Let $X$ a linear space with two norms $\|x\|_1\cdot\|x\|_2$. Also $X$ is complete with respect to both norms. If $\exists M>0$ (a finite constant) such that $\|x\|_2 \leqslant M\|x\|_1$ ,$\forall x \in X$, then $\exists N>0$ such that $\|x\|_1 \leqslant N\|x\|_2$ ,$\forall x \in X$.
This is my proof:
We define the linear operator $T:(X,\|\cdot\|_1) \longrightarrow (X,\|\cdot\|_2)$ to be $Tx=x$.
$T$ is a bigective linear operator.
Also $T$ is continuous,because $\exists M>0$ such that $\|Tx\|_2=\|x\|_2 \leqslant M\|x\|_1$, $\forall x \in X$
From open mapping theorem we see that $T$ is an open map, in other words $T(V)$ is open for every open $V \subseteq X$.
Also $T$ is invertible and $T^{-1}$ is continuous because $(T^{-1})^{-1}=T$ and $T^{-1}(V)$ is open for every open subset $V$ of $X$
Hence $x=T^{-1}x$ is a bounded linear operator ($T^{-1}:(X,\|\cdot\|_2) \longrightarrow (X,\|\cdot\|_1)$)
Is this proof valid or do i have to use a different approach ?(with the open mapping theorem)