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?

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    $\begingroup$ 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. $\endgroup$ Feb 16 at 11:55
  • $\begingroup$ @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 $X$. $\endgroup$
    – SoupMath
    Feb 19 at 14:54
  • $\begingroup$ You are right. I have mixed up things. The mapping does not need to be injective. See my answer. $\endgroup$ Feb 19 at 17:21

1 Answer 1


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 $0.$


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