On the properties of the two-sie Lple trnsform n the Riemnn hypothesis Seong Won Ch, Ph.D. swh@gu.eu Astrt We will show interesting properties of two-sie Lple trnsform, minly of positive even funtions. Further, we will lso prove tht the Lguerre inequlities n generlize Lguerre inequlities re true n finlly, the Riemnn hypothesis is true. Remrk:. Introution We re using n z n, F (z), F (n) (z) n D z n s the ifferentil opertors n hoosing the most suitle nottion for the se. We will egin with the efinition of the two-sie Lple trnsform. The Lple trnsform of rel funtion f(t) is efine s: n the inverse trnsform is: where z = x + iy for x n y rel. F(z) = f(t) e zt t f(t) = x+i iπ F(z) e zt z x i () () If f(t) is even, then F( z) = F(z) n we n write F(z) = f(t) e zt t = f(t) e zt t = f(t) osh(zt) t From (3), we hve the power series expnsion of F(z), nmely = f(t) osh(zt) t (3) where n = (n)! F(z) = n z n = + z + 4 z 4 + (4) n= tn f(t)t. ote tht if f(t) is non-negtive, n is positive for ll n. F(z) n e seprte into the rel n imginry prt, nmely, F(z) = u(x, y) + iv(x, y). Sine z = x + iy, if f(t) is n even funtion, we hve Sine we re only eling with the two-sie Lple trnsform, the term two-sie will e omitte fterwr.
Therefore, F(z) = f(t) osh(xt) os(yt) t + i f(t) sinh(xt) sin(yt) t u(x, y) = f(t) osh(xt) os(yt) t (5) (6) v(x, y) = f(t) sinh(xt) sin(yt) t otie tht u(x, y) is n even funtion n v(x, y) is n o funtion for oth x n y, hene v(, y) = v(x, ) =. Therefore, we hve F(z) on the rel xis, enote F(x), n on the imginry xis, enote F(iy) s follows F(x) = f(t) osh(xt) t = u(x, ) (8) The funtion F(x) is inresing log-onvex n hs unique minimum t x =. F(iy) = f(t) os(yt) t = u(, y) Thus, F(iy) is rel if y is rel n ll erivtives of F(iy) re rel if f(t) is non-negtive n even n sine u(x, ) n u(, y) re even, we hve x u(x, ) x= = n Sine F(z) = u + v n F(z) = u + v, we hve (7) (9) u(, y) =. y y= F(x + iy) = F(x iy) = F( x + iy) = F( x iy) () F(x + iy) = F(x iy) = F( x + iy) = F( x iy) () The equlities () n () re vli if f(t) is even. Another efinition is y repling z to iz in (), hene we hve f(t) e izt t whih n e often seen in the literture. It n e otine n if f(t) is even, from (), we hve F(iz) = f(t) e izt t F(iz) = f(t) os(zt) t = f(t) os(zt) t ote tht F(iz) is only the rotte funtion of F(z) y π/, thus we get F(iz) from F(z) y swpping x n y if f(t) is even. This is lso vli for F(iz). () (3). The positive efiniteness n o-positive efiniteness ow, we onsier the Lple trnsform of f(t) whih is non-negtive s well s even, mening f( t) = f(t) n f(t) for ll t.
Definition : Rel-vlue positive efiniteness n o-positive efiniteness For rel θ, funtion φ(θ) is positive semi-efinite if n only if n k φ(θ n θ k ) for ny whih is non-zero positive integer, ny omplex vlue n n ny rel vlue θ n. The str * enotes the omplex onjugte. Similrly, funtion φ(θ) is o-positive semi-efinite if n only if n k φ(θ n + θ k ) n= k n= k Definition : Complex-vlue positive efiniteness n o-positive efiniteness A funtion φ(z) is omplex-vlue positive semi-efinite if n only if n k φ(z n z k ) for ny whih is non-zero positive integer, ny omplex vlue n n ny omplex vlue z n. Similrly, funtion φ(z) is o-positive semi-efinite if n only if n k φ(z n + z k ) For the ske of onveniene, we efine the ifferentil opertor D θ, thus D θ θ n φ(θ) = n n φ(θ). The positive semi-efinite n o-positive semi-efinite funtions hve mny properties. Some of them re: For the positive-efinite funtions: i. If φ(θ) is positive semi-efinite, then ( ) n D θ n φ(θ) is lso positive semi-efinite. ii. φ() n φ(θ) φ(). For the o-positive-efinite funtions: i. If φ(θ) is o-positive semi-efinite, then D θ n φ(θ) is lso o-positive semi-efinite. ii. φ(θ) n hene φ(). From the property of φ(θ), the rel-vlue o-positive semi-efinite funtions re rel when θ rel. The properties of omplex-vlue (o-)positive semi-efinite funtions re similr to of the relvlue ones. Oserving eq. (), F(z) is lerly omplex-vlue o-positive semi-efinite if f(t), sine k n k F(z n + z k ) = f(t) n k e (z n+z k)t t = f(t) n e z nt t. n= k n= Similrly, from (), it n e shown tht F(iz) is omplex-vlue positive semi-efinite when f(t). n= n= k k n= θ Theorem : Assuming f(t) is non-negtive, not neessrily even, F(z) is entire if F(z) hs no pole on the rel xis. Proof From (), we hve F(x + iy) = f(t) e xt e iyt t, whih is the Fourier trnsform of f(t) e xt. Sine f(t) e xt when f(t), For fixe x, F(x + iy) is positive semi-efinite for y y the Bohner s theorem. Hene F(x) F(x + iy) n if F(x) oes not hve ny pole, F(x + iy) oes not hve ny pole. Therefore F(z) is entire.
