Let $X$ be a vector space equipped with a norm $p$ and a seminorm $q$. Denote the completion of $X$ with respect to $p$ with $X\_p$ and with respect to $p+q$ by $X\_{p+q}$. Then the induced map $\iota : X\_{p+q} \to X\_p$ is well-defined and continuous but not necessarily injective as can be seen in analogy to [this answer](https://math.stackexchange.com/a/901717) to [Kernel of the Extension of a Bounded Linear Operator](https://math.stackexchange.com/questions/901579/kernel-of-the-extension-of-a-bounded-linear-operator) by taking $X = l^2$,
\begin{equation}
p(x) = \sqrt{\sum\_{n = 1}^\infty \left( \frac{x\_n - x\_{n+1}}{n} \right)^2}, \qquad
q(x) = \sqrt{\sum\_{n = 1}^\infty \left( \frac{x\_n}{n} \right)^2} \, .
\end{equation}
Then setting $x^m\_n = m/(m+n)$ we have $\lim\_{m \to \infty} p(x^m) = 0$, $x^m$ being Cauchy in $q$ and $\lim\_{n \to \infty} q(x^m) = \sqrt{\pi^2/6}$ such that the equivalence class of $x^m$ in $X\_{p+q}$ lies in the kernel of $\iota$ while being nonzero in $X\_{p+q}$.


On the other hand we can take the Schwartz space $X = \mathcal{S}(\mathbb{R})$ and let $p = \lVert \cdot \rVert\_{L^1}$ and $q = \lVert \cdot \rVert\_{L^2}$.
Then $\iota$ is certainly injective because any sequence $f\_n$ in $\mathcal{S}(\mathbb{R})$ with $\lim\_{n \to \infty} \lVert f\_n \rVert\_{L^1} = 0$ has a subsequence converging to zero almost everywhere such that if $f\_n$ is Cauchy in $L^2 (\mathbb{R})$, the corresponding limit $f$ in $L^2 (\mathbb{R})$ (which exists by completeness) has to be the zero function. Thus $\lim\_{n \to \infty} \lVert f\_n \rVert\_{L^2} = 0$ as well proving that the equivalence class of $f\_n$ is zero in $X\_{p+q}$.


In the latter example we have much more structure: $X$ is a reflexive, complete and metrisable nuclear space and $p$ and $q$ are continuous norms on $X$.
But I suspect that these conditions are not sufficient.


What are some necessary and sufficient conditions for $\iota$ to be injective?



When is a natural map between completions injective?
A simple sufficient condition for two norms $p\le r$ on $X$ to induce an *injective* continuous linear map $i:X\_r\to X\_p$ between the completions is that the unit ball $B\_r=\{x\in X: r(x)\le 1\}$ is $p$-closed.


Indeed, for $x\in X\_r$ with $i(x)=0$ choose a sequence $x\_n\in X$ with $x\_n\to x$ in $X\_r$. This sequence is $r$-Cauchy so that, for every $\varepsilon>0$, there is $n\_\varepsilon\in \mathbb N$ with $x\_n-x\_m\in\varepsilon B\_r$ for all $m,n\ge n\_\varepsilon$. The continuity of $i$ and $i|\_X=id\_X$ yield $x\_m=i(x\_m)\to i(x)=0$ in $X\_p$ and hence in $(X,p)$. For every $n\ge n\_\varepsilon$, this implies $$x\_n=\text{$p$-}\lim\limits\_{m\to\infty} x\_n-x\_m+x\_m\in \overline{\varepsilon B\_r}^p= \varepsilon B\_r.$$ This shows that $x\_n$ converges to $0$ in $(X,r)$ so that $x=0$.




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You can relax the assumption to $\overline{B\_r}^p \subseteq cB\_r$ for some constant. Moreover, this principle can be generalized to general uniform spaces. In one form or another you find it in most text books about locally convex spaces, for instance in Grothendieck's *Topological Vector Spaces*.


