TITLE: Norms are equivalent if one hand side of inequality holds
QUESTION [5 upvotes]: Let $(X,\|.\|_1)$ and $(X,\|.\|_2)$ are Banach spaces and $\forall x \in X$. Show that if  $\|.\|_1 \le k \|.\|_2 $ for $\exists k \gt 0$ then $\|.\|_1$ and $ \|.\|_2 $ are equivalent.
I could think only belows 
$(x_n)$ is an arbitrary Cauchy Sequence in $X$ and since it is Banach wrt $ \|.\|_2 $ thus $\exists x \in X$ such that
$\|x_n-x\|_2 \lt \varepsilon /k$
From inequeality we have
$\|x_n-x\|_1 \le k\|x_n-x\|_2 \lt \varepsilon$  hence $x_n \to x$ wrt $\|.\|_1$ norm. 
I cannot continue and I think there are some mistakes and deficiencies my writtens above. 
I will be apreciated for any help

I have used a version of Banach Isomorphism Theorem thanks to your comments :
Let $I:(X,\|.\|_2) \to (X,\|.\|_1)$ identity map. It is (1-1) and onto. 
$\|I(x)\|_1 = \|x\|_1 \le k\|x\|_1 $ thus $I$ is bounded and equivalently continuous.
By Banach Isomorphism Theorem it is a homeomorphism hence $I^{-1}$ is continuous and equivalently bounded.
We can find $\exists c \gt 0$ such that
$\|I^{-1}(x)\|_2 = \|x\|_2 \le c\|x\|_1 $
Finally we get the norms are equivalent 
Could you please check it?

REPLY [2 votes]: You just need to prove that $id: (X,\lVert \bullet\rVert_2)\to (X,\lVert \bullet\rVert_1)$ is continuous. In other words, by the closed graph theorem, that the set $\Delta^2_1:=\{(x,x)\,:\, x\in X\}$ is a closed subset of the Banach space $(X\times X,\lVert \bullet\rVert_{1\otimes 2})$, where $\lVert (x_1,x_2)\rVert_{1\otimes 2}=\lVert x_1\rVert_1+\lVert x_2\rVert_2$. This is the case if and only if the normed space $(\Delta^2_1,\lVert \bullet\rVert_{1\otimes 2})$ is a Banach space.
The fact that $id': (X,\lVert \bullet\rVert_1)\to (X,\lVert \bullet\rVert_2)$ is continuous gives you that $(\Delta^1_2,\lVert \bullet\rVert_{2\otimes 1})$ is a Banach space. But the tautological map $\Delta^1_2\ni (x,x)\mapsto (x,x)\in\Delta^2_1$ is a bijective isometry. Thus, $\Delta_1^2$ is Banach as well.