Theorem : Assuming f(t) is non-negtive, not neessrily even, F(iz) is entire if F(iz) hs no pole on the imginry xis. Proof Sine F(iz) is the rotte funtion of F(z) y π/, the proof is strightforwr from theorem. We showe tht F(x + iy) is positive semi-efinite for y. Aout x, however, F(x + iy) is generlly neither positive semi-efinite nor o-positive semi-efinite. Sine F(iy) n e negtive, F(x + iy) nnot e positive semi-efinite for x, n F(x + iy) is generlly omplex-vlue, hene F(x + iy) nnot e o-positive semi-efinite for x s well. But how re F(x + iy) n F(x + iy)? Sine they re non-negtive, they oul e positive semi-efinite or o-positive semi-efinite. Firstly, oserving of F(x) is neee, sine if F(x + iy) is (o-)positive semi-efinite for x, it must e (o-)positive semi-efinite t ny fixe y. Sine F(x) is speil se of F(x + iy) when y =, if F(x) is neither positive semi-efinite nor o-positive semi-efinite, F(x + iy) n F(x + iy) nnot e (o-)positive semi-efinite. From (), we hve F(x) = f(t) e xt t. This is not positive semi-efinite, ut it is o-positive semi-efinite if f(t). Therefore, F(x + iy) n F(x + iy) my e o-positive semi-efinite for x, if f(t) is non-negtive. Theorem 3: If f(t) is non-negtive n even, F(z) = F(re iθ ) on the irle entere t the origin hs mxim t x-xis, tht is, θ = n π. Moreover, F(iz) = F(ire iθ ) hs mxim t iy-xis. Proof Sine F(x + iy) is positive semi-efinite for y, F(x + iy) F(x) on the vertil line n hene, F(re iθ ) F[r os (θ)] n sine f(t) is non-negtive n even, F(x) hs unique minimum t x = n is inresing when x >, therefore F(re iθ ) hs mximum t θ =. Similrly, if x <, the mximum of F(re iθ ) is θ = π. F(iz) is positive semi-efinite for x on the horizontl line, n in the sme wy, we n prove tht F(ire iθ ) hs mxim t iy-xis. 3. The minim of F(z) If f(t) is even, from (8) we hve F(z) on the rel xis Let e hve f(t)t F(x) = f(t) osh (xt)t n efine G(x) = F(x)/ to normlize F(x), i.e. F() =.Therefore, we
where = f(t)t. G(x) = f(t) osh(xt) t Further, if f(t) is lso non-negtive, y the men-vlue theorem of integrtion, we hve (4) G(x) = osh[p(x)] f(t)t = osh[p(x)] (5) where P(x) is funtion epening on x n f(t) whih is rosh[g(x)]. By letting x z, we hve where P(z) = rosh[g(z)]. G(z) = osh[p(z)] (6) Sine osh (z) hs the glol minimum on the imginl xis of the horizontl line s well s on irle entere t the origin, osh [P(z)] hs, t lest, lol minim when P(z) is purely imginry. Sine P(z) = rosh[g(z)] = ln [G(z) + G (z) ], P(z) is purely imginry if G(z) + G(z) =. The Tylor series of ln [G(z) + G (z) ] is ln [G(z) + G (z) ] = i [ π (n + ) 4 n n= (n)! (n!) Gn+ (z)] (7) From (7), to e P(z) purely imginry, G(z) must e rel. Clerly, if P(z) is purely imginry, osh [P(z)] is rel, n therefore G(z) is rel. Sine F(z) = u(x, y) + iv(x, y), v(x, y) must e zero to e P(z) purely imginry. From the equtions G(z) = F(z)/ = u(x, y) + i v(x, y) n v(x, y) =, we hve u(x, y) G(z) + G (z) = + u (x, y) = (8) n fin out u(x, y) oring to (8). We will onsier three ses, i.e. u(x, y) <, u(x, y) > n u(x, y) <. Clerly, u(x, y)/ + u (x, y)/ > when u(x, y) < n u(x, y) >. In the se of u(x, y) <, sine u (x, y)/ <, we hve u(x, y)/ + u (x, y)/ = u(x, y)/ + i u (x, y)/ =. Hene the neessry onition to e F(z) minimum is v(x, y) = n u(x, y) <. However, they re only the neessry onition, not suffiient sine P(z) is series of z n therefore there n exist some points where v(x, y) = n u(x, y) <. Thus, we nee to nother onition, tht is, u(x, y) =. To sum up, F(z) is minimum (glol or lol) if n only if x v(x, y) =, u(x, y) < n x u(x, y) =. If oth u(x, y) n v(x, y) re zero, F(z) is not ifferentile ut sine F(z) is zero, it is glol minimum nywy.
