# Completion with respect to stronger norm is no subset?

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. $$\forall x \in M: \| x \|_2 \leq c \|x\|_1$$ 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: $$(x_n) \sim_i (y_n) \Leftrightarrow \lim_{n \rightarrow \infty} \| x_n - y_n \|_i = 0$$ 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?

## 1 Answer

You are not missing anything. What you want to prove is just not true in general.

Here is a somewhat different perspective of what you want to show: The identity map $\iota : M_1 \to M_2 \subset \overline{M_2}$ is clearly a well-defined bounded linear map. Thus, it is not hard to see that there is a unique continuous extension $\overline{\iota} : \overline{M_1} \to \overline{M_2}$ of $\iota$. What you are asking is essentially whether $\overline{\iota}$ is always injective.

This is not so: To see this, consider the set $M := C^1([0,1])$, with $\|\bullet\|_2 = \|\bullet\|_\sup$ and $\|f\|_1 := \|f\|_2 + |f'(0)|$. Now, fix some $g \in C^1([0,\infty))$ with $g'(0) = 1$ and such that $g,g'$ are bounded. Set $g_n (x) := n^{-1} \cdot g(nx)$. We then have $\|g_n\|_{L^\infty([0,1])} \to 0$ (by boundedness of $g$), but $g_n'(x) = g'(nx)$, so that $|g_n'(0)| = 1$ for all $n \in \Bbb{N}$. Therefore, $$\| g_n - g_m \|_1 = \|g_n - g_m\|_{L^\infty([0,1])} + |g_n'(0) - g_m'(0)| \leq \|g_n\|_{L^\infty} + \|g_m\|_{L^\infty} \to 0,$$ so that $(g_n)_n$ is Cauchy with respect to both norms. By completeness of $\overline{M_i}$, we thus see $g_n \to g$ with respect to $\|\bullet\|_1$, and it is easy to see $g_n \to 0$ with respect to $\|\bullet\|_2$. Since we have $\|g_n\|_1 \geq |g_n'(0)| = 1$, we see $\|g\|_1 \geq 1$, i.e., $g \neq 0$. But $\overline{\iota}(g) = \overline{\iota}(\lim_n g_n) = \lim_n \iota(g_n) = \lim_n g_n = 0$, so that $\overline{\iota}$ is not injective.

The usual way to ensure that $\iota$ is injective is to already know that $\overline{M_i} \hookrightarrow X$ for a suitable Hausdorff space $X$ (typically, $X$ could be the space of measurable functions with convergence in measure, or the space of distributions, etc.). Note that one needs to know this for the completion to hold, not just for the spaces $M_i$ themselves. Indeed, in the above example, we have $M_i \hookrightarrow C([0,1])$, but this was not enough to conclude the injectivity of $\overline{\iota}$.

• I don't understand your last paragraph. Apart from that, good example! Dec 8, 2020 at 12:48
• @JossevanDobbendeBruyn: What I am trying to indicate there is a possible argument for showing (under additional assumptions) that $\overline{M_1} \subset \overline{M_2}$. Let us assume that there is some Hausdorff space $X$ for which you can show $\overline{M_i} \hookrightarrow X$ for $i = 1,2$. Then let $x,y \in \overline{M_1}$ with $\overline{\iota}(x) = \overline{\iota}(y)$. Dec 9, 2020 at 10:58
• There are sequences $(x_n)_n, (y_n)_n \subset M_1$ with $x_n \to x$, $y_n \to y$, with convergence in $\overline{M_1}$ and hence in $X$. By continuity of $\iota$, also $x_n = \iota(x_n) \to \overline{\iota}(x)$ and $y_n = \iota(y_n) \to \overline{\iota}(y)$, with convergence in $\overline{M_2}$ and hence in $X$. Since $X$ is Hausdorff, limits are unique, so that $\overline{\iota} (x) = \lim_n x_n = x$ and $y = \lim_n y_n = \overline{\iota} (y) = \overline{\iota} (x) = x$, which proves injectivity of $\overline{\iota}$. Essentially, this argument shows that $\overline{\iota}$ is the identity. Dec 9, 2020 at 10:58