# Which rate of growth of the Sobolev norms guarantees analyticity?

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 $\alpha>0$.

-
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

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.

-
Could you provide a russian reference? –  smyrlis Dec 10 '13 at 8:10
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