Suppose $(B,\|\cdot\|)$ is a Banach space, $V\subset B$ a dense subspace, and $V$ is equipped with a norm $\|\cdot\|\_V$ such that $\|x\|\_V = \|x\|$ for all $x\in V$.


Is $(B,\|\cdot\|)$ a completion of $(V,\|\cdot\|\_V)$ with respect to the $\|\cdot\|\_V$ topology? I.e., can the spaces be considered the same/identical up to some isometry?



Norms agreeing on dense subspace
This is standard, and the answer has been been indicated in the comments. Recall that in a normed space $X,$ the triangle inequality easily yields that $| \|x\| - \| y \| |\leq \|x-y\|$
for all $x,y \in X.$


Now let's turn to your dense subspace $V$ of the Banach space $B.$ Take an element $b \in B.$ There is a sequence $(v\_{n})$ of elements of $V$ such that $v\_{n} \to b$ (with respect to the norm on $B).$ Then by the above remark, the sequence $( \|v\_{n} \|)$ is a Cauchy sequence of real numbers whose limit is the real number $\| b \|$. Hence $\| b \|$ is uniquely specified in terms of $ \| \|\_{V} $ since you assume that $\| \|$ and $\| \|\_{V}$ agree on $V.$ Also, $\|b\|$
is the same as would be assigned if considering $b$ as an element of the completion of $V.$
Hence $B$ embeds isometrically as a susbpace of the completion of $V.$ However, it is clearly dense in that completion, as $V$ already is, and it is a closed subspace of that completion, since any Cauchy sequence of elements of $V$ has a limit which lies in $B.$ Hence $B$ is indeed isometrically isomorphic to the completion of $V.$


