Smooth function becomes analytic

Question

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?

Answer 1

Yes.

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 $s=\infty$.

For proof see G. Akrivis, D. T. Papageorgiou, and Y.-S. Smyrlis On the analyticity of certain dissipative–dispersive systems, Bull. London Math. Soc. (2013) 45 (1): 52-60.

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 $y=0$.