Let $M$ be a normed vector space equipped with two norms $\| \cdot \|_1$ and $\| \cdot \|_2$, where $\| \cdot \|_1$ is stronger than $\| \cdot \|_2$, i.e. \begin{equation} \forall x \in M: \| x \|_2 \leq c \|x\|_1 \end{equation} Let now $\bar{M}_i$ denote the completion (cp. Reed Simon, Thm. I.3) of $M$ with respect to $\| \cdot \|_i$.
Now I wanted to show, that $\bar{M}_1 \subset \bar{M}_2$ holds.
But somehow, the proof does not work. Therefore recall, how the completion works:
The completion of $M$ w.r.t. $\| \cdot \|_i$ is defined as the set $B_i$ of all sequences that are Cauchy in $M$ w.r.t. $\| \cdot \|_i$ modulo the equivalence relation: \begin{equation} (x_n) \sim_i (y_n) \Leftrightarrow \lim_{n \rightarrow \infty} \| x_n - y_n \|_i = 0 \end{equation} I.e. $\bar{M}_i = B_i /\sim_i$.
We now have $B_1 \subset B_2$, since $\| \cdot \|_2$ is weaker than $\| \cdot \|_1$. Let $i: B_1 \hookrightarrow B_2$ be the corresponding injective embedding. Now $i$ does descend to an injective map $j: B_1 / \sim_1 \hookrightarrow B_2 / \sim_2$ if and only if $\forall (x_n), (y_n) \in B_1: (x_n) \sim_1 (y_n) \Leftrightarrow i( (x_n)) \sim_2 i((y_n))$ ($\Rightarrow$ is needed for the existence and $\Leftarrow$ for injectivity).
But since $\| \cdot \|_2$ is weaker than $\| \cdot \|_1$, $ \Leftarrow$ does not hold in general. Hence in general $\bar M_1 \subset \bar M_2$ does not hold.
This seems for me quite strange, especially, since the closure of a dense subspace in a complete space is a completion, and here the assertion holds (i.e.: Let X be a normed space, which is complete w.r.t. $\| \cdot \|_i$ and $\| \cdot \|_1$ is stronger than $\| \cdot \|_2$. Let $Y \subset X$. Then the closure of Y in X w.r.t. $\| \cdot \|_2$ contains the closure of Y in X w.r.t. $\| \cdot \|_1$). What am I missing?