Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $u\in C^\infty(\mathbb T^k)$, where $\mathbb T^k$ is the $k$-dimensional torus. (Equivalently, $u\in\mathbb R^k$ and $u$ is $2\pi$-periodic with respect of each argument.) We define the semi-norm $$ \|u\|_s=\|(-\Delta)^{s/2}u\|_{L^2(\mathbb T^k)}=\Big(\sum_{\ell_1,\ldots,\ell_k\in\mathbb Z}(\ell_1^2+\cdots+\ell^2_k)^{s/2}\big|\hat{u}_{\ell_1,\ldots,\ell_k}\big|^2\Big)^{1/2}, $$ where $s>0$ and $u(x_1,\ldots,x_k)=\sum_{\ell_1,\ldots,\ell_k\in\mathbb Z}\mathrm{e}^{2\pi i(\ell_1x_1+\cdots\ell_kx_k)}\hat{u}_{\ell_1,\ldots,\ell_k}$.

My question is the following: Which rate of growth of $\|u\|_s$, as $s\to\infty$, implies that $u$ extends holomorphically to an open neighborhood of $\mathbb R^k$ in $\mathbb C^k$?

More specifically: Which rate of growth of $\|u\|_s$, as $s\to\infty$, guarantees that $u$ extends holomorphically to $$ \Omega_\alpha= \{(x_1+iy_1,\ldots,x_k+iy_k): x_1,y_1,\ldots,x_k,y_k\in\mathbb R\,\&\,|y_1|,\ldots,|y_k|<\alpha\} \subset \mathbb C^k, $$ for a given $\alpha>0$.

share|improve this question
    
potentially related: en.wikipedia.org/wiki/Hardy_space –  Otis Chodosh Dec 9 '13 at 15:37
    
Not the kind of answer I am looking for. I do know that the rate of growth is expected to be over-exponential, as exponential growth corresponds only to trigonometric polynomials. –  smyrlis Dec 9 '13 at 15:42
    
Have a loot at volume 3 in Lions-Magenes' book Non-Homogeneous Boundary Value Problems and Applications –  Liviu Nicolaescu Dec 9 '13 at 16:22

1 Answer 1

up vote 3 down vote accepted

The rate of growth must be $(cs)^s$ for some $c>0$. In my sketch of the proof I assume for simplicity that $k=1$ and Fourier coefficients $a_n$ are zero for $n<0$.

The function has an analytic extension in a neighborhood of the unit circle if $|a_n|$ decrease faster than a geometric progression, that is $\log |a_n|^2\leq-\delta n$ for some $\delta>0$. Then your norm squared is $$\Phi(s)=\sum_{1}^\infty n^s|a_n|^2\leq\sum e^{s\log n-\delta n}.$$ We consider this as a Dirichlet series of one complex variable $s$. Let $m(s)$ be the maximal term of the sum in the RHS. It is known from the theory of Dirichlet series that for every $\epsilon$ $$\Phi(s)\leq A(\epsilon)m(s+1+\epsilon).$$ The maximim term on the right hand side is found by calculus, by maximizing with respect to $n$, and we get $$\Phi(s)\leq (cs)^s$$ for some $c>0.$

In the opposite direction, suppose that $\Phi(s)\leq (cs)^s$. Then $$|a_n|^2n^s\leq (cs)^s.$$ This gives an estimate for $|a_n|$ with parameter $s$. Minimizing with respect to $s$ we obtain the desired $\log|a_n|\leq -\delta n$, for some $\delta>0$.

For the inequality from the theory of Dirichlet series (this is also no more than calculus) I refer to the proof of Theorem III.2.1 of S. Mandelbrojt, Series de Dirichlet. Principes et methodes, Paris Gauthier-Villars, 1969. The only references that I know are in French, Russian and Ukrainian, sorry.

share|improve this answer
    
Could you provide a russian reference? –  smyrlis Dec 10 '13 at 8:10
1  
MR0584943 Leontʹev, A. F. Ryady eksponent. –  Alexandre Eremenko Dec 10 '13 at 14:42
    
The formula is not explicitly written there, but the estimate is contained in page 177 of Leontiev. –  Alexandre Eremenko Dec 10 '13 at 18:57
    
How is $c$ related to $\alpha$? –  smyrlis Dec 10 '13 at 20:38
    
Small $\alpha$ gives large $c$. According to my computation, $\alpha c$ is bounded above and below. –  Alexandre Eremenko Dec 10 '13 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.