)(p+3) (p+n) t p+n ( t ) p+n/ U n [ f (t, x)]dt Using () with p+n=p λ i. e.λ= (n+), we get d t p+n+ t =(p+n+) p+n+, dt ( t n+ ) p+ ( t n+3 ) p+ 0

37 . Cntinuatin f the peratr B p 3 s that B p [ f (x) ] = ( ) n b p (p+) (p+n+) we shall nw set t p+n+ 0 t p+n+ ( t n+ ) p+ } D t {( t ) 3 Un f (t, x) dt [ ] [ ] n Bp f (x) = (n)bp U ( t n+ n f (t, x) dt ) p+ (6) ( ) n b p (n) bp = (p+)(p+3) (p+n+) (7) (n) bp is a mermrphic functin f p, with ples at the pints p= 3, 5,... Prpsitin 4. Suppse that p satisfies n < Rep< (8) and des nt assume any f the values 3, 5,... (9) Then 40 a) n B p f E fr each f E b) The mapping f n B p f is linear cntinuus frm E int E. c) The functin p n B p is mermrphic in the strip defined by (8) with ples situated at the pints given by (9). We mit the prf f a) and b) since it is exactly similar t that f Prpsitin. The prf f c) reduces as in Prpsitin, t shwing that the functin p (n) b p 0 t p+n+ [ ] U ( t n+ n f (t, x) dt ) p+

38 3 3. Transmutatin in the Irregular Case is mermrphic in the strip (8) with ples at the pints (9), which is bvius since the integral cnverges in (8) and is therefre analytic in (8) and n bp is mermrphic in (8) with ples given by (9). 4 Crllary. If m, n are tw distinct psitive integers, the tw functins m Bp, n Bp cincide in the cmmn part f their dmains f definitin. This crllary is immediate like the crllary f Prpsitin. Cnsequently we have a unique functin B p defined fr Rep<. Prpsitins () and (4) give finally the cntinuatin therem fr the peratr B p. Therem. The functin p B p defined initially in < Rep< with values in L (E,E ) endwed with the tplgy f simple cnvergence can be cntinued in the whle plane int a mermrphic functin. The ples f this functin are situated at the pints 3, 5, 7,... Remark. The ntin f an analytic functin with values in a lcally cnvex vectr space E depends nly n the system f bunded subsets f E. E being cmplete it is easy t see (in view f Therem, page, Ch.III, Espaces vectrieles Tplgiques by N. Burbaki) that the space L (E,E ) when furnished with the tplgy f simple cnvergence has the same system f bunded sets as when furnished with the tplgy f unifrm cnvergence n the system f bunded sets f E. Hence in the therem we can replace the tplgy f simple cnvergence by the tplgy f unifrm cnvergence n bunded subsets f E r by any ther lcally cnvex tplgy which lies between these tw tplgies. Let E and D be subspaces f E (x 0) defined by E = { f f E (x 0), f n+ (0)=0 fr n 0 } D = { f f (x 0), f n (0)=0 fr n 0 } 4 When <Rep<, we have D r [ ] B p f (x) = bp t p++r ( t ) p 3 f r (tx)dt 0

39 . Cntinuatin f the peratr B p 33 s that where D r B p [ f (0) ] = bp,r f r (0) (0) b q,r = ( πγ p+ r ) + Γ ( ) r+ Γ(p+) This shws that when r is even, p b p,r is an entire functin f p and when r is dd it is a mermrphic functin with ples at 3, 5,... (0) is therefre true fr all p nt equal t the exceptinal values 3, 5,... Then B p[ f ] is in E r in D accrding as f is in E r D. Therefre fr p 3, 5,..., B p L (ε,ε )( r L (D, D )) and p B p is mermrphic with ples at p= 3,... Actually the fllwing strnger result hlds. Therem. The functin p B p is an entire analytic functin with values in L (ε,ε ) (als in (D, D )). We have seen mre than nce, that in rder t investigate the analyticity f the functin p B p, it is sufficient t d the same fr the functin p B p [ f (x)], where f E and x 0 are arbitrarily chsen and are fixed. In this case we study the behaviur f the functin 43 p B p [ f (x)] where f E and x 0, in the neighburhd f the pint p = m+, (a suppsed singularity f the functin). By Taylr s frmula, f (tx)= N 0 x n t n t n! f n (0)+ xn+ f (tξ) f N+ (ξx)dξ, N N being an arbitrary integer. Suppse first that <Rep<. Then B p [ f (x)]= π Γ(p+) N Γ ( p+ n + ) Γ ( n + ) x n f n (0) n!