ow, we will onsier F(z) on the imginry xis whih is F(iy). From (9), we hve F(iy) = u(, y) n therefore v(x, y) =. Moreover, sine F(z) is symmetri y iy-xis, x F(x + iy) x= = n lso F(iy), therefore F(iy) is t lest lol minimum for ll y on the horizontl line. Sine F(x + iy) hs lol minim t x =, F(x + iy) hs lso lol minim t x =. F(x + iy) is ifferentile y x even if F(x + iy) = unlike F(x + iy). Sine F(x + iy) hs lol minim t x =, its seon erivte t x = must e positive, i.e. D x F(x + iy) x= > where D x enotes x n sine F(x + iy) = u (x, y) + v (x, y), we hve D x F(x + iy) x= = u(, y) D x u(, y) + (D x u(, y)) + v(, y) D x v(, y) + (D x v(, y)) > (9) n v(, y) =, D x u(, y) =, thus we hve u(, y) D x u(, y) + (D x v(, y)) > () n sine D x v(, y) = D y u(, y), D x u(, y) = D y u(, y) n F(iy) = F(y) = u(, y) 3, we hve (D y F(y)) F(y) D y F(y) > () The inequlity () implies tht F(y) is log-onve, mening F (y) D y [ln(f(y))] <. Sine F(y) is log-onve, D y [ln(f(y))] = D yf(y) F(y) () n D y [ln(f(y))] = y [D yf(y) F(y) ] = F(y) D y F(y) (D y F(y)) F < (3) (y) Therefore, if F(y) hs zeros on iy-xis, D yf(y) is monotonilly eresing n F(y) hs unique F(y) extremum etween the two ontiguous zeros. 4. The o-positive efiniteness of F(x + iy) We hve shown tht F(x + iy) hs minim t x = ut it is unknow tht the minim re lol or glol on the horizontl line where y is fixe. We will show tht F(x + iy) is o-positive semiefinite for x. Theorem 4: Let Ψ(θ) = φ(θ, t)t, then Ψ(θ) is (o-)positive semi-efinite for θ if n only if φ(θ, t) is (o-)positive semi-efinite for θ for ll t. 3 Sine F(iy) is rel n ll the erivtives of F(iy) re lso rel, fterwr, we omit i from F(iy) unless neee. Hene F(y) refers to F(iy).
Proof The proof is strightforwr. If Ψ(θ) is positive semi-efinite, then n k Ψ(θ n θ k ) for ny ( ), n n θ n, n sine n= k n k Ψ(θ n θ k ) = n k φ(θ n θ k, t) t, to e n= k n k Ψ(θ n θ k ), n= k n k φ(θ n θ k, t) must e non-negtive for ll t. Hene φ(θ, t) must e positive semi-efinite for θ for ll t. The proof of o-positive efiniteness is sme. ote tht it oes not hol for the oule integrl. n= n= k k From () n z = x + iy, we hve F(x + iy) = f(t) e xt e iyt t whih is the Fourier trnsform of f(t) e xt. Hene y the Prsevl theorem, we hve Sine (f(t) e xt ) t = f (t) e xt t = F(x + iy) y π f (t) e xt t is lerly o-positive semi-efinite for x, (4) (5) F(x + iy) y must e lso o-positive semi-efinite for x, n therefore, F(x + iy) is o-positive semi-efinite for x for ll y y the theorem 4. However, F(x + iy) n F(x + iy) re only o-positive semi-efinite if F(x) is o-positive semi-efinite. Hene the neessry n suffiient onition to e F(x + iy) o-positive semiefinite is f(t) is non-negtive. Assuming f(t) is n even funtion, F(x + iy) = n= n (x + iy) n from (4) n we hve F(x + iy) = F(x + iy) F (x + iy) = ( n (x + iy) n ) ( n (x iy) n ) (6) n= n sine F(x + iy) hs only even powers of x, the power series expnsion of F(x + iy) is n= F(x + iy) = A n x n where A n epens on y n sine D x n F(x + iy) x=, A n is non-negtive for ll n. Sine F(x + iy) = F(x + iy) F(x iy), we hve n= (7) F(x + iy) = ( f(t) e xt e iyt t ) ( f(t) e xt e iyt t ) (8) n from (8) F(x + iy) = f(t ) f(t ) e x(t +t ) e iy(t t ) t t (9) By letting t = t + t n τ = t, we hve F(x + iy) = f(τ) f(t τ) e xt e iyτ e iy(t τ) τt (3)
