Proof that Frechet-metric generates same topology as the semi-norms

Question

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?

Answer 1

To prove equivalence of the topologies you can use the fact that both topologies are characterized by their convergent sequences. This is true in any first countable space. Metric spaces are first countable, and the topology induced by a countable family of seminorms is first countable. Here, since you have translation invariance, it is enough to check sequences converging to 0 (although this simplification is not essential). That is, you can show that if $x_1,x_2,\ldots$ is a sequence in your vector space, then $d(x_n,0)\to 0$ as $n\to\infty$ if and only if for all $i$, $p_i(x_n)\to0$ as $n\to\infty$. For the left to right direction, you can use the fact that $\frac{p_i}{1+p_i}\leq 2^id$. For the other direction you can first bound the tail, then work with the remaining finitely many terms.

There is no actual need to consider sequences. You basically want to show that the identity map is a homeomorphism, and this reduces to showing it is continuous at 0 in both directions, which in any case involves the same estimates you would make when working with sequences. (In other words, considering first countability a priori isn't necessary. If you really want to, you could instead work with nets.) Note that Jonas T's answer suggests a complementary approach (that was given around the same time), working directly with neighborhoods of $0$ instead of with sequences.

Answer 2

Assuming that you forgot a $2^{-i}$ in the definition of $d$ we argue as follows:

We know that a Hausdorff TVS is metrizable iff zero has a countable neighborhood basis. If this is true then the topology is even generated by a translation invariant metric.

Another theorem states that a topology on a TVS is locally convex iff it is generated by a family of seminorms.

So, if we combine these and use that your metric is translation invariant, we obtain the result.

However, this is indirect, probably there is a direct proof.

Edit: Oh, I don't think the direct proof is so hard, I will see if I have time to add it. You should just observe that the neighborhood $\{x : p_n(x) \leq 2^{-k}\}$ contains the set $\{x : |x| \leq 2^{-m-k-1} \}$ and that the set $\{x : |x| \leq 2^{-k} \}$ contains the neighborhood $\{x : p_{k + 1}(x) \leq 2^{-k - 2} \}$. Where $|x| := d(x,0)$.