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Suppose we have two normed linear space $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$. Also, the norms are equivalent to each other. Do we have same completion of $X$ with respect to $1$ and $2$ norms? In other words, do we have an isometric isomorphism between between the completion of $(X,\|\cdot\|_1)$ and the completion of $(X,\|\cdot\|_2)$? Thanks in advance for your help.

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    $\begingroup$ Hint: how are the "Cauchyness" and "converge to the same point" relation induced by these two norms related? $\endgroup$ Aug 1, 2014 at 19:08
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    $\begingroup$ If the two norms on $X$ are equivalent but not identical, the canonical topological isomorphism between the completions is not an isometry (the identity on $X$ is already not). But there is a unique isomorphism between the completions which is the identity on $X$. $\endgroup$ Aug 1, 2014 at 19:13

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Thank you for the hints. I really mean, norms are equivalent but not identical. Denoting completions of $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$ as $V_1$ and $V_2,$I am trying to get the isomorphism between $V_1$ and $V_2$. Let $x\in V_1$, then we have a sequence $x_n$ in $(X,\|\cdot\|_1)$ with $x_n\to x$ . But, norms are equivalent, so $x_n$ is cauchy in $(X,\|\cdot\|_2)$ and there is an $y\in V_2$ such that $x_n \to y$ in $V_2$. So, for every $x\in V_1$ , we have $y\in V_2$. In this way, we can form the isomorphism. If $x \in X \subset V_1$, then we can take the constant sequence and $x \in V_2$. So, on $X$, it is identity. This isomorphism is never isometric.

I really appreciate your comments.

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