40 34 3. Transmutatin in the Irregular Case + b p x N+ N! t p+ ( t ) p 3 dt (t ξ) N f N+ (ξx)dξ. The right hand side f this frmula is well defined when N+ 3 < Rep< and depends analytically n p and therefre cincides with the cntinuatin f B already btained. Chsing N such that N+ 3 < p, the integral n the right is analytic at the pint p. Hence we have nly t cnsider the finite sum at p. The functinγ(p+ n +) has ples at the pints p such that p+ n + = µ,µ a psitive integer. It has a ple at p 0 if n = µ p 0= m µ 44 i.e. n=(m µ). But when n is dd, f n (0)=0 since f E. The finite sum is therefre analytic at p 0 and B p [ f (x)] has false singularity at m+ and the prf f the therem is cmplete. Therem. The frmula D B p f= B p L p f hlds fr every f E and fr every cmplex number p. L p f is in E if f E s that L p L (E, E ) and it is easy t verify that p L p is an entire functin with value is L (E, E ). The tw entire functins p B p L p and p D B p cincide by Prpsitin, in <Rep < and are therefre identical in the whle plane. Remark. It is necessary t suppse that f E. If f is nly in E,L p f is nt in E (always). 3 Cntinuatin f B p We nw cnsider the extensin f the peratr B p initially defined fr Rep> by B p [ f (x)]=β p ( t ) p f (tx)dt. (I)

41 3. Cntinuatin f B p 35 where β p = Γ(p+) Γ(p+ ) Changing the variable t = sin Θ, 45 B p [ f (x)]=β p f (x sinθ) cs p θdθ. and and Let M p[ f (x,θ)]= f (x sinθ) d + {sinθ ddθ } (p+)(p+) dθ [sinθf (x sinθ)] N p [ f (x,θ)]= d {sinθf (x sinθ)} p+ dθ We determine by inductin, the functins M n p d [ f (x,θ)]= Mn p [ f (x,θ)]+ p+n dθ { sinθ d } d dθ sinθnn p [ f (x,θ] + (p+n )(p+n) dθ { sinθ d } dθ [sinθmn p ( f (x,θ)) N n p[ f (x,θ)]=n n p [ f (x,θ)]+ d p+n dθ { sinθm n p [ f (x,θ)] } Prpsitin. Fr Rep>, f E (x 0) and fr any n, B p [ f (x)]=β p π cs p+n θm n p[ f (x,θ)]dθ π +β p cs p+n+ θn n p f (x,θ)dθ () 0

42 36 3. Transmutatin in the Irregular Case The prf f this prpsitin is elementary and is based n inductin 46 n n and the prcess f integratin by parts. B p f (x)=β p π 0 f (x sinθ) cs p+ θdθ +β p π f (x sinθ) sin θ cs p θdθ. But as sinθcs p θ= d dθ ( csp+ θ ) integrating by parts the secnd (p+) integral, we get, B p f (x)=β p π f (x sinθ) cs p+ θdθ +β p π 0 d p+ dθ (sinθf (x sinθ)) csp+ θdθ Nw the secnd integral in this equatin equals 47 π 0 s that d p+ dθ (sinθf (x sinθ)) csp+3 θdθ π π 0 π = d p+ dθ (sinθf (x sinθ)) csp+ θ sin θdθ 0 B p f (x)=β p π d p+ dθ (sinθf (x sinθ)) csp+3 θdθ (p+)(p+) csp+ θ d dθ 0 { cs p+ θ f (x sinθ) { sinθ d dθ (sinθf (x sinθ)) } dθ d + (p+)(p+) dθ (sinθ d } dθ (sinθf (x sinθ))) dθ π +β p cs p+3 d θ (sinθf (x sinθ))dθ p+ dθ 0

