Suwa, Masanori

Yoshino, Kunio

Kyoto University Research Information Repository

TitleA new proof of Carlson's theorem by Plana's summationformula (Asymptotic Analysis and Microlocal Analysis ofPDE)Author(s)Yoshino, Kunio; Suwa, MasanoriCitation数理解析研究所講究録 (2001), 1211: 166-174Issue Date2001-06URL http://hdl.handle.net/2433/41136RightType Departmental Bulletin PaperTextversionpublisherKyoto UniversityA new proof of Carlson’s theorem by P1ana’ssummation formula(Kunio Yoshino)(Masanori Suwa)Department of Mathematics, Sophia UniversityAbstractIn this paper we will prove Carlson’s theorem by using Plana’ssummation formula.In \S 1, we will recall the definitions and transformations of analyticfunctionals with unbounded carrier.In \S 2, we will give aproof of Carlson’s theorem by Plana’s sum-mation formula.1The definitions of analytic functionals withunbounded carrier and their transforma-tions.Let $L$ and $L_{\epsilon}$ be following strip regions:$L$ $=$ $(-\infty, a]+\iota[-b, b]$ ,For $\epsilon$ $>0$ , $L_{\epsilon}$ $=$ $(-\infty, a+\epsilon]+\iota[-b-\epsilon, b+\epsilon]$ .For $\epsilon>0$ , $\epsilon’>0$ and $k’\in \mathbb{R}$, we put$\zeta=\xi+\iota\eta\}$$Q_{b}(L_{\epsilon} : k’+\epsilon’)$$:=$ $\{f(\zeta)\in H(L_{\epsilon})\cap C\circ(L_{\epsilon})$ : $\sup_{\zeta\in L_{\epsilon}}|f(\zeta)e^{k’\xi+\epsilon’|\xi|}|<\infty$ ,1211 2001 166-174166$7\mathrm{f}(L.)$ is the space of holomorphic functions defined on $L_{g)}$, (interior of L.).$C(L.)$ is the space of continuous functions defined on $L_{e}$ . We put$Q(L : k’)= \lim_{\epsilonarrow\vec{0,\epsilon’}arrow 0}Q_{b}(L_{\epsilon} : k’+\epsilon’)$ ,where $\lim_{arrow}$ means inductive limit. If $z\in(-k’, \infty)+\iota \mathbb{R}$, then the function $e^{\zeta z}$of $\langle$ belongs to $Q(L:k’)$ . We denote by $Q’(L : k’)$ the dual space of $Q(L : k’)$ .The elements of $Q’(L:k’)$ is called analytic functional$\mathrm{s}$ with carrier $L$ andof type $k’$ .We define the Fourier-Borel transform $\tilde{T}(z)$ of $T\in Q’(L : k’)$ as follows:$\tilde{T}(z)=<T_{\zeta}$ , $e^{\zeta z}>$ .$\tilde{T}(z)$ is holomorphic function on the right half plane $(-k’, \infty)+i\mathbb{R}$ and sat-isfies following estimate :$\forall\epsilon>0$ , $\epsilon’>0$ , $\exists C_{\epsilon,\epsilon’}\geq 0$ such that$|\tilde{T}(z)|\leq C_{\epsilon,\epsilon’}e^{ax+b|y|+\epsilon|z|}$ , $({\rm Re} z\geq-k’+\epsilon’, z=x+\iota y)$ . (1)$\mathrm{E}\mathrm{x}\mathrm{p}((-k’, \infty)+i\mathbb{R}$ : $L$ ) denotes the space of holomorphic functions definedon the right half plane $(-k’, \infty)+$-ill and satisfy the estimates (1). Followingtheorem characterizes the Fourier-Borel transform of $Q’(L:k’)$ .Theorem 1.1 $([2],[9])$ . Fourier-Borel transfom