This is the definition which we need for the theorem:
This is the theorem $5.9$:
There is $5.10$:
If $X$ and $Y$ are Banach spaces and if $\Lambda$ is a bounded linear transformation of $X$ onto $Y$ which is also one-to-ne, then there is a $\delta$ $\gt$ $0$ such that
$||\Lambda x||$ $\ge$ $\delta$$||x||$. $(x\in X)$. ( mark this by $(1)$).
In other words, $\Lambda^{-1}$ is a bounded linear transformation of $Y$ onto $X$.
There is the proof:
If $\delta$ is chosen as in the statement of Theorem $5.9$, the conclusion of that theorem, combined with the fact that $\Lambda$ is now one-to-ne, shows that $||\Lambda x||$ $\lt$ $\delta$ implies $||x||$ $\lt$ $1$. Hence $||x||$ $\ge$ $1$ implies $||\Lambda x ||$ $\ge$ $\delta$, and $(1)$ is proved.
The transformation $\Lambda^{-1}$ os defined on $Y$ by the requirement that $\Lambda^{-1}$$y$ $=$ $x$ if $y$ $=$ $\Lambda x$. A trivial verification shows that $\Lambda^{-1}$ is linear, and $(1)$ implies that $||\Lambda^{-1}||$ $\leq$ $\frac {1}{\delta}$.
I don't understand how does $||\Lambda x||$ $\lt$ $\delta$ imply that $||x||$ $\lt$ $1$.
I also don't understand how does the $(1)$ imply that $||\Lambda^{-1}||$ $\leq$ $\frac {1}{\delta}$ ?
Any help would be appreciated.
