Given a countable family of semi-norms $p_i$, we can define a metric
$d(f,g) = \sum \limits_{i=0}^{\infty} 2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$
We have the locally convex topology induced by the semi-norms as above, as well as the topology induced by the metric.
How does the proof work to show their equality?
I am known to the proof of Rudin (Functional analysis), but utilizes a different metric:
$d(f,g) = \max \limits_{i \in \mathbb N} 2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$
Whereas the proof for this metric is fairly easy - you can handle value of the sequence on its own - i do not see how a similar proof might work for the first metric.
One guess would be to show equivalence of both metrics, but I don't see even that, as on $l^1$, the sum-norm-topology is strictly finer than the max-norm-topology.
Can you help me?