Completion of a subspace of a Banach space Let $(X,\|\cdot\|_X)$ be a Banach space and let $A\subset X$ be a subspace. Define a norm $\|\cdot\|_1$, which is stronger than $\|\cdot\|_X$ on $A$, that is,
$$\forall x\in A,\quad\|x\|_X\leq C\|x\|_1$$
Let $\hat{A}$ be the completion of $A$ with respect to $\|\cdot\|_1$. Can we say $\hat{A}\subset X$?
Similar to Completion with respect to stronger norm is no subset?, we set the $\hat{A}$ (the completion of $A$ with respect to $\|\cdot\|_1$) to be the set $C(A)$ of all $\|\cdot\|_1$ Cauchy sequences in $A$ modulo the equivalence relation
$$(x_n)\sim_1(y_n)\Leftrightarrow\lim_{n\rightarrow\infty}\|x_n-y_n\|_{1}=0$$
That is $\hat A=C(A)/\sim_1$. Now one could identify $\hat X=C(X)/\sim_X$ with $X$. Clearly $C(A)\subset C(X)$, since all $\|\cdot\|_1$ Cauchy sequences in $A$ are $\|\cdot\|_{X}$ Cauchy sequences in $X$, but one would have to extend this injective mapping
$$i: (C(A),\|\cdot\|_1)\rightarrow (C(X),\|\cdot\|_X)$$
to an injective map (why injective btw?)
$$j: C(A)/\sim_1 \rightarrow C(X)/\sim_X,$$
and it does not seem like we can guarantee injectivitiy, that is, $i((x_n))\sim_X i((y_n)) \Rightarrow (x_n)\sim_1(y_n)$ for all $(x_n),(y_n) \in A$. Am I missing something obvious?
 A: Assume $X$ is an infinite dimensional Banach space. Then there exists an unbounded linear functional $\varphi$ on $X.$ This means there is a sequence $y_n\in X$ such that $\|y_n\|=1$ and $|\varphi(y_n)|\to \infty.$ Let $x_n={y_n\over \varphi(y_n)}.$ Then $$\varphi(x_n)=1,\qquad \|x_n\|\to 0.$$ We introduce the new norm $\|\cdot \|_1$ on $X$ $$\|x\|_1=\|x\|+|\varphi(x)|.$$
The space $X$ with norm $\|\cdot\|_1$ is not complete. Indeed, the sequence $\{x_n\}$ satisfies the Cauchy condition with respect to this norm $$\|x_n-x_m\|_1=\|x_n-x_m\|\le \|x_n\|+\|x_m\|\to 0.$$ But this sequence is not convergent. Indeed, assume, for a contradiction, that  $\|x_n-x\|_1\to 0.$ Then $\|x_n-x\|\to 0,$ i.e. $x=0.$ Therefore $\|x_n\|_1\to 0.$ However $\|x_n\|_1=\|x_n\|+1\ge 1,$ a contradiction.
Therefore the completion $\widehat{X}$ of $X$ with respect to $\|\cdot\|_1$ is not contained in $X.$
We can define a bounded linear map from $\widehat{X}$ to $X$ as follows. The space $\widehat{X}$ consists of equivalence classes of  sequences $(u_n)_{\sim},$ $u_n\in X,$ such that $\|u_n-u_m\|_1\to 0.$ For $(u_n)_{\sim}\in \widehat{X}$ we have $\|u_n-u_m\|\to 0,$ hence $u_n$ is convergent in $(X,\|\cdot \|)$ to some $u\in X,$ independent of the choice of the sequence in the equivalence class. The map $\widehat{X}\ni (u_n)_{\sim}\longmapsto u\in X$ is linear and bounded. However it is not injective. Indeed the element $(x_n)_\sim$ is mapped to $0.$