43 3. Cntinuatin f B p 37 Hence frmula () hlds fr n=. Assuming it fr n, it can be verified fr n by integratin by parts. We define fr every integer n>0, π B n p f (x)=β p cs p+n θm n p f (x,θ)dθ+ 0 π β p cs p+n+ θn n p f (x,θ)dθ The tw integrals cnverge fr Rep> n. Prpsitin. If p satisfies 48 Rep> n () and p,, 3,... (3) Then a) Fr f E (x 0)=E,B n p f E. b) The mapping f B n p f is linear cntinuus f E int itself. c) The functin p B n p is mermrphic in the half plane defined by () with values in L (E,E ) the ples being situated at the pints (3). We shall give nly the utline f the prf since it is analgus t that f Prpsitin (),. The functinβ p = Γ(p+) πγ(p+ ) is a mermrphic functin f p which vanishes at, 3,..., and ples at,, 3,... and fr x,θ fixed, M n p f (x,θ), and Nn p f (x,θ) are mermrphic functin with ples at,, 3,,... Hence fr p fixed and,,..., the functin (x,θ) β p M n p (x,θ) and (x,θ) β p N n p f (x,θ) are indefinitely differentiable in [0, ) (0,π )

44 38 3. Transmutatin in the Irregular Case s that (a) is true. The prf fr (b) is similar t that f (b) f prpsitin,. Fr f, x andθfixed, M n p f (x,θ) and N n p f (x,θ) are mermrphic functins with ples at,,... s that p B p f (x) is 49 mermrphic Rep> n with ples at (3). Further since p B p is cntinuus in the regin given by () and (3), (c) fllws. Crllary. If m, n are tw distinct psitive integers, the functins B m p and B m p cincide in the intersectin f their dmains f definitins. In fact the tw mermrphic functins cincide with B p in the half plane Rep> which is cmmn t their dmains f definitins and hence they cincide everywhere in the intersectin f their dmains f definitin. We have prved the fllwing Therem. The functin p B p defined in Rep> by (I) with values in L (E, E ) can be cntinued analytically int a functin mermrphic in the whle plane with ples at,, 3,... Remark. B = identity. Remark. Fr Rep>, we have D r B p f (x)=β p t r ( t ) p f r (tx)dt and [D r B p f (x)] x=0 =β p,r f r (0) (4) ( ) r+ Γ(p+)Γ whereβ p,r = ( r ) is a mermrphic functin f p with πγ + p+ 50 ples at,,... By analytic cntinuatin, the equatin (I) hlds fr p,,... Then if f E (resp. D ), B p f E (resp. D ) and p B p is mermrphic with values in L (E, E )(resp. L (D, D )).

45 3. Cntinuatin f B p 39 Therem. Fr any p,,... we have B p D f= D B p f, f E. By Prpsitin,, the given equatin hlds fr Rep> and hence fr all p,,... by analytic cntinuatin. Therem. Fr any cmplex p,,... the peratrs B p and B p are ismrphisms f E (resp D ) nt itself which are inverses f each ther. We have seen that D B p B p = I(identity) fr < Rep<. Als fr p in the same regin D B p = B p s that B p B p = I fr < Rep< and the same hlds fr all p,, 3, by analytic cntinuatin. Similarly B p B p = I fr p,,...

