I am trying to prove the following statement from the book Spectral Theory and Quantum Mechanics by V. Moretti:
A sequence $\{x_n\}_{n\in N} \subset X$ is Cauchy for a distance $d$ in a locally convex metrisable space $X$ if and only if it is Cauchy for every seminorm $p$ generating the topology: for every $\epsilon > 0$ there is $N_\epsilon^{(p)} \in \mathbb N$ such that $p(x_n −x_m ) < \epsilon$ whenever $n,m > N_\epsilon^{(p)} $. Consequently, completeness does not actually depend on the distance used to generate the locally convex topology.
How can we prove this?
If a sequence is Cauchy for $d$, then it will eventually lie in some ball $B_{d,\delta}(x)$ for any $\delta>0.$ Somehow we need to use this fact to show that it will eventually lie in some ball $B_{p,\epsilon}(y)$ for any fixed $p\in P,\epsilon>0.$ I'm sure the result will rely on $d$ and $P$ generating the same topologies, but I don't see how to link the two. We can always nest a metric-open set in a seminorm-open set, and vice versa, but this still doesn't lead me to an obvious solution.
This post contains a proof that completeness of any metric generating the same topology as $P$ guarantees completeness of all such metrics. But the statement here involves seminorms, so it is not an equivalent claim, from what I can tell.
I am assuming that the metric is translation invariant. Let $(x_n)$ be Cauchy in the metric and $\epsilon >0$. If $p$ is a semi norm generating the topology then $\{x: p(x) <\epsilon\}$ contains an open all $B_d(0,\delta)$. For $n,m$ sufficiently large $x_n-x_m \in B_d(0,\delta)$ and hence $p(x_n-x_m) <\epsilon$.
Converse follows from the fact that $B_d(0,\epsilon)$ contains a set of the type $\{x:p_i(x) <\epsilon_i, 1\leq i \leq N\}$ for some positive integer $N$, some posotive numbers $\epsilon_i$ and some $p_i$'s.