# Smooth function becomes analytic

Let $f$ be a smooth function ,defined on unit interval $[0,1]$.Moreover $\Vert f^{(k)}\Vert_2\leq \alpha,\:\forall k\in\mathbb{N}_o$. Can we conclude that $f$ is analytic. More generally when $C^{\infty}([0,1])\cap W^{\infty, 2}([0,1])$ contains only analytic functions?

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More specifically, if $$\limsup_{n\to\infty} \frac {\|f^{(n)}\|^{1/n}_{L^2}} {n}=s<\infty,$$ then $f$ can be extended analytically in $$\Omega=\{x+iy: x\in(0,1),\,\lvert y\rvert<1/(s\mathrm{e})\}\subset\mathbb C.$$ In the case $\|f^{(n)}\|^{1/n}_{L^2}\le a$, it extends analytically in $(0,1)\times\mathbb R$.
It does not extend however to an analytic function in the case $s=\infty$.
Sketch of proof. First observe that if $\|f^{(n)}\|_{L^2}=a$, for every $n$, then $\|f^{(n)}\|_{L^\infty}\le b$, for a suitable $b>0$. Then define the power series $$\varphi(x,y)=\sum_{n=0}^\infty \frac{f^{(n)}(x)(iy)^n}{n!}.$$ This series converge, for $|y|<1/{\mathrm e}(s)$, satisfies Cauchy-Riemann equations and agrees with $f$, for $y=0$.
I saw your paper and now have a question.Let say $s=\infty$ then it seems to me that we get an extension to the the whole complex plane.(Because radius of convergence is infinite) Is this true? do we end up with an entire extension? – BigM Mar 6 '14 at 4:37
If $s=0$, then the function extends to an entire analytic. If $s=\infty$ it does not extend to analytic. – Yiorgos S. Smyrlis Mar 6 '14 at 7:48