Doubt in one step of the proof in order to show $u(X)$ is closed in $Y$, where $X$ and $Y$ are Banach spaces and $u$ is bounded

Question

In the following theorem :

Let $X, Y$ be Banach spaces and $u \in B(X, Y) .$ Suppose that there is a closed vector subspace $Z$ of $Y$ such that $u(X) \oplus Z=Y$. Then $u(X)$ is closed in $Y$.

The following proof is given :
Proof : The bounded linear map $$ X / \operatorname{ker}(u) \rightarrow Y, x+\operatorname{ker}(u) \mapsto u(x) $$ has the same range as $u$ and is injective, so we may suppose without loss of generality that $u$ is injective. The map $$ v: X \oplus Z \rightarrow Y,(x, z) \mapsto u(x)+z $$ is a continuous linear isomorphism between Banach spaces, so by the open mapping theorem, $v^{-1}$ is also continuous. If $x \in X$, then $\bf \|x\|=\left\|v^{-1} u(x)\right\|$ $\leq\left\|v^{-1}\right\|\|u(x)\|$, so $\|u(x)\| \geq\left\|v^{-1}\right\|^{-1}\|x\| .$ Thus, $u$ is bounded below, and therefore $u(X)$ is closed in $Y$. $\square$

I understood most of the proof but one step I am finding difficult to understand. How to justify the following equation used in the proof. $$\|x\|=\left\|v^{-1} u(x)\right\|$$

Thank you for your help.

Answer 1

You need a norm on $X\oplus Z$. An appropriate norm for the proof is $\|(x,z)\|=\sqrt {\|x\|^{2}+\|z\|^{2}}$.

$v(x,0)=u(x)$ by definition of $v$. Hence $v^{-1} (u(x))=(x,0)$ and $\|(x,0)\|=\|x\|$.

Answer 2

Note that $X$ embeds in $X\oplus Y$ isometrically, $x\mapsto(x,0)$. So $\|x\|=\|(x,0)\|$ but $v(x,0)=u(x)$, so $(x,0)=v^{-1}u(x)$ and thus $\|x\|=\|(x,0)\|=\|v^{-1}u(x)\|$.