46

47 Chapter 4 Transmutatin in the Irregular Case [ Case f the general peratr : D + ( p+3 x ) ] + M(x)D+N(x) Let M and N be tw indefinitely differentiable functins, M dd (i.e., 5 M (n) (0)=0 fr every n 0) and N even (i.e., N n+ (0)=0 fr every n 0i.e., N E ). We cnsider the tw fllwing prblems. Prblem. T find u(x, y), indefinitely differentiable even in x and y which is a slutin f ( ) u p+ u x + + M(x) x x + u N(x)u y= 0 () with u(0, y)=g(y), g E () Prblem. T find v(x, y) indefinitely differentiable, even in x and y, slutin f the same equatin () with 4

48 4 4. Transmutatin in the Irregular Case We intend t prve the fllwing v(x, 0)= f (x), f E (3) Therem. Fr p,,..., each f the tw prblems stated abve admits a unique slutin, which depends cntinuusly n g (r f ). 5 Once we prve this therem, we can define Operatrs X p andx p. Definitin. Fr g E, and p,,... X p [g(x)]=u(x, 0), where u is the slutin f the Prblem. Definitin. Fr f E, and p,,... X p [ f (y)]=v(0, y) where v is the slutin f the secnd prblem. Assuming Therem, definitins () and () give immediately Therem. Fr p,,..., X p andx p are in L(E, E ). We nw prve Therem 3. Fr p,,..., we have D X p = X p Λ p (4) X p D =Λ p X p (5) where Λ p = D + p+ D+ M(x)D+ N(x) x (6) Applying the peratr t the tw sides f y ( ) v p+ v x + + M(x) x x + v N(x)v y = 0

49 and setting v y= V, we have (Λ p ) x V V y = 0, and V(x, 0)= v y (x, 0)=(Λ p) x (v(x, 0))=Λ p [ f (x)]; and by the definitin f X p, V(0, y)=x p Λ p [ f (x)]. But we have als V y (0, y)=d y[x p f (y)]. Thus (4) is prved. Similarly (5) can be prved. But (5) fllws frm (4) and the Therem 4. Fr p,,..., the peratrs X p,x p are ismrphisms f E nt itself and X p X p =X p X p = the identity. Let g E and u be the slutin f Prblem. Then u(x, 0) = X p [g(x)]. In rder t find X p X p [g(x)] it is sufficient t determine the slutin v(x, y) f v y+ (p++ M(x)) v x x + v N(x)v y = 0 with v(x, 0)=X p [g(x)]=u(x, 0). Then, in view f Therem, u(x, y)=v(x, y). Hence X p X p [g(y)]= u(0, y)=g(y) similarlyx p X p = the identity. 54 The Therem. We nw prceed t slve Prblem. Let (B p ) x [u(x, y)]=w(x, y) where (B p ) x dentes the peratr B p relative t the variable x. Applying (B p ) x t the tw sides f equatin (), i. e., f (L p ) x [u(x, y)]+ M(x) u x + N(x)u u y = 0 and using D B p = B p L p, we btain w x w y + (B p) x [M(x) u x + N(x)u]=0 and we have als w(0, y)=g(y)=u(0, y) since B p f (0)= f (0). We intend t put this equatin which is equivalent t equatin () in a prper frm, We first prve

50 44 4. Transmutatin in the Irregular Case Prpsitin. B p AB p f (x)=a(x) f (x)+ x x T p[a](x, y) f (y)dy where A(x) and f (x) E and T p is a map defined in E with values in the space f even indefinitely differentiable functins f the variables x and y. Fr <Rep<, x B p f (x)= b p (x y ) p y p+ f (y)dy π where b p = Γ(p+)Γ( p+, B p f (0)=0, D B p = B p ) 55 and fr Rep>, x B p g(x)=β p x p (x y ) p f (y)dy where β p = Γ(p+). πγ(p+ ) Hence fr < Rep<, we cmpute B p AB p f (x)= b p β p x (x z ) p z p+ A(z) z ] [z p (z y ) p f (y)dy dz x [ x = b p β p f (y) (x z ) p (z y ) p A(z)zdz ]dy. Setting z = x sin θ+y cs θ, we get B p AB p f (x)= b p β p φ p (x, y) f (y)dy where, φ p (x, y)= π y [ ] sin p θ cs p θa x sin θ+y cs θ dθ

