# Banach space with respect to two norms must be Banach wrt the sum of the norms?

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.

Let $f: X_1\to X_2$ be the identity map. The statement that for any sequence $x_n$ so that $x_n\to_1 x_1$ and $f(x_n)\to_2 x_2$ converges one has $x_2=f(x_1)=x_1$ is the same as the statement that $f$ is a closed operator.
There exist however vector spaces that have two inequivalent Banach norms. Note that $\ell^1(\Bbb N)$ and $\ell^2(\Bbb N)$ both have cardinality $\mathfrak c$, thus the cardinalities of Hamel basis they have are bounded by $\mathfrak c$. However both are infinite and the minimal dimension of infinite Banach spaces is $\mathfrak c$, so there is a bijection between the Hamel basises. A bijection between the two basises induces a linear isomorphism between the two spaces, so view them as the same linear space given two different norms.
There however exists no continuous isomorphism between $\ell^1$ and $\ell^2$.
So there are sequences that have different limits, even though they are Cauchy in both norms. Then there cannot be any $x^*$ so that $\|x_n-x^*\|_1 + \|x_n-x^*\|_2\to 0$.