Let $f_n : [0, 1] \to \mathbb R$, a sequence of $C^\infty$ functions, and $f : [0, 1] \to \mathbb R$.
If $f_n$ converges pointwise/uniformly to $f$ then $f$ is also $C^\infty$.
I think that if $f_n$ converges pointwise only to $f$ then $f$ is not $C^\infty$ : counter-example : $x^n$ to converges pointwise to $0$ if $x \neq 1$ and $1$ if $x=1$. All $x^n$ are $C^\infty$ but not $f$.
For the uniformly part, I think it is true but I do not know how to deal with all the derivatives and show they are continuous.
Best Answer
No, it is not true. Take $f_n(x)=\sqrt{\frac1n+\left(x-\frac12\right)^2}$. Then each $f_n$ is a $C^\infty$ function. But $(f_n)_{n\in\Bbb N}$ converges uniformly to $f\colon[0,1]\longrightarrow\Bbb R$ with $f(x)=\left|x-\frac12\right|$, which is not differentiable.