whih n e written F(x + iy) = [ f(τ) f(t τ) e iyτ e iy(t τ) τ] e xt t (3) Letting τ τ, n ssuming f(t) is even, we hve or simply, where F(x + iy) = [ f(τ) f(t + τ) e iyτ e iy(t+τ) τ] e xt t (3) F(x + iy) = r y (t) e xt t (33) r y (t) = f(τ) f(t τ) e iyτ e iy(t τ) τ The onjugte of r y (t) = f(τ) f(t + τ) e iyτ e iy(t+τ) τ (34) r y (t) = f(τ) f(t + τ) e iyτ e iy(t+τ) τ n y letting τ τ t, n ssuming f(t) is even, we hve (35) r y (t) = f(τ t) f(τ) e iy(τ t) e iyτ τ = f(τ) f(t τ) e iyτ e iy(t τ) τ whih is the eqution (34), therefore, r y (t) is rel. Moreover, r y (t) is n even funtion whih n e esily prove. The funtion r y (t) is rel n even ut oes not hol the positivity, nmely, It n e negtive. The oeffiients A n in (7) n e esrie with the funtion r y (t), tht is, A n = (n)! tn r y (t) t (36) Sine A is F(iy), A n e zero, ut n A e lso zero? By twie ifferentiting F(x + iy) n letting x =, we get A, tht is A = D x F(x + iy) x= (37) We hve erive D x F(x + iy) x= in (9) n the result ws tht (D y F(y)) F(y) D y F(y) > in (). If A is zero, then it implies tht (D y F(y)) F(y) D y F(y) =, i.e. ( x F(y)) F(y) F(y) = (38) x The ifferentil eqution (38) is esily solvle y letting P(y) = y F(y) n we get F(y) = e y where the onstnts n re to e etermine y the initil onitions. Sine F(y) is n even
funtion, y F(y) y= = n F() = in (4). Thus, we get = n =, n therefore F(y) =, whih mens F(y) is onstnt. Sine F(y) = f(t) e iyt t =, f(t) = δ(t) where δ(t) enotes the Dir elt funtion. If f(t) = δ(t), F(z) is onstnt in the whole zplne. However, sine F(z) is ifferent from onstnt, A nnot e zero n it shoul e stritly positive. As result, A > n A n (n ), n hene, if y is fixe, F(x + iy) hs unique glol minimum t x = n monotonilly inresing when x >. Therefore, F(x + iy) n hve zeros only t x = if it hs ny. Sine x = r os (θ), F(x + iy) is o-positive semi-efinite for r os(θ). However, how is it when θ is fixe or r is fixe. Firstly, we onsier when θ is fixe ssuming os(θ) is non-zero. Sine θ is fixe, os(θ) is onstnt. Letting = os(θ), where is non-zero onstnt, n hene, x = r. Putting it to (7), we hve F(r + iy) = B n r n n= (39) where B n = A n n. Sine is non-zero, n >, n sine A >, B is stritly positive n other oeffiients B n re non-negtive. Thus, when θ is fixe, F(r + iy) ehves like F(x + iy), mening, it hs unique minimum t r = n monotonilly inresing when r >. Seonly, we will onsier the se when r is fixe n θ vries. It is the ehvior of F(x + iy) on irle entere t the origin. By hnging the vrile in the eq. (7), nmely, x = r os(θ), we hve where C n = A n r n. F(r os(θ) + iy) = C n os n (θ) (4) Assuming tht r is non-zero, then C is stritly positive n other oeffiients C n re nonnegtive. By ifferentiting eq. (4) y θ, we hve sin(θ) n C n os n (θ) = sin(θ) n C n os n (θ) n= Sine C is stritly positive n n C n os n (θ) is non-negtive when n, the slop is only epening on sin(θ). In the intervl of < θ < π/, sin(θ) is negtive n hene, the slop is negtive, therefore, F(θ + iy) is monotonilly eresing. In the intervl of π/ < θ < π, sin(θ) n the slop re positive n thus, F(θ + iy) is monotonilly inresing. ote tht the slop is zero when θ = n θ = π/. n= n= By the Pley-Wiener theorem, to e F(z) entire, f(t) must e eresing rpily n F(z) oes not hve ny poles on x-xis s mentione. Moreover, if F(z) hs zeros only on the iy-xis, tht is, F(z) hs only rel zeros, F(z) must e entire. We onlue with the theorem.