51 . 45 which cnverges fr < Rep<. It fllws that D B p AB p f (x)= b p β p φ p (x, x) f (x)+ b p β p x 0 x φ p(x, y) f (y)dy But we have 56 φ p (x, x)=a(x) Γ(p+ ( )Γ p+ ) s that b p β p φ p (x, x)=a(x). Hence B p AB p f (x)=a(x) f (x)+γ p x 0 γ p = Γ(p+ )Γ( p+ ). x φ p(x, y) f (y)dy where Nw x φ p(x, y)= π 0 sin p+ θ cs p θ A (z) xdθ s that if we give z Definitin 3. Fr f E π T p [ f ](x, y)=γ p sin p+ θ cs p θ f (z)d 0 where f (z)= f (z), z= [ x sin θ+y cs θ ],γ p = b p β p, then we z can write B p AB f (x)=a(x) f (x)+ x (7) is valid fr Rep < and f E. x T p [A](x, y) f (y)dy (7) Remark. We haveγ p = 0 fr p=±,±3,... s that T = 0. Again 57 fr p =, B p = B p = the identity. Hence (7) is valid fr p =. Nw w(x, y) = (B p) x u(x, y) s that u(x, y) = (B p ) x w(x, y) and

52 46 4. Transmutatin in the Irregular Case (B p ) x M(x) u x = (B p) x M(x)D x (B) x w(x, y). We shall dente byφ(x), the functin w(x, y) cnsidered as a functin f x; let (B p ) x w(x, y)=b p φ= ψ. Then B p ψ=φ, B p MDB p φ=b p MDψ= B p (D(Mψ))B p (M ψ)= B p [D(xM ψ)] B p (M ψ) where M (x)= M(x) E and B p [MDψ]= x B p [xdm ψ]+ B p [(M M )ψ]. But fr <Rep<, B p [xdf]=b p x t p+ ( t ) p 3/ F (tx)dt= xd(b p [F]) Hence B p (MDψ)= xdb p M ψ+ B p [(M M)ψ] Nw by (7), { B p [MDψ]= xd M φ+ x x + (M M )φ+ x = x T p [M ](x, x)φ(x)+ x } T p [M ](x, y)φ(y)dy x x T p [M M](x, y)φ(y)dy. S p [M](x, y)φ(y)dy+ M(x)φ (x) 58 where S p [M](x, y)=t p [M M](x, y)+ x x T p[m ](x, y) As T p [M ](x, x)=γ p M (x) x B p MDB p [φ]= M(x)φ (x)+ + x x π sin p+ θ cs p θdθ ( = p+ ) M (x), x ( p+ S p [M](x, y)φ(y)dy ) [M (x) M (x)]φ(x) Als B p [N(x)ψ]=B p NB p φ=n(x)φ(x)+ x x T p[n](x, y)φ(y)dy

53 3. Cntinuatin f T p 47 Hence the differential equatin in w has the frm ( w w x y+ M(x) w x + P+ ) [M (x) M (x)]w x { + N(x)w+ x S p [M](x,ξ)+T p [N](x,ξ) } w(ξ, y)dξ= 0 ( Let Q p (x)=n(x)+ p+ )( M (x) M(x) ) ξ and R p (x,ξ)= x S p [M](x,ξ)+T p [N](x,ξ). We see that Prblem is equivalent t the determinatin f the indefinitely differentiable slutin even in x and y the integr differentiable equatin w w x y M(x) w x + Q p(x)w+ x x R p (x,ξ, )w(ξ, y)dξ= 0 with the cnditin w(0, y) = g(y). It is easy and classical t transfrm this prblem t a purely integral 59 equatin f Valterra type and the slutin is btained by the methd f successive apprximatin. It can be verified that all the cnditins f w are verified. A prcess cmpletely analgus gives the slutin f Prblem. 3 Cntinuatin f T p Fr f E fixed we define by inductin the functins g n (x, y,θ) as fllws g (x, y,θ)= f (z)= f (z) z In general where z= sin p θ csθg (x, y,θ)= θ x sin θ+y cs θ [ sin p+ θg (x, y,θ) ] sin p n+ θ csθ g n (x, y,θ)= θ[ sin p n+3 θg n (x, y,θ)

