Existence of smooth extension of a function defined on a closed interval Suppose $ f: [0,1] \rightarrow \mathbb{R}$ is a function such that derivatives of all orders exist ( at the end points of the interval the appropriate one-sided derivatives exist) and are continuous $ \forall x \in [0,1] $ . How to prove that $f$ is smooth, in the sense that it admits a $C^\infty$  extension to an open interval containing the interval $[0,1]$ ?
 A: I am only three years late, so I am probably just wasting time.... but it is an interesting  question, so....
I will only prove that $f$ can be extended to $(-\infty,0)$. The case
$(1,\infty)$ is similar. Construct a $C_{c}^{\infty}(\mathbb{R})$ function
$\psi$ such that $\psi=1$ in $[0,\frac{1}{2}]$ and $\psi(x)=0$ for
$x\geq\frac{3}{4}$ and define the function $g(x):=f(x)\psi(x)$. By
extending $g$ to be zero for $x\geq1$, we have that $g$ is $C^{\infty}$ in
$[0,\infty)$ and $g=f$ in $[0,\frac{1}{2}]$. Let $\phi:[1,\infty
)\rightarrow\mathbb{R}$ be a continuous function be such that
$$
\int_{1}^{\infty}\phi(t)\,dt=1,\quad\int_{1}^{\infty}t^{n}\phi(t)\,dt=0,\quad
\lim_{t\rightarrow\infty}t^{n}\phi(t)=0
$$
for all $n\in\mathbb{N}$. For $x<0$ define
$$
h(x)=\int_{1}^{\infty}\phi(t)g(x(1-t))\,dt.
$$
Note that since $t\geq1$, $x-tx>0$ and so $h$ is well-defined. Since $g$ is
bounded, by the Lebesgue dominated convergence theorem,
$$
\lim_{x\rightarrow0^{\_}}h(x)=\int_{1}^{\infty}\phi(t)\lim_{x\rightarrow
0^{\_}}g(x(1-t))\,dt=g(0)\int_{1}^{\infty}\phi(t)\,dt=1g(0)=f(0).
$$
Since all the derivatives of $g$ are bounded, by the Lebesgue dominated
convergence theorem, we can differentiate under the integral sign to get that
for $x<0$,
$$
h^{(n)}(x)=\int_{1}^{\infty}\phi(t)(1-t)^{n}g^{(n)}(x(1-t))\,dt.
$$
Again by the Lebesgue dominated convergence theorem,
\begin{align*}
\lim_{x\rightarrow0^{\_}}h^{(n)}(x)  & =\int_{1}^{\infty}\phi(t)(1-t)^{n}%
\lim_{x\rightarrow0^{\_}}g^{(n)}(x(1-t))\,dt=g^{(n)}(0)\int_{1}^{\infty
}(1-t)^{n}\phi(t)\,dt\\
& =1g^{(n)}(0)=f^{(n)}(0),
\end{align*}
where by the binomial theorem
$$
\int_{1}^{\infty}(1-t)^{n}\phi(t)\,dt=\int_{1}^{\infty}\phi(t)\,dt+\sum
_{k}c_{k}\int_{1}^{\infty}t^{k}\phi(t)\,dt=1+0.
$$
This shows that $h$ is a $C^{\infty}$ extension of $f$ to $(-\infty,0)$.
