# Rudin's RCA $5.10$ theorem

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.

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