If $V$ is completely normable, then is every norm complete? Here is a theorem that motivated my question.

Let $(V,||\cdot||_V)$ be a normed space over $\mathbb{K}$. Then, there exists a Banach space $(X,||\cdot||_X)$ such that $V$ is dense in $X$ and $||\cdot||_X$ is an extension of $||\cdot||_V$.

Let $(V,\tau)$ be a normable space over $\mathbb{K}$.
Assume that there exists a norm $||\cdot||$ on $V$ such that $(V,||\cdot||)$ is a Banach space and $||\cdot||$ induces $\tau$.
Now, let $||\cdot||_0$ be an arbitrary norm on $V$ that induces $\tau$. Then, is $(V,||\cdot||_0)$ a Banach space? Or if not, is there an extension $X$ of $V$ such that $(X,||\cdot||_X)$ is a Banach space for every norm $||\cdot||_X$ on $X$?
Just like there exists an algebraic closure of a field (which cannot be algebraically extended further), I'm curious whethere there is a Banach-closure of a normable space.
 A: Let $||\cdot||_0$ and $||\cdot||_1$ induce the same topology on a vector space $V$. Let $r>0$, since $B_r(0)_0$ is open in the topology induced by $||\cdot||_0$, it must be open in the one induced by $||\cdot||_1$. This means there exists a $k>0$ so that $B_k(0)_1 \subset B_r(0)_0$.
Now any $x \in V, x \neq 0$ can be written as $x=||x||_0\frac{x}{||x||_0}$. Take $r=2$ and $||x||_1=||x||_0\cdot ||\frac{x}{||x||_0}||_1$. The term in the $||\cdot||_1$ norm lies in $B_2(0)_0$, so it also lies in $B_k(0)_1$ for some $k$ independent of the choice of $x$. So we have:
$$||x||_1 ≤ k\cdot||x||_0$$
for all $x \in V$. It follows from the same argument that there exists a $h>0$ so that $||x||_0≤h\cdot||x||_1$ for all $x \in V$.
These two inequalities have as a consequence that the question of convergence/Cauchy property of a sequence are identical for both norms, ie if $V$ is complete wrt $||\cdot||_0$ it is complete wrt $||\cdot||_1$.
A: No. Take $X=C([0,1])$. Give $X$ the sup norm to make it a Banach space. But if we give $X$ the norm from $L^2([0,1])$ and $X$ is incomplete with this norm.
