Theorem 5.10 in Rudin's Real & Complex Analysis

Question

I am stuck on theorem 5.10 on Big Rudin.
The theorem is here:
If $X$ and $Y$ are Banach spaces and if $\Lambda$ is a bounded linear transformation of $X$ onto $Y$ which is also one-to-one, then there is a $\delta >0$ such that $$||\Lambda x||\geq \delta ||x||$$ for all $x\in X$.
The proof is here:
If $\delta$ is chosen as in theorem 5.9 (the open mapping theorem) in big Rudin, that is such that to every $y\in Y$ with $||y||<\delta$ there corresponds an $x\in X$ with $||x||<1$ so that $\Lambda x=y$, the conclusion of that theorem, combined with the fact that $\Lambda$ is now one-to-one, shows that $||\Lambda x||<\delta$ implies $||x||<1$. Hence $||x||\geq 1$ implies $||\Lambda x||\geq \delta$, and $$||\Lambda x||\geq \delta ||x||$$ holds for all $x\in X$.

I understand that
$||x||\geq 1$ implies $||\Lambda x||\geq \delta$, but I couldn't fill gap between it and $||\Lambda x||\geq \delta ||x||$ holding for all $x\in X$. If $||x||=1$, it is clear. How fill this gap?

Answer 1

Let $x\neq0\in X$ and $y:=\dfrac{x}{\|x\|}$. Then $\|y\|=1$. So $\|\Lambda y\|\geq \delta$. Can you complete now?

Answer 2

It is clear for $x = 0$. So let's suppose that $x \neq 0$. $y = x/\Vert x \Vert$ is having norm equal to $1$. Hence

$$\Vert\Lambda y\Vert = \left\Vert \frac{\Lambda x}{\Vert x \Vert} \right\Vert \le \delta.$$

Multiply RHS inequality by $\Vert x \Vert$ to get the desired result.