Norm equivalence and Banach spaces If $(X,\|\cdot\|_1)$ is a banach space, and $\|\cdot\|_1$ is equivalent to $\|\cdot\|_2$, then $(X,\|\cdot\|_2)$ is a banach space.
Does it also hold that if $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$ are both banach spaces that the norms are equivalent?
I suspect so, but I can't find it in the literature.
 A: Yes, it's clear that it's normed. I'll assume that it's also a vector space (ie that it's equipped with the same linear operator). What's left to show is that it's also complete.
But since the norms are equivalent the criterion on being a Cauchy sequence is the same, and the criterion for a sequence having a specific limit is the same.
To be more precise if $x_n$ is a Cauchy-sequence in $(X, ||\cdot||_2)$ it means that given $\epsilon>0$ we have $||x_j-x_k||_2<\epsilon$ if $j$ and $k$ is large enough, but that will mean that $||x_j-x_k||_1<K\epsilon$ for some constant and therefore you have that $x_n$ is convergent in $(X, ||\cdot||_1)$ (call it's limit $x$) which means that for every $\eta>0$ you have that $||x_j-x||_1<\eta$ which means that $||x_j-x||_2<L\eta$ for some constant $L$. Therefore $\lim x_n = n$ in $(X, ||\cdot||_2)$ as well.
However it does not hold that if two spaces are Banach spaces that their norm would be equivalent. Take for example the space of absolutely summable sequences and take $\sum |x_n|$ as one norm, since it's absolutely summable also the sum $\sum |x_n|/n$ would be convergent, but now you can consider the sequence $\delta_j$ (that's zero, except it's $j$th term is $1$). You have that $||\delta_j||_1=1$, but $||\delta_j||_b = 1/j$ you can't have the equivalence between these norms.