is a linear topologictil is0-morphism from $Q’(L:k’)$ onto $\mathrm{E}\mathrm{x}\mathrm{p}((-k’, \infty)+i\mathbb{R}:L)$ .Definition 1.2. $\epsilon’$-Cauchy transform $\check{T}(w, \epsilon’)$ of $T\in Q’(L : k’)$ is defined asfollows :$\check{T}(w, \epsilon’)$ $=$ $\frac{-1}{2\pi \mathrm{z}}\langle T_{\zeta}$ , $\frac{e^{(-k’+\epsilon’)(\zeta-w)}}{\zeta-w}\rangle$ .Proposition 1.3 $([2],[9])$ . The following integral representation holds :$<T$ , $\varphi>=\frac{1}{2\pi i}\int_{\partial L_{6}}\check{T}(\zeta, \epsilon’)\varphi(\zeta)d\zeta$, $(\forall\varphi\in Q_{b}(L_{\epsilon} : k’+\epsilon_{1}’), \epsilon’<\epsilon_{1}’)$ .Put $\varphi(\zeta)=e^{\zeta z}$ , then toe have$\tilde{T}(z)=\frac{1}{2\pi i}\int_{\partial L_{\epsilon}}\check{T}(\zeta, \epsilon’)e^{\zeta z}d\zeta$.167Conversely, we can express $\check{T}$ by $\tilde{T}$ in as follows :$\check{T}(w, \epsilon’)=\int_{-k’+\epsilon’}^{\infty}\tilde{T}(z)e^{-wz}dz$.Definition 1.4. For $S\in \mathrm{H}’(\mathrm{K})$ , $K$ is acompact set and $T\in \mathrm{Q}’(\mathrm{L}:k’)$ , wedefine $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}*\mathrm{b}\mathrm{y}$$T*S=S*T=<T_{\zeta},$ $<S_{\tau}$ , $\varphi(\tau+\zeta)>>$ , $\varphi\in Q’(L+K : k’)$ .Proposition 1.5. We have the following equalities:(i) $T*S\in Q’(L+K : k’)$ ,(ii) $(\overline{T*S})(z)=\tilde{T}(z)*\tilde{S}(z)$ .In [15] we derived Plana’s summation formula for holomorphic functionsof exponential type by using the theory of analytic functional$\mathrm{s}$ with un-bounded carrier.Proposition 1.6 (Plana’s summation formula [6], [13],[15]). Let $T\in$$Q’(L:k’)$ . If $k’>0$ , $0\leq b<2\pi$ , ${\rm Res}>a$ and $|{\rm Im} s|+b<2\pi$ , then we have$\sum_{n=0}^{\infty}\tilde{T}(n)e^{-\epsilon n}$$= \frac{1}{2}\tilde{T}(0)+\int_{0}^{\infty}\tilde{T}(x)e^{-\epsilon x}dx+\iota\int_{0}^{\infty}\frac{\tilde{T}(\iota x)e^{-lsx}-\tilde{T}(-\iota x)e^{\prime sx}}{e^{2\pi x}-1}dx$ .2Applications.In this section we determine the form of holomorphic functions $f(z)$ of.ex-ponential type with $f(\mathrm{N})=$ {0}.Lemma 2.1. Suppose that $T\in \mathrm{Q}’(\mathrm{L}:\mathrm{k}’)$ , $L=(-\infty,\alpha]+\dot{x}[-b, b]$ , $0<k’<$$1,0\leq b<2\pi$ . If $\tilde{T}(n)=0$ , $(n=0,1,2, \cdots)$ , then there eist holomorphicfunctions $a(s)$ and $b(s)$ such that$\check{T}(s, \epsilon’)=a(s)+b(s)$ ,where $a(s)\in H(\{s\in \mathbb{C}:{\rm Im} s<2\pi-b\})$ , $b(s)\in H(\{s\in \mathbb{C} : {\rm Im} s>b-2\pi\})$and satisfy the following estimate :$|a(s)|$ $\leq$ $C_{\epsilon’}e^{(k’-\epsilon’){\rm Res}}$ ,$|b(s)|$ $\leq$ $C_{\epsilon’}e^{(k’-\epsilon’){\rm Res}}$ .168Corollary 2.2. Let $T\in Q’(L : k’)$ . If $\tilde{T}(n)=0(n=0,1,2, \cdots)$ , thei$\check{T}(s, \epsilon’)$ is