The Bilateral Laplace Transform of the Positive Even Functions and a Proof of Riemann Hypothesis

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1 The Bilateral Laplace Trasform of the Positive Eve Fuctios ad a Proof of Riema Hypothesis Seog Wo Cha Ph.D. swcha@dgu.edu Abstract We show that some iterestig properties of the bilateral Laplace trasform of eve ad positive fuctios both o the lie z 0 = x + iy 0 ad o a circle. We also show the Riema hypothesis is true usig these properties. We do ot prove well-kow theorems ad ecourage readers to refer to the literatures. 1. Itroductio We begi with the defiitio of the bilateral Laplace trasform. 1 The defiitio of Laplace trasform of a real fuctio f(t) is as follows: ad the iverse trasform: F(z) = f(t) e zt dt (1) where z = x + iy. f(t) = 1 x+i iπ F(z) ezt dz x i () If f(t) is eve, the F(z) = f(t) e zt dt = f(t) e zt dt = f(t) cosh (zt)dt (3) ad sice F( z) = F(z), F(iy) is real-valued for all y. If f(t) is eve ad also positive for all t, its Laplace trasform F(z) is trascedetal.. Power Series expasio It is ot always easy to fid the Laplace trasform of a fuctio i the closed form, but we ca get the power series usig the defiitio of the Laplace trasform as follows: F(z) = a z =0 = a 0 + a 1 z + a z + (4) To fid the coefficiets of F(z), we examie the derivatives of (1). 1 Sice we are oly dealig with the bilateral Laplace trasform, The term bilateral will be omitted.

2 a = 1! t f(t)dt (5) Moreover, if the fuctio f(t) > 0 for all t ad eve, all the odd terms are vaished ad the power series will be like that: F(z) = a z = a 0 + a z + a 4 z 4 + =0 (6) where a = 1 ()! t f(t)dt ad a > 0 for all. Sice f(t) is eve ad positive, F(z) is also eve ad F( z ) = F(z ) = F (z). Hece we have F(z) = F( z) = F(z ) = F( z ). 3. The covex fuctios For a real fuctio f(x) ad x 1, x R ad λ [0,1], the f(x) is covex if ad oly if f[λx 1 + (1 λ)x ] λf(x 1 ) + (1 λ)f(x ). ad similarly, f(x) is strictly covex if ad oly if f[λx 1 + (1 λ)x ] < λf(x 1 ) + (1 λ)f(x ). The fuctio f(x) is a midpoit covex if f ( x 1 + x ) f(x 1) + f(x ) Further, a cotiuous fuctio f(x) is multiplicatively covex if ad oly if f( x 1 x ) f(x 1 )f(x ) A multiplicatively covex fuctio is which is icreasig covex. This iequality ca be obtaied from the defiitio of the midpoit covex. Hardy-Littlewood Theorem Every polyomial f(x) = c k x k with o-egative coefficiets is multiplicatively covex o (0, ). Moreover f(x) = c k x k for c k 0 is strictly multiplicatively covex which is also icreasig ad strictly covex. Let F(z) be the Laplace trasform of a eve ad positive fuctio f(t), the by Hardy- Littlewood, F(z) is icreasig ad strictly covex o the x-axis of the iterval (0, ) if F(z) is etire. If F(z) is ot etire, F(x) is icreasig ad strictly covex i the radius of covergece. Sice F( x) = F(x), F(x) is symmetric at x = 0, ad therefore F(x) has a uique miimum at x = 0.

3 F(z) is complex-valued co-semipositive defiite if 4. The co-positive defiiteess c c k f(z + z ) k 0 k=1 (7) ad similarly, f(z) is complex-valued co-positive defiite if for all c C ad z = x + iy. c c k f(z + z ) k > 0 k=1 (8) Clearly, if a complex fuctio f(z) is complex-valued co-(semi)positive defiite, the the realvalued fuctio f(x) is also co-(semi)positive defiite. Lemma 1 A real fuctio f(t) > 0 for all t, the its Laplace trasform is complex-valued co-positive defiite. Proof From the defiitio F(z) = f(t) e zt dt c c k F(z + z ) k = c c k k=1 = f(t) c c k k=1 = f(t) c e zt e (z +z k dt > 0 k=1 )t dt = f(t) e (z +z k )t dt f(t) c e zt c k k=1 e z t k dt Lemma If f 1 (z) ad f (z) are co-positive defiite the they are also co-positive defiite: 1. f 1 (z) ad f (z). f 1 ( z) ad f ( z) 3. f 1 (z) ad f (z) 4. f 1 (z) f (z) 5. a 1 f 1 (z) + a f (z) for a 1, a > 0 These properties ca be easily proved usig the defiitio of the Laplace trasform (1).