Theorem 5: If funtion f(t) is non-negtive n even n its two-sie Lple trnsform is enote y F(z), then F(x + iy) n F(x + iy) hve unique minimum t x = on the horizontl line when y is fixe n monotonilly inresing in the region of onvergene, n therefore, F(x + iy) n F(x + iy), n thus F(x + iy) n hve zeros only t x =, n sine F(iy) is rel when y is rel, F(x + iy) hve only rel zeros. Moreover, F(iz) hs zeros only on the x-xis n onsequently, only rel zeros, if f(t) is rpily eresing non-negtive even funtion. 5. The Lguerre inequlities n generlize Lguerre inequlities We hve shown tht the two-sie Lple trnsform of n even n non-negtive funtions hve only rel zeros. We will show tht the Lguerre inequlities n generlize Lguerre inequlities re true for the two-sie Lple trnsform of n even n non-negtive funtions. Theorem 6: The onvexity ) A funtion f(x) is onvex if n only if f (x) for ll x. ) The suffiient ut not neessry onition to e funtion f(x) stritly onvex is f (x) > for ll x. Theorem 7: The onvity ) A funtion f(x) is onve if n only if f (x) for ll x. ) The suffiient ut not neessry onition to e funtion f(x) stritly onve is f (x) < for ll x. Theorem 8: The log- onvexity n log-onvity A funtion f(x) is log-onvex if ln[f(x)] is onvex. Similrly, funtion f(x) is log-onve if ln[f(x)] is onve. ) A funtion f(x) is log-onvex, if f(λx + μx ) [f(x )] λ [f(x )] μ where λ, μ > n λ + μ =. ) A funtion f(x) is log-onve, if f(λx + μx ) [f(x )] λ [f(x )] μ where λ, μ > n λ + μ =. 3) If f(x) n g(x) re oth log-onvex, then f(x) g(x) is lso log-onvex. Similrly, if f(x) n g(x) re oth log-onve, then f(x) g(x) is lso log-onve. The neessry ut not suffiient onition to hve F(y) only rel zeros is tht F(y) n ll the erivtives of F(y) re log-onve, hene we hve the theorem. Theorem 9: The Lguerre inequlities F(y) elongs to the Lguerre-Póly lss if
where n =,, 3, n for ll y R. [F (n) (y)] F (n ) (y) F (n+) (y) (4) ote tht the theorem 9 is the neessry onitions, not suffiient. Theorem : Let e F(x) = f(t) e xt t where f(t) is non-negtive n x is rel, then F(x) is log-onvex. If f(t) is non-negtive, then t n f(t) is lso non-negtive. Consequently, (n) th erivtive of F(x), tht is, F (n) (x) = t n f(t) e xt t is lso log-onvex for n =,,, It n e esily prove using theorem 8 n the Höler inequlity. Sine F (n) (x) is log-onvex, we hve F (n) (x) F (n+) (x) [F (n+) (x)] (4) ote tht (n + ) th erivtive of F(x), tht is, F (n+) (x) = not log-onvex. t n+ f(t) e xt t is generlly Theorem : Let e F(x) = then we hve the inequlities: Proof Sine F (n+m) (x) = f(t) e xt t where f(t) is ontinuous non-negtive funtion n x is rel, [F (n+m) (x)] = ( [F (n+m) (x)] F (n) (x) F (m) (x) (43) t n+m f(t) e xt t, we hve t n+m f(t) e xt t By the Cuhy-Shwrz inequlity, we hve ) ( t n f(t) e xt t m f(t) e xt t) Hene, [F (n+m) (x)] F (n) (x) F (m) (x). = ( t n f(t) e xt t m f(t) e xt t) ( t n f(t) e xt t ) ( t m f(t) e xt t ) ow, we onsier the theorem (9) in the se of n =. Sine F(x) is log-onvex, n if f(t) is non-negtive n even, then we hve F(x) F (x) [F (x)] (44) F(x) = f(t) osh(xt) t F (x) = t f(t) sinh(xt) t
From eq. (9), we hve By letting x iy, we hve F (x) = t f(t) osh(xt) t F(iy) = f(t) os(yt) t F (iy) = t f(t) sin(yt) t F (iy) = t f(t) os(yt) t F(x) x=iy = f(t) os(yt) t = F(iy) F (x) x=iy = i t f(t) sin(yt) t = i F (iy) F (x) x=iy = t f(t) os(yt) t = F (iy) Therefore, y letting x iy, we get the inequlity (43) with the funtion of F(iy), tht is, From this inequlity, we hve (F(x) F (x) [F (x)] ) x=iy = F(iy) F (iy) [i F (iy)] More intuitively, the inequlity (44) is [F (iy)] F(iy) F (iy) (45) F(x) n y hnging the vrile x iy we hve n the inequlity (46) is x F(x) [ x F(x)] F(iy) (iy) F(iy) [ (iy) F(iy)] (46) (i) F(iy) y F(iy) (i) [ y F(iy)] = F(iy) y F(iy) + [ y F(iy)] n sine F(iy) n ll the erivtives of F(iy) re rel, we hve whih is the sme result in (45). [F (iy)] F(iy) F (iy)
Generlly, we get the theorem: Theorem : If F(x) oes not hve ny pole on the x-xis, tht is, F(z) is entire, n F(iy) is rel when y rel, then where n n m re non-negtive integer. Proof By hnging the vrile x to iy, we hve F (n) (x) F (m) (x) x=iy = (i) n+m F(n) (iy) F (m) (iy) (47) n m F(x) xn x m F(x) = n F(iy) x=iy (iy) n n therefore, eq. (47) is vli. In generl, if F (n) (x) F (m) (x) = n x n F(x) m F(x) xm m F(iy) = (iy) m (i) n n y n F(iy) m (i) m F(iy) ym then F (n k ) (x) F (m k ) (x) k= F (n k ) (x) F (m k ) (x) k= x=iy = (i) n F (n k ) (y) F (m k ) (y) (48) k+m k The inequlity oes not hnge sine we hve only hnge the vrile. k= The theorem is very useful, sine we n prove some inequlities on the x-xis more esily thn on the iy-xis. Theorem 3: We get the inequlity elow y pplying (47) to (4) 4 [F (n+) (y)] F (n) (y) F (n+) (y) (49) whih re the even terms of the Lguerre inequlities. Theorem 4: We get the inequlity elow y pplying eq. (47) to (43) 4 Sine F(iy) n ll the erivtives of F(iy) re rel when y is rel, for the ske of onveniene, i is omitte fterwr, i.e. F(y) mens F(iy).