holomorphic in $|{\rm Im} s|<2\pi-b$, and $|\check{T}(s, \epsilon’)|\leq C_{\epsilon’}e^{(k’-\epsilon’){\rm Res}}$ .Proof of lemma 2.1. By Plana’s summation formula and assumptioi$\tilde{T}(n)=0$ , $(n=0,1,2, \cdots)$ , we have0 $=$ $\int_{0}^{\infty}\tilde{T}(x)e^{-xs}dx+i\int_{0}^{\infty}\frac{\tilde{T}(ix)e^{-\iota xs}-\tilde{T}(-ix)e^{\iota xs}}{e^{2\pi x}-1}dx$$=$ $\int_{0}^{\infty}\tilde{T}(x)e^{-xs}dx+i\int_{0}^{\infty}\frac{\tilde{T}(ix)e^{-\iota xs}}{e^{2\pi x}-1}dx-i\int_{0}^{\infty}\frac{\tilde{T}(-ix)e^{\iota xs}}{e^{2\pi x}-1}dx$$=$ $\int_{0}^{\infty}\tilde{T}(x)e^{-xs}dx+\int_{0}^{l\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz$$- \int_{0}^{-\iota\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz-\int_{0}^{-\iota\infty}\tilde{T}(z)e^{-sz}dz$$=$ $\int_{0}^{\infty}\tilde{T}(x)e^{-xs}dx+\int_{-k+\epsilon}^{0},,\tilde{T}(x)e^{-xs}dx+\int_{0}^{l\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi lz}-1}dz$$- \int_{0}^{-\iota\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz-\int_{0}^{-\iota\infty}\tilde{T}(z)e^{-sz}dz-\int_{-k+\epsilon}^{0},,\tilde{T}(x)e^{-xs}dx$$=$ $\int_{-k’+\epsilon’}^{\infty}\tilde{T}(x)e^{-xs}dx+\int_{-k+\epsilon’}^{\iota\infty},\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz$$- \int_{-k’+\epsilon’}^{-\iota\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz-\int_{-k’+\epsilon’}^{-\iota\infty}\tilde{T}(z)e^{-sz}dz$ .Hence$\int_{-k’+\epsilon’}^{\infty}\tilde{T}(x)e^{-xs}dx$$=$ $- \int_{-k+\epsilon’}^{l\infty},\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz+\int_{-k’+\epsilon’}^{-\iota\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz+\int_{-k’+\epsilon’}^{-\iota\infty}\tilde{T}(z)e^{-sz}dz$.Put$a(s)$ $=$ $- \int_{-k+\epsilon’}^{\iota\infty},\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz$,$b(s)$ $=$ $\int_{-k’+\epsilon’}^{-t\infty}\frac{\tilde{T}(z)e^{-sz}}{e^{-2\pi\iota z}-1}dz+\int_{-k’+\epsilon’}^{-\iota\infty}\tilde{T}(z)e^{-sz}dz$ .169Since by proposition 1.3, $T \vee(s, \epsilon’)=\int_{-k+\epsilon}^{\infty},,\tilde{T}(z)e^{-sz}dz$ , we have $\check{T}(s, \epsilon’)=$$a(s)+b(s)$ . Then $a(s)$ and $b(s)$ satisfy the conditions in lemma 2.1. 1Proposition 2.3. Suppose that $T\in Q’(L : k’),$ $L=(-\infty, a]+i[-b, b]\subset$$\{\zeta\in \mathbb{C} : |{\rm Im}\zeta|<2\pi\}$ , $0<k’<1.\tilde{T}(n)=0$ , $(n=0,1,2, \cdots)$ .(i) If $0\leq b<\pi$ , Then $T\equiv 0$ .