4 Lemma 3 If F(z) is co-positive defiite, the F(0) is real ad F(0) > 0. Proof From (8), We let =1, the c 1 c 1 F(x 1 + x 1 ) > 0. Lettig x = x 1 + x 1, we have c 1 F(x) > 0. Hece F(x) is real ad F(x) > 0 for all real x ad therefore F(0) is real ad F(0) > 0. Lemma 4 If F(z) is co-positive defiite, the F () (z) is also co-positive defiite for o-egative iteger. Proof From the defiitio of the Laplace trasform (1), takig derivative times with respect to z, we have t f(t) e zt dt Sice f(t) > 0 for all t ad eve, ad therefore t f(t) > 0 ad t f(t) is eve ad hece copositive defiite. Lemma 5 If y is fixed, say y = y 0 ad z = x + iy 0, the ay complex-valued co-positive defiite fuctio F(z) is co-positive defiite for x. Proof c c k F(z + z ) k = k=1 This is clear sice = f(t) c c k k=1 e (z +z )t k dt = f(t) c c k e (x +iy 0 +x k iy 0 )t dt = k=1 = f(t) c e xt dt > 0 f(t) c c k c c k F(z + z ) k = c c k F(x + x k ) > 0 k=1 k=1 k=1 that is, y 0 is cacelled out ad we have a real-valued co-positive defiite fuctio. e (x +x k )t dt We ote that this also is valid for F(z). If y is fixed, F(x + iy 0 ) is real-valued co-positive defiite for x.

5 5. Expasio of F(z) From (6), we have F(z) = f(t) e zt dt = a z =0 (9) ad F(z) = F(z) F(z) = F(z) F(z ) = [ =0 a z ] [ =0 a z ]. The power series expasio of [ =0 a z ] [ a z ] is =0 F(z) = C 0 + C (z + z ) where C 0 = a k z 4k, C = a k a k+ z 4k. Sice a k 0, C 0 > 0 ad C > 0 for all. Sice there are oly eve terms of x, we let (10) To determie A 0 ad A m, C 0 = a k z 4k F(z) = A 0 + A m x m = a k (x + y ) k = a k ( k j ) y4k j x j j=0 k=j (11) (1) ad sice we have ad C (z + z ) = By the Leibiz rule, C = (z + z ) = p=0 ( 1) p ( p ) y p x p (13) ( 1) y C for p = 0 ( 1) ( { p ) y p x p C for p 1 p=1 =p a k a k+ z 4k = a k a k+ (x + y ) k m x m [(x + y ) k x m ] x=0 = m x m [(x + y ) k ] x=0 = [ [(m j)(k j)] ] y (k m) (16) where m 1 ad k m. Takig derivatives (m) times to Eq. (11) ad Eq. (1), (14) ad lettig x = 0, we have m 1 j=0 (14) (15) where C m 1 3 A m = C m + C m + C m (17) 1 = a k ( k m ) y4k m

