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 $[0,1]$ ?

Solution 1:

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 $(-\infty,0)$.