(ii) $Ifrr\leq b<2\pi$ , Then iAere exists an analytic functional $S$ with unboundedcarrier such that$T=(\delta_{l\pi}-\delta_{-l\pi})*S$ , $S\in Q’((-\infty, a]+\iota[\pi-b, b-\pi]$ : $k’$ ), $where*is$convolution.Proof, (i)$<T$, $\varphi>$ $=$ $\frac{1}{2\pi\iota}\int_{\partial L_{e}}\check{T}(\zeta, \epsilon’)\varphi(\zeta)d\zeta$, $(\varphi\in Q_{b}(L_{\epsilon} : k’+\epsilon_{1}’), \epsilon’<\epsilon_{1}’)$ ,and $\check{T}(\zeta, \epsilon’)\in H(\mathbb{C}\backslash L)$ . By lemma 2.1, $\check{T}(\zeta, \epsilon’)\in H$({ $\zeta\in \mathbb{C}$ : $|{\rm Im}\zeta|<$$2\pi-b\})$ . Prom the assumption $0\leq b<\pi$ , $L$ is contained in { $\zeta\in \mathbb{C}$ : $|{\rm Im}\zeta|<$$2\pi-b\}$ and $|\check{T}(\zeta, \epsilon’)|\leq Ce^{(k’-\epsilon’)\zeta}$ . Hence $<T$, $\varphi>=0$ .(ii) In this case $\check{T}(\zeta, \epsilon’)\in H(\mathbb{C}\backslash \{(-\infty, a]+\iota[2\pi-b, b]\cup(-\infty, a]+$$\iota[-b, b-2\pi]\})$ . We define analytic functionals with unbounded carrier Si, $S_{2}$as follows :$<S_{1}$ , $\varphi>$ $=$ $\frac{1}{2\pi\iota}\int_{\partial L_{\acute{\epsilon}}}\check{T}(\zeta+\pi\iota)\varphi(\zeta)d\zeta$ ,$<S_{2}$ , $\varphi>$ $=$ $\frac{1}{2\pi\iota}\int_{\partial L_{\acute{e}}}\check{T}(\zeta-\pi\iota)\varphi(\zeta)d\zeta$ ,where $\varphi\in Q_{b}((-\infty, a]+\iota[\pi-b, b-\pi]$ : $k’+\epsilon’$), $L_{\epsilon}’=(-\infty, a+\epsilon]+\iota[\pi-b-$$\epsilon$ , $b-\pi+\epsilon]$ . By lemma 2.1, Si, $S_{2}\in Q’((-\infty, a]+\iota[\pi-b, b-\pi]$ : $k’$ ) and wehave $T=S_{1}*\delta_{\pi},+S_{2}*\delta_{-m}$ . Prom the assumption $\tilde{T}(n)=0$ , $(n=0,1,2, \cdots)$ ,0 $=$ $\tilde{T}(n)$$=$ $\tilde{S}_{1}(n)e^{n\pi\iota}+\tilde{S}_{2}(n)e^{-n\pi l}$.Therefore $\tilde{S}_{1}(n)=-\tilde{S}_{2}(n)$ . By$b-\pi|$$-(\pi-b)=2(b-\pi)<2\pi$ and (i) inthis proposition, $S_{1}=-S_{2}$ .170Corollary 2.4. Let $f(z)$ E $\mathrm{E}\mathrm{x}\mathrm{p}((-\mathrm{A}’,$oo) $+\mathrm{z}\mathrm{R}\ovalbox{\tt\small REJECT}$ L), $0<k’<1$ , $f^{\ovalbox{\tt\small REJECT}}(n)\ovalbox{\tt\small REJECT}$0, (n $\ovalbox{\tt\small REJECT}$0,1,2, \cdots ). Then(i) $0\leq b<\pi\Rightarrow f(z)=0$ .(ii) $\pi\leq b<2\pi\Rightarrow f(z)=\sin\pi z\tilde{S}(z)$ , uthere $S\in Q’((-\infty, a]+\iota[\pi-$$b$ , $b-\pi]$ : $k’$ ).Remark 2.5. Corollary 2.4(i) are called “Carlson’s theorem”.([3])Corollary 2.6.(i) $0\leq b<\pi\Rightarrow\{e^{n\zeta}\}_{n=0}^{\infty}$ is total in $Q(L:k’)$ .(ii) $\pi\leq b<2\pi--$, $\{e^{m\zeta}, \zeta e^{n\zeta}\}_{m}^{\infty}n=0=0$, is total in $Q(L:k’)$ .2.1 ExamplesCarlson’s theorem is used in several branches of mathematical physics. Forexample,1. Auniqueness of analytic continuation of scattering amplitude to com-plex angular momentum plane ([7], [11],[12]).2. Dyson Conjecture in statistical mechanics ([5], [8]).3. Calculation of Selberg Integral $([1],[8])$ .4. Auniqueness of Ramanujan Resummation ([4]).For the history of Carlson’s theorem, we refer the reader to [14]. Here wehave an example of proposition 2.3 (ii) :Example 2.7 (Functional equation of Riemann zeta