[F (n+m) (y)] F (n) (y) F (m) (y) [F (n+m) (y)] F (n) (y) F (m) (y) when (n + m) is o when (n + m) is even (5) We shll show the o terms of the Lguerre inequlities, ut efore tht, we prove Höler inequlity for the oule integrl. Theorem 5: Höler inequlity for the oule integrl f(x, y) g (x, y)xy ( f(x, y) p xy) p ( g(x, y) q xy) q (5) where p, q > n p + q =. Proof We efine U(x, y) n V(x, y) suh tht n U(x, y) = V(x, y) = f(x, y) ( f(x, y) p xy g(x, y) ( g(x, y) q xy Sine U(x, y) n V(x, y), y Young s inequlity for prouts, we hve whih is ( f(x, y) p xy U(x, y) V(x, y) Up (x, y) p f(x, y) g(x, y) ) p ( g(x, y) q xy) + Vq (x, y) q q p ) q ) f(x, y) p + g(x, y) q p f(x, y) p xy q g(x, y) q xy Sine oth sies of the inequlity re non-negtive, the inequlity oes not hnge y integrting. By integrting twie, we hve ( f(x, y) p xy f(x, y) g(x, y) xy ) Sine + = we hve p q p ( g(x, y) q xy) f(x, y) g (x, y)xy ( f(x, y) p xy) q f(x, y) p xy p + g(x, y) q xy f(x, y) p xy q g(x, y) q xy p ( g(x, y) q xy) q Sine F(x) = f(t) osh(xt) t, y ifferentiting (n + ) times, we hve
n we n write (5) F (n+) (x) = t n+ f(t) sinh(xt) t (5) n sine F (n+) (x) = t n+ f(t) sinh(xt) t = x t n+ f(t) sinh(xt) t xt osh(xtτ) τ = sinh(xt) xt We hve F (n+) (x) = x t n+ f(t) osh(xtτ) tτ = x t n+ f(t) e xtτ tτ We efine P(x) suh tht P(x) = t n+ f(t) osh(xtτ) tτ = t n+ f(t) e xtτ tτ The purpose of hnging osh(xtτ) to e xtτ is to prove the log-onvexity of P(x) esily. Sine t n+ f(t) is even, the hnge is vli. Theorem 6: The funtion P(x), whih is efine s is log-onvex. P(x) = t n+ f(t) e xtτ tτ Proof t n+ f(t) e (λx +μx ) tτ tτ = t n+ f(t) e λxtτ e μx tτ tτ where λ + μ =. Sine f(t), we hve t n+ f(t) e λxtτ e μx tτ tτ = (t n+ f(t) e xtτ ) λ (t n+ f(t) e xtτ ) μ tτ By the Höler inequlity for the oule integrl (theorem 5), we hve whih is (t n+ f(t) e xtτ ) λ (t n+ f(t) e xtτ ) μ tτ ( t n+ f(t) e x tτ λ tτ) ( t n+ f(t) e x tτ μ tτ)
t n+ f(t) e (λx +μx ) tτ tτ This inequlity yiels ( t n+ f(t) e x tτ tτ) n y the theorem 9, P(x) is log-onvex. Sine P(x) is log-onvex, we hve From (9), we hve F(y) = where n =,,, Sine we hve n sine P(x) x=iy = P(y) = n from (5), we get thus, P(y) is log-onve. λ ( t n+ f(t) e x tτ tτ) P(λx + μx ) [P(x )] λ [P(x )] μ P(x) P (x) [P (x)] (5) f(t) os(yt) t n its (n + ) th erivtive is F (n+) (y) = ( ) n+ t n+ f(t) sin(yt) t os(xtτ) τ = sin(xt) xt F (n+) (y) = ( ) n+ y t n+ f(t) os(ytτ) tτ Letting g(y) = ( ) n+ y, we hve t n+ f(t) os(ytτ) tτ, we hve F (n+) (y) = ( ) n+ y P(y) [P (y)] P(y) P (y) [g (y)] g(x) g (x) = > n therefore, ( ) n+ y is log-onve. Sine oth ( ) n+ y n P(y) re log-onve, F (n+) (y) is log-onve. Thus, F (n) (y) is log-onve for ll non-negtive integer n. Consequently, the Lguerre inequlities re vli for ll n. Theorem 6: The generlize Lguerre inequlities We efine L n (y) s follows: μ n L n (y) = ( ) n n! ( )k ( n k ) F(k) (y) F (n k) (y) k= (53)
