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

Question

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.

Answer 1

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.

A closed globally defined operator is automatically continuous (closed graph theorem). You can do the same with the inverse map (also the identity but with domain and image reversed). Thus if the limits of mutually Cauchy sequences agree, the two norms must be equivalent.

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 $\|x_n-x^*\|_1 + \|x_n-x^*\|_2\to 0$.