Let $X $ be an infinite dimensional $R$-vector space, suppose that $||\cdot||_1$ and $||\cdot||_2$ are two norms that makes $X$ into a Banach space. Let $||\cdot||_3 = ||\cdot||_1 + ||\cdot||_2 $ ,this is another norm on $X$.
Is $X$ complete wrt $||\cdot||_3 $?
It is obvious that if $\{x_n\}_{n\in \mathbb{N}} \subset X$ is $||\cdot||_3 $-Cauchy sequence then it is also $||\cdot||_1 $-Cauchy and $||\cdot||_2 $-Cauchy. Therefore there exist $x_1^*, x_2^*\in X$ such that $x_n \overset{||\cdot||_1}{\to} x_1^*$ and $x_n \overset{||\cdot||_2}{\to} x_2^*$ but a priori $x_1^*$ can be different from $x_2^*$.
Can we prove that $x_1^* = x_2^*$ in general? Otherwise can we find a counterexample?
I only managed to show this if $||\cdot||_1 < C ||\cdot||_2$, thanks to the fact that $X$ is T2 (or in another manner this condition implies that the two Banach norms are equivalent). Infact if $x_1^*\neq x_2^*$ then for $n$ big enough, $x_n $ must belong to a neighbourhood $U(x_1^*)$ of $x_1^*$ (neighbourhood in both the topologies) disjoint from another neighbourhood (in both topologies) of $x_2^*$. Unfortunately I cannot generalize this proof since in the general case the two topologies are not comparable.