where n =,,, F(y) hs only rel zeros if n only if L n (y) for ll n. We return to eq. (33), whih is where n reform (55), so tht F(x + iy) = r y (t) e xt t (54) r y (t) = f(τ) f(t + τ) e iyτ e iy(t+τ) τ r y (t) = f(τ) e iyτ f(t + τ) e iy(t+τ) τ (55) (56) n y letting g(τ) = f(τ) e iyτ, r y (t) is the ross-orreltion funtion of g(τ) n g (τ) where g (τ) = f(τ) e iyτ. Let F(ω) e the Fourier trnsform of f(τ), then the Fourier trnsform of g(τ) is F(ω y) n the Fourier trnsform of g (τ) is F(ω + y). By the ross-orreltion theorem, we hve n sine F(y) is even, we hve r y (t) = π F(ω y) F(ω + y) eitω ω r y (t) = π F(y ω) F(y + ω) eitω ω whih is similr to the Wigner-Ville istriution funtion. By hnging vrile x = iθ, from (54), we hve (57) n F(iθ + iy) = π F(y ω) F(y + ω) eiωt e iθt ω t (58) F(y ω) F(y + ω) e iωt e iθt ω t = F(y ω) F(y ω) [ e iωt e iθt t] ω n thus, we hve e iωt e iθt t = π δ(θ ω) F(iθ + iy) = F(y ω) F(y + ω) δ(ω θ) ω n y omitting i for the onveniene, we hve, F(θ + y) = F(y θ) F(y + θ) (59) whih is the hrteristi eqution of F(x + iy) where x = iθ, hene, from (54)
F(θ + y) = r y (t) e iθt t (6) The n th moment of F(x + iy), whih is enote s M n, is efine s follows or M n (y) = t n r y (t) t (6) M n (y) = ( ) n D x n F(x + iy) x= (6) Another metho to get M n (y) is ifferentiting (6), tht is, M n (y) = ( i) n D θ n F(θ + y) θ= or y (59), M n (y) is D ( i) n θ n [F(θ y) F(θ + y)] θ= whih n e ompute using the Leiniz rule, tht is, M n (y) = ( i) n D θ n [F(y θ) F(y + θ)] θ= = n ( i) n ( )k ( n k ) F(k) (y) F (n k) (y) k= (63) However, sine r y (t) is n even funtion, M n (y) vnishes when n is o n we nee to ompute only for even n, hene, M n (y) = D θ n [F(y θ) F(y + θ)] θ= = ( ) n ( ) k ( n k ) F(k) (y) F (n k) (y) n we hve n k= (64) where L n (y) is efine in (53). F(x + iy) = (n)! M n(y) x n = L n (y) x n (65) n= If L n (y) for ll n, F(x + iy) hs unique minimum t x = when y is fixe n hene, F(y) hs only rel zeros. n= ow, we will prove the generlize Lguerre inequlities. In ft, F(y + θ) = F(iy + iθ) = F[i(y + θ)], i.e. this funtion lies on the iy-xis. F(y θ) is the sme. We will mp F(y θ) F(y + θ) on x-xis, i.e. F(x θ) F(x + θ), n we hve F(x θ) = f(t) e xt e θt t
n k F(x (θ n + θ k )) = f(t) e xt n k e (θ n+θ k )t t n= k = f(t) e xt n e θ nt n= n= t Therefore, F(x θ) is o-positive semi-efinite. In the sme wy, we n prove tht F(x + θ) is o-positive semi-efinite. Sine oth F(x θ) n F(x + θ) re o-positive semi-efinite, F(x θ) F(x + θ) is o-positive semi-efinite for θ. Sine F(x θ) F(x + θ) is o-positive semi-efinite for θ, D θ n [F(x θ) F(x + θ)] θ= n we hve n M n (x) = D n θ [F(x θ) F(x + θ)] θ= = ( ) k ( n k ) F(k) (x) F (n k) (x) (66) By letting x = iy, we hve n M n (iy) = D n θ [F(y θ) F(y + θ)] θ= = ( ) k ( n k ) (i) k F(k) (y) (i) n k F(n k) (y) whih implies n hene L n (y) for ll n. n k= k= M n (y) = ( ) n ( ) k ( n k ) F(k) (y) F (n k) (y) k= k The funtion F(iy) = f(t) os(yt) t = f(t) os(yt) t elongs to the Lguerre-Póly lss n hs only rel zeros if f(t) is non-negtive even funtion n rpily eresing so tht F(z) is entire where F(z) is efine s We n lso efine F(iz), tht is, F(z) = f(t) osh(zt) t F(iz) = f(t) os(zt) t whih is the fmous form in the literture euse of the Riemnn s ig-xi funtion Ξ(z), then F(x) efine s F(x) = f(t) os(xt) t elongs to the Lguerre-Póly lss n hs only rel zeros. Bsilly, the two efinitions re sme.