54 48 4. Transmutatin in the Irregular Case i.e., where g (x, y,θ)=(p+) f (z)+(x y ) sin θ f (z) f (z)= f (z) E. z In fact we can prve immediately by inductin n n the fllwing 60 Lemma. g n (x, y,θ) is a linear cmbinatin f f, f,..., f n+ with cefficients which are plynmials in x, y and sin θ where f n (z) = z f n (z) E. Crllary. The functins g n (x, y,θ) are indefinitely differentiable in x, y, θ and even in x and y fr (x, y) R andθ [0, π ]. Definitin. T p (n) [ f ](x, y)= ( ) n π γ p (p )(p 3) (p n+) 0 cs p+n θ sin p n+ θg n (x, y,θ)dθ The integral cnverges fr n 3 < Rep < n+ and T p (n) L (E, E) where E is the space f indefinitely differentiable functins f tw variables x, y which are als even in x and y with the usual tplgy f unifrm cnvergence n every cmpact subset f R f functins tgether with their derivatives. It can be verified that p T p (n) is an analytic functin fr p in the strip in view f the fact that γ p p is an entire functin. (p ) (p n+) Lemma. In each strip n 3 = T p (n ) T (n) p T p (n ) [ f ](x, y)= ( ) n γ p (p )(p 3) (p n+3) π 0 < Rep < n, we have cs p+n θ sin p n+4 θg n (x, y,θ)dθ

55 3. Cntinuatin f T p 49 6 The integral can be written as π = 0 cs p+n θ sinθsin p n+3 θg n (x, y,θ)dθ (p n+) π 0 cs p+n θ θ { sin +p n+3 θg n (x, y,θ) } dθ (integrating by parts, the integrated part being zer fr n 3 < Rep< n ). The lemma is nw evident by the recurrence frmula fr g n. Thus we have Prpsitin. The functin p T p defined fr 3 < Rep< admits an analytic cntinuatin in the half plane Rep> 3, with values in L (ε, E). The explicit definitin f T p is given by the frmula fr T (n) P fr suitable n. Analytic cntinuatin f T p in the half plane Rep< 3 is btained by exactly similar prcess. We intrduce functins and by inductin, h (x, y,θ)= f (z), z=(x sin θ+y cs θ), cs p θ sinθh (x, y,θ)= [ ] cs p θh (x, y,θ), θ cs p n θ sinθh n (x, y,θ)= θ [ ] cs p n+ θh n (x, y,θ), It is easy t see by inductin n n that h n (x, y,θ) is a linear cm- 6 binatin f f, f,..., f n+, with cefficients which are plynmials in x, y, cs θ s that the functins h n (x, y,θ) are indefinitely differentiable in x, y,θ and are even in x and y. If we set (n)t p f (x, y)= ( ) n γ p (p+3)(p+5) (p+n+)

56 50 4. Transmutatin in the Irregular Case π 0 sin p+n+ θ cs p n θh n (x, y,θ)dθ fr n, then (n) T p L (ε, E) fr n 3 < Rep< n+ and p (n) T p is analytic in this strip with values in L (ε, E) and that fr n < Rep< n+, (n ) T p = (n) T p. Finally we have the Therem. The functin p T p initially defined fr 3 < Rep<, admits an analytic cntinuatin int an entire functin with values in L (ε, E) and we have the explicit frmula fr this cntinuatin. In particular, we have, by analytic cntinuatin, T p [ f ](x, x)=(p+ ) f (x) fr any p. Crllary. Fr p,,..., the frmula is valid. B p AB p f (x)=a(x) f (x)+ x x T p [A](x, y) f (y)dy 63 The first member B p AB p f (x) is defined and is analytic except fr p=,,... The secnd member is an entire functin f p, and the tw members are equal fr Rep < s that the crllary fllws. Remark. If A(x) is a cnstant, we have T p [A](x, y)=0 and the frmula f the crllary is equivalent t B p B p = the identity.

