In the test space $K$ of all infinitely differentiable compact supported functions, let $N_{0}$ be the neighborhood base at zero consisting of all sets of the form
$$U_{\gamma_{0}, \ldots , \gamma_{n}} = \{ \phi: \phi \in K, |\phi (x)| < \gamma_{0}(x), \ldots , |\phi^{(n)}(x)| < \gamma_{n}(x) \text{ for all x} \}$$
for some positive functions $\gamma_{0}, \ldots, \gamma_{n}$ continuous on $(- \infty, \infty)$. Prove that the topology generated in $K$ by $N_{0}$ leads to the same kind of convergence in $K$ as in Definition 1.
Definition 1. A sequence $(\phi_{n})$ of functions in $K$ is said to converge to a function $\phi \in K$ iff.
(i) there exists an interval outside which all the functions $\phi_{n}$ vanish; and
(ii) The sequence $(\phi_{n}^{(k)})$ of the kth derivative converges uniformly on this interval to $\phi^{(k)}$ for every $k = 0, 1, 2, \ldots$ .
I am lost in how to approach this problem. What I know is how a base generates a topology, namely a set $U$ is open iff. for each $x \in U$ there is a base element $B$ such that $x \in B$ and $B \subseteq U$.
Edit: So I am not sure whether I have taken the right approach to the problem but what I have done is to assume that $(\phi_{n}) \rightarrow \phi$ in the topology $\tau$ generated by $N_{0}$. And now I want to show that this will be the same as Definition 1 above.
By definition of convergence in a topological space this means that for every open neighborhood $U$ of $\phi$ there is an $N$ such that for each $n > N$ we have $\phi_{n} \in U$.
But this means that $U$ is open and hence by the base $N_{0}$ there is a base element $U_{\gamma_{0}, \ldots , \gamma_{n}}$ s. t. all $\phi, \phi_{n} \in U_{\gamma_{0}, \ldots , \gamma_{n}}$.
This means that $\phi \in K$ and $|\phi (x)| < \gamma_{0}(x), \ldots , |\phi^{(k)}(x)| < \gamma_{k}(x) \text{ for all x}$.
Similarly, $\phi_{n} \in K$ and $|\phi_{n} (x)| < \gamma_{n, 0}(x), \ldots , |\phi_{n}^{(k)}(x)| < \gamma_{n, k}(x) \text{ for all x}$,
for some positive functions $\gamma$ continuous on $(- \infty, \infty)$.
But since all $\phi_{n} \in K$, means that all $\phi_{n}$ are compactly supported and hence they vanish in some interval. This is condition (i) in Definition 1.
To show (ii) we must show that $(\phi_{n}^{(k)}) \rightarrow \phi^{(k)}$ uniformly. In other words, for all $\epsilon > 0$ there exists $N$ such that for all $n > N$ and all $x$ we have $|\phi_{n}^{(k)} - \phi^{(k)}| < \epsilon$.
And I am lost here but it does seem that I can somehow use the inequalities bounded by the functions $\gamma$ to show uniform convergence perhaps by the triangle inequality but I get lost in the inequalities.
Hopefully my approach and understanding of the problem is correct so far. Any suggestions and hints are welcomed.
Edit 2: I also thought about what if the functions $\gamma$ were uniform continuous since they are continuous. But continuity does not imply uniform continuity in general. However, continuity on a closed bounded interval does indeed imply uniform continuity. I thought about using the compact support of the functions $\phi$ and then perhaps restricting the functions $\gamma$ to this closed and bounded interval. Not sure if this helps.