Consider the space $C^\infty(0,\infty)$ of functions $f: (0,\infty) \rightarrow \mathbb{R}$ which have continuous derivatives of all orders. Equip $C^\infty(0,\infty)$ with the topology induced by the semi-norms $$p_n(f) = \max_{t \in [\frac{1}{n}, n]}| f^{(n-1)}(t)|, \quad n \in \mathbb{N}.$$ I would like to show that convergence of a net in $C^\infty(0,\infty)$, implies pointwise convergence. I have produced the following. Given a net $f_\tau$ in $C^\infty(0,\infty)$ that converges to $f$, it is true that $p_1(f_\tau - f) \rightarrow 0$. This is exactly the pointwise convergence in 1. By translation, the function converges pointwise also for all $t \geq 1$. But what about the points $t \in (0,1)$? Does someone have a clue?
Thanks very much!!
Let's prove that $f_\tau^{(n-k)}\to f^{(n-k)}$ uniformly on $[1/n,n]$ for $n\in\mathbb N$ and $1\leq k\leq n$ by induction over $k$. Otherwise replacing $f_\tau$ by $f_\tau-f$, we can assume $f=0$.
The case $k=1$ follows from $p_n(f_\tau)\to 0$. Assume for the induction step that $f_\tau^{(n-k)}\to 0$ uniformly on $[1/n,n]$. By the fundamental theorem of calculus, $$ \lvert f_\tau^{(n-1-k)}(x)\rvert=\left\lvert f_\tau^{(n-1-k)}(1)+\int_1^x f_\tau^{(n-k)}(y)\,dy\right\rvert\leq p_{n-k}(f_\tau)+(n-1)\sup_{y\in [1/n,n]}\lvert f_\tau^{(n-k)}(y)\rvert $$ for $x\in [1/n,n]$. The right side is clearly independent from $x$ and converges to $0$ by induction hypothesis.