57 Part II Tpics In Mean-Peridic Functins 5

58

59 Chapter Expansin f a Mean-peridic Functin in Series Intrductin. There is n cnnectin between Part I and Part II f this lecture 64 curse. But bth these parts are essential fr the tw-radius therem which will be prved in the last part. The thery f Mean peridic functins was funded in 935 (refer t Functins Myenne-peridiques by J. Delsarte in Jurnal de Mathematique pures et appliques, Vl. 4, 935). A mean peridic functin was then defined as the slutin f f the integral equatin b a K(ξ) f (x+ξ)dξ= 0 () where K is density given by a cntinuus functin K(x), given n a bunded interval [a, b]. It was bvius that the study f an rdinary differential equatin with cnstant cefficients r f peridic functins with perid b a was a prblem exactly f the same type as the study f equatin (). It was prved in the paper mentined abve that if K satisfied certain cnditins and if f were a slutin f () i.e. f were mean peridic with respect t K then f is develpable in a series f expnential functins which cnverges twards f unifrmly n each interval n 65 53

60 54. Expansin f a Mean-peridic Functin in Series which f is cntinuus. Equatin () in mdern ntatin is essentially a cnvlutin equatin K f= 0 if we replace f by f ( f (x)= f ( x)). L.Schwartz in 947 applied the thery f distributins t the thery f Mean peridic functins. He calls a mean peridic functin a cntinuus functin f which is a slutin f µ f= 0 whereµis measure with cmpact supprt. Mre generally a cntinuus functin f is mean peridic with respect t the distributin T with cmpact supprt if T f = 0. He has als given a new definitin f mean peridic functin which is imprtant frm the tplgical pint f view and the thery develped in his paper (Annals f Mathematics, 947) is cmplete in the case f ne variable. J.P. Kahane has given a special and delicate develpment f the thery which gives cnnectin between Mean peridicity and almst peridicity. The extensin t the case f several variables is certainly difficult and the first knwn result in R n is due t B. Malgrange. Fr x R,λ C we have, T (e λx )=e λx M(λ) 66 where M(λ)= T, e λξ is the Furier - Laplace transfrm f the distributin T. When T is a density K r a measure, µ, M(λ)= b a K(ξ)e λξ dξ r M(λ)= b a e λξ dµ(ξ) respectively and fr peridic functin f perid a, we have M(λ) = e aλ. In the case f the differential peratr with cnstant cefficients, M(λ) is the characteristic plynmial f the peratr. In all these cases M(λ) is an entire functin fλ, f expnential type and behaves like a plynmial if Reλis bunded and the (simple) zers f this functin give the expnential functins (r the expnential plynmials in

61 . Determinatin f the cefficients in the frmal series 55 the case f multiple zers) which are mean peridic with respect t the distributin T. It is clear that any linear cmbinatin f mean peridic expnentials is als mean peridic and the first prblem is t determine, fr any mean - peridic functin f, the cefficients f the mean peridic expnentials in the expansin f the functin. Determinatin f the cefficients in the frmal series Let T be a distributin with cmpact supprt and M(λ) its Furier- Laplace transfrm. The set f zers f M(λ) will be called the spectrum f M(λ) and will be dented by (σ). Fr the sake f simplicity we suppse that these zers are simple. 67 Let F be a functin sufficiently regular fr the validity f the scalar prduct T, x e λ(x ξ) F(ξ)dξ If T is a measure with cmpact supprt, F need be nly an integrable functin and if T is a distributin with cmpact supprt and rder m, F can be any (m ) times cntinuusly differentiable functin. Let F(x)=e αx whereα C is fixed s that T, x x e λ(x ξ) e αξ dξ= eλx e αx λ α e λ(x ξ) e αξ M(λ) M(α) dξ = =τ α (λ) τ α (λ) λ α is an integral functin f expnential type and fr α, β (σ), τ α (β)=0 if β α = M (α) if α=β. Let t α (λ)= M (α) τ α(λ) s that t α (β)=δ β = ifα=β α = 0 ifα β Then t α (λ) is an entire functin f expnential type and therefre by the therem f Paley-Wiener, there exists a distributin T α with cmpact 68