6 ad C m = ( 1) a k a k+ [ (m j)(k j) ] y y (k m) C m k=m 3 = ( 1) m ( m ) y m =m k=m m 1 j=0 m 1 a k a k+ [ (m j)(k j) ] y (k m) j=0 A 0 = a k y 4k + ( 1) a k a k+ y +4k (18) If y is fixed, A 0 ad A m are costats ad F(z) are a fuctio of oly x, that is, we have a fuctio F(z) which is lyig o the horizotal lie of z = x + iy 0 where y 0 is a fixed value. By Lemma 3 ad lettig x = 0, we ote A 0 > 0. Likewise, from Eq. (11), takig derivative times with respect to x ad lettig x = 0, we ote A m > 0 for all m 1 by Lemma 3 ad 4. Sice all the coefficiets of Eq. (11) positive, by Hardy-Littlewood, F(z) is a strictly multiplicatively covex fuctio o the horizotal lie z = x + y 0 ad i the iterval of x (0, ). Moreover, sice F(z) is symmetric at x = 0, F(z) has a uique miimum at x = 0. Further, sice y 0 is arbitrary, we coclude the theorem: Theorem 1. Let F(z) be Laplace trasform of a positive ad eve fuctio, the if y is fixed, F(z) is strictly multiplicatively covex for 0 x < o the lie z = x + iy 0. If F(z) is ot etire, F(z) is multiplicatively covex i the radius of covergece.. Sice F(z) is symmetric by iy-axis, F(z) has a uique miimum at x = Sice F(z) has a uique miimum at x = 0, all zeros of F(z) locate oly at iy-axis. It caot have zeros o other places sice F(iy 0 ) 0 ad F(x + iy 0 ) > F(iy 0 ) for all x 0. Clearly, this is also valid for F(z). 4. Sice the zeros of F(z) locate at iy-axis, the zeros of F(z) locate oly at iy-axis, if F(z) has ay zeros. If F(z) had zeros o other places tha o iy-axis, F(z) should have zeros o the same places, but F(z) has zeros oly o iy-axis as show. The theorem is valid oly for the Laplace trasform of fuctios that are eve ad positive for all t. It is ot valid for o-eve fuctios. However, the Laplace trasform of the positive fuctios are co-positive defiite ad the power series ca be foud which is similar to (11) but cotais odd powers. We do ot have ay iformatio about the coefficiets of odd powers but the coefficiets of eve powers should be positive if the power series coverges. There are some fuctios whose Laplace trasform does ot have ay zero. For example, πe z is the Laplace trasform of e t /4, but πe z has either poles or zeros.

7 6. The positive defiiteess By replacig z iz ad for some fuctio f(t) > 0, the Laplace trasform will be G(z) = f(t) e izt dt (19) It ca be show that G(z) is complex-valued positive defiite, that is c c k G(z z ) k k=1 > 0 (0) which is the geeralized Bocher s theorem. The Properties of positive defiiteess are similar to the co-positive defiiteess. For example, if G(z) is positive defiite, ( 1) G () (z) is also positive defiite ad G(0) > 0 ad therefore G(0) is real ad so o. If f(t) > 0 for all t ad is eve, we have: G(z) = f(t) e izt dt = f(t) cos (zt)dt 0 = f(t) cos (zt)dt (1) Usig the power series expasio ad properties of the positive defiite fuctios, it ca be show that G(z) from Eq. (1) has oly real zeros. I fact, we do ot eed to prove it. Because of mappig z iz, G(z) is a rotated fuctio by π/ of F(z) i Eq. (9), ad therefore, sice F(z) i Eq. (9) has zeros oly o the iy-axis, G(z) should have zeros oly o x-axis, that is, oly real zeros. 7. Behavior o a circle We ow cosider F(z) o a circle cetered at the origi with radius r. Lettig z = r e iθ, from Eq. (10) we have F(r e iθ ) = C 0 + C r cos (θ) () Sice F(x) is co-positive defiite as show, lettig x = r cos (θ) ad if r is fixed, F(x) is co-positive defiite for cos (θ), thus F(r e iθ ) is also co-positive defiite for cos (θ). This ca be easily proved usig the defiitio of the Laplace trasform Eq. (1). Cautio must be take. F(r e iθ ) is co-positive defiite for cos (θ) itself, ot for θ. If F(r e iθ ) is co-positive defiite for θ, it must be that [ F(r θ eiθ ) ] > 0. Takig derivative twice to Eq. () ad lettig θ = 0, θ=0 we have () C r. Sice C > 0 for all, () C r < 0, thus F(r e iθ ) is ot co-positive defiite for θ. I fact, F(r e iθ ) is positive defiite for θ because cos (θ) is positive defiite ad F(r e iθ ) is summig of cos (θ) with positive coefficiets, hece F(r e iθ ) is positive defiite for θ. Lettig α = cos (θ) ad for fixed r, we have

