This is the definition which we need for the theorem:

enter image description here This is the theorem $5.9$: enter image description here

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.

  • $\begingroup$ I cannot see any mistake from the answer below. The solution should answer exactly the question. $\endgroup$
    – Zorualyh
    Mar 18 at 22:02
  • $\begingroup$ @Zorualyh it doesn’t answer on the 1 st question $\endgroup$
    – JohnNash
    Mar 19 at 5:51


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