function, $[6],[13]$ ).We define an analytic functional $T$ as follows :$<T$, $\varphi>=\frac{1}{2\pi\iota}\int_{\partial L_{\epsilon}}\frac{1}{2(e^{-e^{-\mathrm{z}^{\zeta}}}-1)1}\varphi(\zeta)d\zeta$, $\varphi\in Q(L:k’)$ , (2)171where $\yen^{-\mathrm{v}\mathrm{r}-\mathit{6},\tau T}+e$], $\mathit{6}>0$ . In (2), we putt $\ovalbox{\tt\small REJECT}$ eA\langle and $\mathrm{t}(0\ovalbox{\tt\small REJECT}$ $e^{\ovalbox{\tt\small REJECT}}$ . Then we have$\tilde{T}(z)$ $=$ $<T_{\zeta}$ , $e^{\zeta z}>$$=$ $- \frac{1}{2\pi\iota}\int_{(+\infty)}^{(+0)}\frac{(-t)^{-2z-1}}{e^{t}-1}dt$$=$ $\frac{\zeta(-2z)}{\Gamma(2z+1)}$ . (3)Remark that $\tilde{T}(n)=\frac{\zeta(-2n)}{\Gamma(2n+1)}=0$. On the other hand, we have$| \frac{1}{2\pi\iota}\int_{\partial L_{e}}\frac{e^{\zeta z}}{2(e^{-e^{-\}\zeta}}-1)}d\zeta|arrow 0$ , $(\xiarrow-\infty, {\rm Re} z>0)$ .Therefore by residue theorem,$\frac{1}{2\pi\iota}\int_{\partial L_{e}}\frac{e^{\zeta z}}{2(e^{-e^{-:\zeta}}-1)}d\zeta$ $=$ $- \sum_{n=0}^{\infty}\frac{(2\pi n)^{-2z}e^{\pi lz}}{2\pi\iota n}+\sum_{n=0}^{\infty}\frac{(2\pi n)^{-2z}e^{-\pi\iota z}}{2\pi\iota n}$$=$ $- \frac{(2\pi)^{-2z}}{\pi}\sin\pi z\zeta(2z+1)$ . (4)Therefore by (3) and (4), we have$\frac{\zeta(-2z)}{\Gamma(2z+1)}$ $=$ $- \frac{(2\pi)^{-2z}}{\pi}\sin\pi z\zeta(2z+1)$ .This means $\tilde{T}(z)=\sin\pi z\tilde{S}(z),\tilde{S}=-\frac{1}{\pi}(2\pi)^{-2z}\zeta(2z+1)$ . Namely, $T=$$(\delta_{\pi l}-\delta_{-m})*S$ , where $S=$ . Now we replace $z$ to $\frac{z-1}{2}$ . Then$\frac{\dot{\zeta}(1-z)}{\Gamma(z)}$ $=$ $- \frac{(2\pi)^{-z+1}}{\pi}\sin(\frac{\pi}{2}z-\frac{\pi}{2})\zeta(z)$$=$ $2(2 \pi)^{-z}\cos(\frac{\pi}{2}z)\zeta(z)$$\zeta(1-z)$ $=$ $2(2 \pi)^{-z}\cos(\frac{\pi}{2}z)\Gamma(z)\zeta(z)$ . (5)Since (5), we obtain “Functional equation of Riemann zeta function”172References[1] AomotO-Kita :Theory of Hypergeometrr.c Functions, Appendix, (Sel-berg Integral and BC type Hypergeometric Functions), (in Japanese),Springer-Verlag.[2] C.A.Berenstein and R.Gay :Complex Analysis and Special Topics inHarmonic Analysis, Springer-Verlag (1991).[3] R.P.Boas :Entire Functions, Academic Press, (1954).[4] B.Candelpergher, M.A.Coppo and E.Delabaere :La sommation de Ra-manujan, L’Enseignement Math\’ematique, (1997), 93-132.[5] F.Dyson :Statistical theory of the Energy level of complex system I,J.Math.Phy. 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A new proof of Carlson's theorem by Plana's summation formula (Asymptotic Analysis and Microlocal Analysis of PDE)

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A new proof of Carlson's theorem by Plana's summation formula (Asymptotic Analysis and Microlocal Analysis of PDE)

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