Consider a linear transformation $\Lambda$ from a normed linear space $X$ into a normed linear space $Y$, and define its norm by
$||\Lambda||$ $=$ sup { $||\Lambda x ||$ : $x$ $\in$ $X$, $||x||$ $\leq$ $1$ }. mark this by $(1)$.
if ||$\Lambda$|| $\lt$ $\infty$, then $\Lambda$ is called a bounded linear transformation.
in $(1)$, $||x||$ is the norm of $x$ in $X$, $||\Lambda x ||$ is the norm of $\Lambda x $ in $Y$.
Observe that we could restrict ourselves to unit vectors $x$ in $(1)$, i.e., to $x$ with $||x||$ $=$ $1$, without changing the supremum, since
$||\Lambda(\alpha x ) ||$ $=$ $||\alpha \Lambda x ||$ $=$ $|\alpha|$ $||\Lambda x ||$.
Observe also that $||\Lambda||$ is the smallest number such that the inequality
$||\Lambda x||$ $\leq$ $||\Lambda||$ $||x||$ holds for every $x$ $\in$ $X$.
Firstly I don't understand what does it mean that we could restrict ourselves to unit vectors $x$ in $(1)$ and how does it come from this $||\Lambda(\alpha x ) ||$ $=$ $||\alpha \Lambda x ||$ $=$ $|\alpha|$ $||\Lambda x ||$?
I also don't understand how does $||\Lambda x||$ $\leq$ $||\Lambda||$ $||x||$ holds for every $x$ $\in$ $X$ ?
Any help would be appreciated.
For the first question: the point Rudin makes is that having vectors with norm smaller than 1 gives smaller results, so for supremum it's enough to check vectors of norm 1. More precisely: for non-zero vector $x \in X$, whose norm is $\|x\| =\alpha \leq 1$, we have $y = \frac x {\|x\|}$ has norm 1 and $$ \|\Lambda x\| = \|\Lambda (\alpha y)\| = |\alpha|\|\Lambda y\|\leq \|\Lambda y\| $$ so because $y$ has norm 1, $x$ does not contribute to the supremum.
For the second question: If $x = 0$, then $\Lambda x = 0$ and we have easily $\|\Lambda x\| = 0 \leq \|\Lambda\|$. So pick non-zero $x\in X$. Now $$ \|\Lambda x\| = \left\|\|x\|\Lambda \frac x {\|x\|} \right\| = \|x\| \left\|\Lambda \frac x {\|x\|} \right\| \leq \|x\| \sup_{\|y\| \leq 1} \|\Lambda y\| = \|x\| \|\Lambda\|. $$