Normed linear space with two norms that are not equivalent, one is complete, what about the other? I have been searching for an answer to the following question:

Given a normed linear space $V$ and two norms that are not equivalent, but $\exists K\in\mathbf{R}$ such that $\|v\|_1\leq K\|v\|_2$ for all $v\in V$. We know that $(V,\|\cdot\|_2)$ is complete, i.e. a Banach space. Is $(V,\|\cdot\|_1)$ also complete?

Some related questions posted so far are:
(i) Example of two norms on same space, non-equivalent, with one dominating the other
(ii) If two normed spaces are Lipschitz equivalent, then one if complete iff the other is
In (i) there is a good example from julien, but I am unable to relate it to Cauchy sequences. 
Could someone please shine some light on this, please?
 A: Part 1: (There are examples in which is complete under $||\cdot||_2$ and not under $||\cdot||_1$) 
Take $V:=C[0,1]$, $||f||_1:=\int_{0}^{1}f^2$, and $||f||_2:=\sup|f|$. Then $||\cdot||_1\leq||\cdot||_2$. While $V$ is complete under $||\cdot||_2$ we can get a sequence of continuous functions converging to $\chi_{[1/2,1]}$ under $||\cdot||_1$.
Part 2: (It is not possible for $V$ to be complete under $||\cdot||_1$) 
Assume that $V$ were complete under $||\cdot||_1$. Consider the identity function $I:(V,||\cdot||_2)\rightarrow (V,||\cdot||_1)$. Since $||I(f)||_1=||f||_1\leq K||f||_2$, we have that $I$ is a continuous linear map. Since it is also surjective (and we are assuming $(V,||\cdot||_2)$ and $(V,||\cdot||_1)$ to be Banach) we get, by the open mapping theorem, that $I$ is open.This means that $I^{-1}:(V,||\cdot||_1)\rightarrow (V,||\cdot||_2)$ is continuous. Since $I^{-1}$ is linear, then is is also bounded: $$||f||_2=||I^{-1}(f)||_2\leq C||f||_1$$ for some constant $C>0$.
Therefore $\frac{1}{C}||\cdot||_2\leq||\cdot||_1\leq K||\cdot||_1$. This means that the norms would be equivalent. But this is a contradiction with the fact that they are not. Therefore $V$ cannot be complete under $||\cdot||_1$.
