To show a norm is finer than other norm Let $V$ and $W$ be two Banach spaces and let $T \in L(V,W)$ be such that $R(T)$ is close and dim $N(T)< \infty$.Let |.| denote another norm in $V$ with $|x|\leq M||x||_V$ for all $x\in V$.Prove that there is a constant $C$ such that 
$||x||_V\leq C(||Tx||_W +|x|)$ for all $x\in V$
I tried to prove it by using Open mapping theorem but I am not able to solve it.
 A: Let $(e_1,...,e_n)$ be a basis of $N(T)$ Hahn Banach implies the existence of $|.|_V$ continuous  forms $(|\alpha_i(v)|\leq c_i|v|_V)$, $\alpha_i$ such that $\alpha_i(e_j)=\delta_{ij}$. Endow $N(T)$ with the norm defined by $\|x_1e_1+..+x_ne_n\|=|x_1|+..+|x_n|$. Let $G:V\rightarrow R(T)\times N(T)$ by $G(v)=(T(v),\alpha_1(v)e_1+...+\alpha_n(v)e_n)$, $G$ is continuous and bijective, we deduce that its inverse is continuous (Open Mapping). We have $\|G^{-1}(T(v)),\alpha_1(v)e_1+..+\alpha_n(v)e_n))\|=\|v\|_V\leq D\|(T(v),\alpha_1(v)e_1+..+\alpha_n(v)e_n)\|=D(\|T(w)\|_W+\|alpha_1(v)|+...+|\alpha_n(v)|)\leq |\|T(v)\|_W+c_1|v|+...+c_n|v|$. 
A: You can fix a new norm on $V$
$||x||_2:=||Tx||+|x|$
but you have that $T$ is continuos so 
$||Tx||\leq \alpha ||x||$
while 
$|x|<M||x||$ 
So 
$||x||_2\leq (\alpha+M)||x||$
but for a consequence of Hahn-Banach theorem (and open map theorem) you have that the two norms $||.||_2$ and $||.||$ must be equivalent so there exists $C>0$ such that 
$||x||\leq C||x||_2$ 
In this case you don’t need of your hypothesis on $T$