8 F(α) = C 0 + C r T (α) where T (α) deotes the Chebyshev polyomials of the first kid. (3) T (α) ca be expressed as the sum of α, which is T (α) = 1 / ( 1)k k k ( k ) (α) k (4) where / deotes the floor fuctio. Lettig, we have ad lettig m = k ad fially we have T (α) = ( 1) k k T (α) = ( 1) m=0 m + m F(α) = C 0 + C r ( 1) m=0 k ( ) (α) k k ( + m m ) (α)m m + m ( + m m ) m α m (5) (6) (7) From Eq. (7), if m = 0, the we get which is the costat part of F(α). With the results, we have F(α) = C 0 + which ca be writte C 0 + ( 1) C r (8) ( 1) C r + ( 1) m + m =m ( + m m ) C r m α m (39) where ad F(α) = B 0 + B α + = B 0 + B m = B 0 = C 0 + =m ( 1) m ( 1) C r B m α m + m ( + m m ) C r m (30) Sice F(α) is co-positive defiite, we have thus F(0) > 0 ad dm dα m [ F(α) ] α=0 > 0

9 Sice α = cos (θ), from Eq. (30) B 0 > 0 ad B m > 0 for m = 1,, ad F(α) = B 0 + B m cos m ( θ) (31) d dθ F(α) = si (θ) m B m cos m 1 ( θ) = si (θ) m B m cos m ( θ) (3) Sice m B m cos m ( θ) = B + m B m cos m ( θ) m= m B m cos m ( θ) caot be zero ad therefore m B m cos m ( θ) is strictly positive, thus si (θ) determies the sig of d dθ F(α). We ote that si(θ) = 0 whe θ = 0 ad θ = π/ as expected. Moreover, si(θ) < 0 i the iterval 0 < θ < π/, ad therefore F(α) is decreasig i the iterval. Sice F(α) 0, F(α) has a uique miimum at θ = π/ ad θ = π/ because F(α) has the same value at θ, θ, π θ ad θ π. Sice the radius r ca be chose arbitrarily, the zeros of F(α) locate oly θ = π/ ad = π/, that is, iy-axis which we have proved. I fact, it ca be show that if F(α) has a uique miimum at = π/, F(z) is strictly multiplicatively covex for x o the lie z = x + iy 0 where y 0 = r. 8. Proof of the Riema hypothesis The Riema zeta fuctio (s) is defied as follows: (s) = 1 s = s s + (33) where s = + iω. The fuctioal equatio of Riema xi-fuctio ξ(s) is where ξ(s) = ξ(1 s) (34) ξ(s) = 1 π s s(s 1)Γ(s/) (s) (35) which is symmetric at = 1 ad therefore ξ (1 + iω) is real-valued. Riema showed that ξ(s) = [ e πt] 1 s(s 1) (t s/ + t (1 s)/ )dt (36) t The Jacobi Theta fuctio ca be expaded usig the Poisso summatio formula ad the Melli trasform, ad fially we have

10 where ξ(s) = φ(t) e (s 1 )t dt φ(t) = e π e t (π 4 e 9 t 3π e 5 t ) (4) (38) It ca be show that φ(t) > 0 for all t ad a eve fuctio. From (37), the power series expasio is ξ(s) = h (s 1 ) =0 (39) Lettig z s 1, that is, shifted by 1 ad therefore the zeros ow locate o the stripe of 1 < x < 1 ad we have ad the power series expasio Φ(z) = φ(t) e zt dt = φ(t) e zt dt (5) Φ(z) = h z =0 (41) which is the Laplace trasform of the positive ad eve fuctio φ(t). We showed that all zeros of the Laplace trasform of a positive ad eve fuctios locate at iy-axis. Hece all the zeros of ξ(s) must locate at = 1 ad therefore (s) too, which meas the Riema hypothesis is true. Riema suggested a fuctio amed big xi-fuctio Ξ(z), which is defied Ξ(z) = φ(t) cos (zt) dt 0 = φ(t) cos (zt) dt (4) ad if Ξ(z) has oly real zeros, the Riema hypothesis is true. This fuctio is othig but a positive defiite fuctio i Eq. (1) ad we showed that these kid of fuctios ca have oly real zeros ad therefore the Riema hypothesis is true. Ackowledge The author would like to express author s gratitude to Mr. Gee-Su Yu who was the author s metor. Refereces [1] Weisstei E. W. CRC Cocise Ecyclopedia of Mathematics CRC Press 008 [] Keiper J. B. Power Series Expasios of Riema s ξ Fuctios Mathematics of Computatio, Volume 58, April 199 [3] Widder D. V. The Laplace Trasform Priceto Uiversity Press 1946 [4] iculescu C. P., Persso L-E Covex Fuctios ad Their Applicatios Spriger 004

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