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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.

EquivalentBanach
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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.

skyking
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    Sorry, my question was: If $(X,|\cdot|_1)$ and $(X,|\cdot|_2)$ are both Banach spaces, are $|\cdot|_1$ and $|\cdot|_2$ equivalent, is that what you answered here? $$$$ I.e. both norms on $X$ make $X$ banach, but are they thus equivalent?

    $$$$

    I asked because of the two norm equivalence theorem saying that if $(X,|\cdot|_1)$ and $(X,|\cdot|_2)$ have that with norm $1$ it is Banach, and the norms are equivalent than $(X,|\cdot|_2)$ is banach

     – EquivalentBanach Dec 17 '15 at 06:55
  • The space of absolutely summable sequences in Banach with both of those norms? – EquivalentBanach Dec 17 '15 at 07:03
  • The original comments said that the problem seems to have no constructive counterexample: http://mathoverflow.net/questions/184464/are-banach-space-norms-up-to-equivalence-unique – EquivalentBanach Dec 17 '15 at 07:04
  • @EquivalentBanach: Your link answers a far stronger question - that for every Banach space X, there exists a nonequivalent norm on X that is also complete. Here you need an unbounded linear functional, whose existence in an arbitrary (infinite dimensional) space requires the axiom of choice. For concrete spaces like $\ell_1,$ you can construct explicit examples. – ktoi Dec 17 '15 at 07:47
  • @ctoi There question is exactly what I intended, so I suppose the answer is, yes, sometimes you can, but not in general. Thanks – EquivalentBanach Dec 17 '15 at 07:49