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?

• I would argue as follows. If $(x_n)\in \widehat{A},$ then $x_n$ is a Cauchy sequence with respect to the norm in $X.$ Hence it is convergent, say, to $x.$ In this way we obtain a mapping from $\widehat{A}$ to $X.$ This mapping is linear. It is also injective, because if $(x_n)=0$ in $\widehat{A},$ then $x_n\to 0$ in $X.$ The correspondence $(x_n)\to x$ is a contraction as $\|x\|=\lim\|x_n\|\le \liminf\|x_n\|_1=\|(x_n)\|_{\widehat{A}}.$ I hope it is correct. Feb 16 at 11:55
• @BazyliZuczek Thank you for your answer. For injectiveness, should we not show if $x=0$, then $(x_n)=0$ in $\widehat{A}$? But I guess this is just to say $(x_n)\sim_{1} 0$ i.e. $\lim_{n\rightarrow \infty} \|x_n\|_1=0$, however this does not follow from $x_n\rightarrow 0$ in $X$. Feb 19 at 14:54
• You are right. I have mixed up things. The mapping does not need to be injective. See my answer. Feb 19 at 17:21

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 $$0.$$