# 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.

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 $$\|G^{-1}(T(v)),\alpha_1(v)e_1+..+\alpha_n(v)e_n))\|=\|v\|_V\leq D\|(T(v),\alpha_1(v)e_1+..+\alpha_n(v)e_n)\|=D(\|T(w)\|_W+\|alpha_1(v)|+...+|\alpha_n(v)|)\leq |\|T(v)\|_W+c_1|v|+...+c_n|v|$$.

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|

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 $$T$$