From eq. (4), y letting z = iy, we hve F(iy) = F(y) = ( ) n n y n = y + 4 y 4 6 y 6 + (67) n= where n n F(iy) hs only rel zeros. Sine the Lguerre inequlities hols for ny y, y letting y =, we hve [F (n) ()] F (n ) () F (n ) () (68) whih is lwys vli if the power expnsion of funtion hs the form of (67). We onsier it in two ses, tht is, when n is o n n is even. i. n is o: F (n) () = F (n ) () n F (n ) () hve the opposite sign n hene, F (n ) () F (n ) (). Therefore, the inequlities (68) re vli. ii. n is even: [F (n) ()] Both F (n ) () n F (n ) () re zero, thus the inequlities (68) re lso vli. Therefore, ll funtions whose power series expnsions hve the form of (66) hol the Lguerre inequlities t y =. However, not ll funtions whose power series expnsions hve the form of (67) hve only rel zeros, therefore, the Lguerre inequlities re the neessry onition, ut not suffiient to hve only rel zeros. By letting u = y in (67), we hve F(u) = ( ) n n u n = u + 4 u 6 u 3 + (69) n= whih lso hve only rel zeros when u >. By pplying (68), we hve n n (n + ) n n+ (7) The equlities hol if n only if F(z) = e αz where α >, whih is the two-sie Lple trnsform of f(t) = πα e t /4α. Therefore, if f(t) is Gussin, its two-sie Lple trnsform F(z) oes not hve ny zero. Hene, if f(t) = e φ(t) or sum of e φ(t) n its two-sie Lple trnsform is F(z), then F(z) hs only rel zeros if the orer of φ(t) is greter thn two. Further, if the orer of φ(t) is less thn two, the two-sie Lple trnsform of e φ(t) oes not hve only rel zeros. The Riemnn zet funtion (s) is efine 6. The Riemnn hypothesis where s = + iω. (s) = n s = + s + 3 s + n=
It is known tht the zeros of (s) re lote only on the strip < <. Riemnn onjeture tht ll the zeros of (s) re lote on the line =, so-lle Riemnn hypothesis. Using the Riemnn s funtionl eqution, n entire n symmetri funtion n e otine whih is lle the xi funtion ξ(s) where n ξ(s) = π s s(s )Γ ( s ) (s) (7) ξ(s) = ξ( s) hene ξ(s) is symmetri t = n the zeros of ξ(s) re lote t the sme position of (s), tht is, on the strip < <. If the Riemnn hypothesis is true, ll the zeros of ξ(s) re lote on the line =. It is well-known tht where ξ(s) = φ(t) e (s )t t (7) φ(t) = π n e πn e t (πn e 9 t 3e 5 t ) n= n it n e shown tht φ(t) > for ll t n n even funtion. (73) By letting z = s where z = x + iy, n φ(t) is even, we hve Φ(z) = φ(t) e zt t = φ(t) osh(zt) t (74) n sine Φ(z) is shifte funtion y of ξ(s), Φ(z) is entire n the zeros of Φ(z) shoul e lote on the strip < x <. From (66), we hve Φ(z) = π 4 π z (z 4 ) Γ (z + 4 ) (z + ) (75) whih is Riemnn s originl efinition of xi-funtion. We onsier the funtion φ(t) efine in (7). It is positive n even. Moreover, it is eresing very rpily (otherwise, ξ(s) n Φ(z) nnot e entire). Therefore, Φ(iy) elongs to the Lguerre- Póly lss n hs only rel zeros. It mens tht ll the zeros of Φ(z) re lote t x =, n hene, ll the zeros of ξ(s) n (s) re lote t =. Thus, the Riemnn hypothesis is true. From eq. (73), we hve n y letting t t, we hve Φ(iz) = φ(t) os(zt) t = φ(t) os(zt) t
Φ(iz) = 4 φ(t) os(zt) t (76) We efine (t) s (t) = π n e πn e t (πn e 9t 3e 5t ) n= then (t) = φ(t) n eq. (7) will e Φ(iz) = 8 (t) os(zt) t n y efining Ξ(z) = 8 Φ(iz/), we hve Ξ(z) = (t) os(zt) t (77) or simply, Ξ(z) = Φ(iz) (78) This funtion is lle the ig-xi or upper-se xi funtion n use to prove the Riemnn hypothesis n to fin the lotion of zeros in most litertures. Sine, (t) is positive even funtion n eresing rpily, Ξ(x) elongs to the Lguerre- Póly lss n hs only rel zeros, whih les tht the Riemnn hypothesis is true.